text:
## DiaHClust : an Iterative Hierarchical Clustering Approach for Identifying Stages in Language Change

2019
##### Publication type
Contribution to a conference collection
Published
##### Published in
Proceedings of the 1st International Workshop on Computational Approaches to Historical Language Change / Tahmasebi, Nina et al. (ed.). - Stroudsburg, PA : Association for Computational Linguistics (ACL), 2019. - pp. 126-135. - ISBN 978-1-950737-31-4
##### Abstract
Language change is often assessed against a set of pre-determined time periods in order to be able to trace its diachronic trajectory. This is problematic, since a pre-determined periodization might obscure significant developments and lead to false assumptions about the data. Moreover, these time periods can be based on factors which are either arbitrary or non-linguistic, e.g., dividing the corpus data into equidistant stages or taking into account language-external events. Addressing this problem, in this paper we present a data-driven approach to periodization: ‘DiaHClust’. DiaHClust is based on iterative hierarchical clustering and offers a multi-layered perspective on change from text-level to broader time periods. We demonstrate the usefulness of DiaHClust via a case study investigating syntactic change in Icelandic, modelling the syntactic system of the language in terms of vectors of syntactic change.
##### Subject (DDC)
400 Philology, Linguistics
##### Conference
The 1st International Workshop on Computational Approaches to Historical Language Change, Aug 2, 2019, Florence, Italy
##### Cite This
ISO 690SCHÄTZLE, Christin, Hannah BOOTH, 2019. DiaHClust : an Iterative Hierarchical Clustering Approach for Identifying Stages in Language Change. The 1st International Workshop on Computational Approaches to Historical Language Change. Florence, Italy, Aug 2, 2019. In: TAHMASEBI, Nina, ed. and others. Proceedings of the 1st International Workshop on Computational Approaches to Historical Language Change. Stroudsburg, PA:Association for Computational Linguistics (ACL), pp. 126-135. ISBN 978-1-950737-31-4. Available under: doi: 10.18653/v1/W19-4716
BibTex
@inproceedings{Schatzle2019DiaHC-50779,
year={2019},
doi={10.18653/v1/W19-4716},
title={DiaHClust : an Iterative Hierarchical Clustering Approach for Identifying Stages in Language Change},
isbn={978-1-950737-31-4},
publisher={Association for Computational Linguistics (ACL)},
booktitle={Proceedings of the 1st International Workshop on Computational Approaches to Historical Language Change},
pages={126--135},
editor={Tahmasebi, Nina},
author={Schätzle, Christin and Booth, Hannah}
}

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text:
Monday
February 20, 2017

# Homework Help: Math: Trigonometry

## Recent Homework Questions About Trigonometry

Math - Calc/Algebra/Trig
I'm trying to find all the points on the graph y=4/3cos^3x-cosx where the tangent line is horizontal. I took the derivative, which gave me: -4cos^2(x)sin(x)+sin(x) Now I need to solve it for x and I'm not sure how. Can you help?
Thursday, February 16, 2017 by Jamie

Calculus Question! ASAP!
Hello! I have this problem: x(dx)/sqrt(9-x^2) I was wondering why I can't use trig substitution and substitute sqrt(9-x^2) for sqrt(1-sec^2) and having: integral x = 3sin(theta) dx = 3cos(theta)d(theata) integral 3sin(theta)(3cos(theta))/3cos(theta) having the 3cos(theta) ...
Wednesday, February 15, 2017 by Lucy

Trig
Solve for t algebraically: inverse cos(t) = inverse sin(t). Where do I start?
Monday, February 6, 2017 by Anonymous

Alg/Trig
If h= -16t^2 + 112t represents the height of a rocket, in feet, t seconds after it was fired, when will the rocket hit the ground? (Hint: The rocket is on the ground when h equals=0).
Sunday, February 5, 2017 by Sydney

Trig Practice
Find the exact value of sinA where a=9 and b=10 and angle C is a right angle. a. sin A= 9/sqrt 181, cos A= sqrt 181/10 b. sin A= sqrt 181/9, cos A= 10/sqrt 181 c. sin A= 9/sqrt 181, cos A= 10/sqrt 181 d. sin A=sqrt181/10, cos A= 9/sqrt 181
Friday, February 3, 2017 by Anonymous

trig
in triangle QRS, q=10.2m, r=20.5m, s12.8m..Solvetriangle QRS by determining the measurements of each degree to the nearest tenth of a degree
Monday, January 30, 2017 by dee

Trig
4x+y=0, x≥0 help finding Sin, Cos, Tan for this
Thursday, January 26, 2017 by Ashley

Trig
can sec θ = negative 5
Thursday, January 26, 2017 by Ashley

trig
sin, cos, and tan for −5x + y =0, x is less than or equal to 0
Thursday, January 26, 2017 by Ashley

Trig
Find the approximate value of cot θ, given that csc θequals=3.5891420 and θ is in quadrant I.
Thursday, January 26, 2017 by Ashley

Geometry/Trig
A painter is placing a ladder to reach the third story window, which is 18 feet above the ground and makes an angle with the ground of 80°. How far out from the building does the base of the ladder need to be positioned? Round your answer to the nearest tenth. The base of the...
Wednesday, January 25, 2017 by Sosa

Geometry / Trig
What is the area of a regular hexagon with a distance from its center to a vertex of 1 cm? (Hint: A regular hexagon can be divided into six equilateral triangles.) hex The area of the regular hexagon is _____ ?
Wednesday, January 25, 2017 by Sosa

trig
Reiny 2+1/3*tan(theta)=2 How do you get 0? Thanks for your help.
Tuesday, January 24, 2017 by Sally

Trig
2+1/3tan(theta)=2 I know that tan=0 0,180 But i don't know how to figure it out. Please help.
Tuesday, January 24, 2017 by Sally

Trig
4=5+cos(theta)
Tuesday, January 24, 2017 by Sally

Trig
Thank you Steve for all your help.
Tuesday, January 24, 2017 by Sally

Trig
Solve equation for 0 less than or equal to theta less than 360. 3-1/5*tan(theta)=16/5
Tuesday, January 24, 2017 by Sally

trig
Brian is riding a Ferris wheel. The wheel has a radius of 25 feet, and at his lowest point, Brian is 8 feet off the ground. Brian times how long it takes to travel from the lowest point to the highest point and finds that it takes 8 seconds. Write a sinusoidal equation to ...
Saturday, January 21, 2017 by bobby

Pre-cal/trig
A dog is attached to a 21-foot rope fastened to the outside corner of a fenced-in garden that measures 18 feet by 22 feet. Assuming that the dog cannot enter the garden, compute the exact area that the dog can wander. I'm confused as to how they want me to solve this...
Wednesday, January 18, 2017 by Aiden

Trig
An isosceles triangle has a height of 12.5 m (measured from the unequal side) and two equal angles that measure 55°. Determine the area of the triangle. Split in half the triangle is a right triangle, so it's three angles are 90%, 55% and 70%. I'm still not sure how to find ...
Tuesday, January 10, 2017 by R

Trig/PreCalc
Find a parametrization for the line segment with endpoints (5,2) and (-2, -4).
Tuesday, January 10, 2017 by William

Trig/PreCalc
Use an algebraic method to eliminate the parameter and identify the graph of the parametric curve. x=5-3t y=2+t -1<-x-<3
Tuesday, January 10, 2017 by Diane

Trig/PreCalc
Find the interior angles of the triangle with vertices (-4,1), (1,-6), and (5,-1)
Tuesday, January 10, 2017 by Maria

Trig/PreCalc
An airplane is flying on a compass heading of 160 degrees at 425 mph. A wind is blowing with the bearing 200 degrees at 75 mph. Find the actual speed and direction of the airplane.
Tuesday, January 10, 2017 by Susan

trig
A person is standing 90 feet away from the Space Needle. The person looks up at the Space Needle at an angle of elevation of 81 degrees. The distance from the floor to the person's eye level is 5 feet. What must be the height of the Space Needle?
Tuesday, January 10, 2017 by Vanni

Trig
Let n be a fixed positive integer. Describe all solutions of sin(n theta) = 1/2 I know that the solutions would be (π/6)/n+(2πx)/n and (5π/6)/n+(2πx)/n ? How do I 'describe'
Sunday, January 8, 2017 by Ryan

Trig
Given: sin x = 4/5, 0 < x < π/2 sin y = 5/13, π/2 < y < π Find the exact value of sin(x + y) I presume I'm supposed to use the sum and difference formulas but I'm not sure how to get the exactly value of cos x or cos y
Wednesday, January 4, 2017 by R

trig prove plz anyone help
Prove that: x - tan^-1(x) = tan^-1{(tanx - x)/(xtanx + 1)} plz help
Tuesday, December 27, 2016 by ...

Math - Trig / Pre - Calc
Standard form of hyperbola: Hyperbola: Vertices: (9,9), (9, -7) Foci: (9, 1 + 4 sqrt 13), (9, 1 - 4 sqrt 13)? Standard form Ellipse: vertices: (2 sqrt 10, -10), (-2 sqrt 10, -10) foci: (sqrt 30, -10), (-sqrt 30, -10)?
Wednesday, December 21, 2016 by Lilly

Math trig
Two towers 50 meters apart from the top of the shorter tower to the top of the taller tower 43 degree elevation, depression from the top of shorter tower to the bottom of the taller tower is 36 degrees what is the height of each tower?
Tuesday, December 20, 2016 by Shania

Trig Question
Find solutions of the following equation which lie in the interval [0,2pi): 6cos(3theta)/(2)+3=0
Tuesday, December 13, 2016 by Amy

Trig identities
Has a negative -5/13 Steve. If angles x and y are in the same quadrant and sin x =3/5 and cos y= -5/13 determine the value of sin (x-y)
Thursday, December 8, 2016 by Luke

Trig
Cicero has a population of 6200 people and is growing at a rate of 8% per year . Mattydale has a population of 8750 and is growing at a rate of 6% per year create and equation solve the problem and in how many years to the nearest year will Cicero have a greater population ...
Tuesday, December 6, 2016 by Christian

trig
The rectangular coordinates of a point are given (0,-6). Find polar coordinates of the point such that r greater than 0 and 0 less than or equals theta less than 2pi. how to solve this damon please help me
Monday, November 28, 2016 by Asfand

trig
The rectangular coordinates of a point are given (-5,-5root3). Find polar coordinates of the point. Express theta in radians.
Monday, November 28, 2016 by Asfand

Math
Consider ∠C such that sin C = 7/25 a) What are the possible quadrants in which ∠C may lie? b) If you know that cos C is negative, how does your answer to part a) change? c) Sketch a diagram to represent ∠C in standard position, given that the condition in part b) is true...
Tuesday, November 22, 2016 by Katie

Trig
What are the focus and directrix of the parabola represented by the equation:8(y-6)=(x+2)squared?
Monday, November 21, 2016 by Emily

Trig
From a point A, 30 meters from the base of a building B, the angle of elevationto the top of the building C is 56 degrees. The angle of elevation to the top of flagpole D on top of the building is 60 degrees. Find the length of flagpole CD
Sunday, November 20, 2016 by Ellie

Trig
Find the exact value of the trigonometric function given that sin u=3/5 and cos v=-8/17. Both u and v are in quadrant II. Tan (u+v)
Monday, November 14, 2016 by Lynn

Trig
find the angle between u=7i+2j and v=-4j
Wednesday, November 9, 2016 by Iris

trig help
simplify cosec40(sin40+sin80+cos120) help step
Monday, November 7, 2016 by Anonymous

trig help
simplify cosec40(sin40+sin80+cos120) help step
Monday, November 7, 2016 by Anonymous

trig
A communication tower is 34m tall. Jane stands somewhere in front of the tower to the right and measures the angle of elevation to be 70 degrees. Bill stands somewhere in front of the tower to the left and measures the angle of elevation to be 50 degrees. How far apart are ...
Sunday, November 6, 2016 by Anonymous

Trig
find all solutions of 2sinx=1-2cosx in the interval from 0 to 360
Tuesday, November 1, 2016 by Anonymous

Pre-Calc/ Trig
Find the sum of the three smallest positive values of theta such that 4(cos^2)(2theta-pi) =3. (Give your answer in radians.) LaTEX: Find the sum of the three smallest positive values of $\theta$ such that $4\cos^2(2\theta-\pi) =3$. (Give your answer in radians.)
Monday, October 31, 2016 by Steve Ho

Physics algebra and trig
A pair of fuzzy dice is hanging from the rearview mirror of a race car. As the car accelerates smoothly, the strings of the dice are tilted slightly toward the rear of the car. From the perspective of the driver, which one of the following statements is true, if the dice are ...
Thursday, October 20, 2016 by Kyle

Trig
Write 3theta - sin 4theta as a product.
Monday, October 17, 2016 by Sade

Trig
Use the half-angle formula to simplify (sin4theta)/(1+cos4theta)
Monday, October 17, 2016 by Sade

math trig
how do you solve arcsin(3/5) without a calculator?
Sunday, October 16, 2016 by help

Trig
If angle A is 45 degrees and angle B is 60 degrees. Find sin(A)cos(B), find cos(A)sin(B), find sin(A)sin(B), and find cos(A)cos(B) The choises for the first are: A. 1/2[sin(105)+sin(345)] B. 1/2[sin(105)-sin(345)] C. 1/2[sin(345)+cos(105)] D. 1/2[sin(345)-cos(105)] You don't ...
Saturday, October 8, 2016 by Sade

Trig
Given tan(A) = 2 and A is in Quadrant I, find tan(2A). A. -1/3 B. 2/3 C. -4/3 D. -2/3
Friday, October 7, 2016 by Sade

Trig
Given tan(A) = 2 and A is in Quadrant I, find cos(2A). A. 0 B. 1 C. 1/2 D. -3/5
Friday, October 7, 2016 by Iris

Trig
Just confused. Given tan(A) = 2 and A is in Quadrant I, find sin(2A). A. 0 B. 1 C. 1/2 D. 4/5
Friday, October 7, 2016 by Iris

Trig
Given Tan(A) = 5 in Quadrant III and Sin(B) = ⅔ in Quadrant II, what is the Quadrant of A-B?
Thursday, October 6, 2016 by Sade

Trig
Given Tan(A) = 5 in Quadrant III and Sin(B) = ⅔ in Quadrant II, find Cos(A-B).
Thursday, October 6, 2016 by Sade

Trig
Given Tan(A) = 5 in Quadrant III and Sin(B) = ⅔ in Quadrant II, find Sin(A-B).
Thursday, October 6, 2016 by Sade

Math (Trig Review)
Evaluate the expression without using a calculator (Make a sketch of a right triangle) tan(arccot 2) Thanks in advance! I'm trying to refresh my memory of these kinds of problems and I'm having trouble on how to solve this problem
Friday, September 30, 2016 by Ray

Trig
Cosine theta=8/17 and 270 degrees is < theta, and theta is< 360
Wednesday, September 28, 2016 by Sarah

math trig
A Ferris wheel with a diameter of 37 meters rotates at a rate of 4 minutes per revolution. Riders board the Ferris wheel 4 meters above the ground at the bottom of the wheel. A couple boards the Ferris wheel and the ride starts. Write a formula for the height of the couple t ...
Wednesday, September 28, 2016 by help

maths trig need help
show that sinx/(1-cotx)+cosx/(1-sinx)=sinx+cosx thanks
Tuesday, September 27, 2016 by rap

maths trig
A flag in the form of an equilateral triangle is connected to the tops of 2 vertical poles. One of the pole has a length of 4 and the other pole has a length of 3. You also know that the third vertex touches the ground perfectly. Calculate the length of a side
Friday, September 23, 2016 by samson

Trig
Can someone check my answers? Find # of triangles possible: <A=44.3, a=11.5, b=7.7 ... 2 triangles, B=~27.9 or B=~152.1 <A=29.3, b=20.5, a=12.8 ... 2 triangles, B=~51.6 or B=~128.4
Wednesday, September 21, 2016 by K

Trig
A ship leaves port and sails at a bearing of 124degrees. Another ship leaves the same port, at the same time, sailing at a bearing of 74degrees. When both ships are 80 miles from the port, how far are they from each other?
Sunday, September 11, 2016 by Adrienne

Trig
1. Given Sin(A) = ⅗ and Cos(B) = 8/17 in Quadrant I, find Sin(A+B). a) 24/80 [b)] 84/85 c) 60/80 d) 60/85 Find Cos(A+B). a) 32/80 b) -45/85 c) -13/80 [d)] -13/85 Find Tan(A+B) a) 0.8 [b)] -1.72 c) -4.21 d) -6.46 I keep getting an 1.61 answer but that isn't an option? ...
Tuesday, September 6, 2016 by Lyss

Algebralgebrad
Trig ientities
Sunday, September 4, 2016 by yaga

Trig
if sin(x)=-0.5 and x is in the third quadrant, find x
Saturday, August 20, 2016 by Jason

trig
if sec theta =sqrt -20/3 state the exact value of tan theta
Tuesday, July 26, 2016 by Anonymous

MAth Trig
Find the diameter of a circle with an arc length of 32 cm and a subtended angle of 72 degrees
Tuesday, July 26, 2016 by Anonymous

MATH
Solve the integral using trig substitution (1/((x)(x^2+1)))dx
Sunday, July 24, 2016 by ANON

MATH
Integrate 1/(x(x^2+1)) dx using trig substitution
Sunday, July 24, 2016 by ANON

Math - Trig
The depth of water, h, in metres at time t, in hours, is given by the formula: h(t)=7.8+3.5sin[pi/6(t−3)]. Its a 24 hours period and t=0 is midnight. Provide an algebraic solution to determine the time(s) of day, the water reaches the depth of 10.29 m i got one time 4.51...
Friday, July 22, 2016 by Sherry

Math - Trig
The equation given below models the average monthly temperature (˚C). t denotes the number of months with t=1 representing January. During which months is the average monthly temperature 12.5˚C? 𝑇(𝑡) = 12.5 + 15.8sin (𝜋/6 (𝑡) −2...
Wednesday, July 20, 2016 by Anonymous

Math - Trig
A hamster enters a wheel at the lowest point and begins to run . The height of point R (in the diagram below) above the ground is given by the function ℎ(𝑡) = −17.5 cos ( 2pi/3 𝑡) + 19, where ℎ is height in centimeters and 𝑡 is time in...
Wednesday, July 20, 2016 by Anonymous

Math - Trig
Solve for exact or to 1d.p of x E [0 ,2pi] 4cos^2x-3
Monday, July 18, 2016 by Anonymous

Pre-Cal
[Note: I'm still having issues with identities with regard to trig] Verify the Trig. identities: (a). cot(x+y)=(cotxcoty-1)/(cotx+coty) (b). sin0(cot0+tan0)=sec0 [Note: 0=theta symbol]
Monday, July 11, 2016 by Amber

trig
what is the exact value of tan 105 degrees
Wednesday, June 29, 2016 by fortue

trig
If a ship leaves port at 9:00 a.m. and sails due south for 3 hours at 14 knots, then turns N 60° E for another 2 hours, how far from port is the ship?
Friday, June 24, 2016 by need help

trig
If a ship leaves port at 9:00 a.m. and sails due north for 3 hours at 12 knots, then turns N 30° E for another hour, how far from port is the ship?
Wednesday, June 22, 2016 by lejla

Math (Trig)
Suppose the point A is on a circle of radius 6 with an angle of 240 degrees. Find the value of the coordinates of this point. Round your results to 3 decimal places. So in this case, x=rcos(theta) y=rsin(theta) x=6cos(240) y=6sin(240) Rounding to 3 decimal places: x=-3 y=-5....
Tuesday, June 7, 2016 by Giovanni

Math (Right Triangle Trig)
Suppose tan(theta)=5/7 and theta belongs to Quadrant III, find the exact value of sec(theta)? I already took the steps to solve this problem but I am confused if whether or not sec(theta)is positive or negative sqrt(74)/7 Any help is greatly appreciated!
Sunday, June 5, 2016 by Giovanni

Trig
6. To win a javelin throwing competition, your last throw must travel a horizontal distance of at least 100 feet. You release the javelin at a 40° angle with an initial speed of 71 feet per second. Do you win the competition? Justify your answer
Friday, June 3, 2016 by Noah Ahmed

Math trig
Hello guys. I am just wondering if tan^2 A (tan squared A) = sin^2 A /cos^2A And 2tan A = 2sin A / 2cos A This has been bothering me when I have to simplify trig. Thanks in advance.
Wednesday, June 1, 2016 by Jack

Trig
If sin(theta) = 3/7, what is sin (pi/2 - theta)?
Thursday, May 19, 2016 by Anonymous

Math (Trig)
Convert the Cartesian coordinate (-1,2) to polar coordinates, 0≤θ<2π I know r is sqrt(5) but how would I find theta between 0 and 2pi? Thanks in advance!
Monday, May 16, 2016 by Giovanni

Math (Trig)
A clock tower has been constructed such that the center of the clock is 40 feet above ground. The minute hand on the clock is 3 feet long. How far above the ground the end of the minute hand at 12:05? 1:00?
Friday, May 13, 2016 by Bree

geometry and trig
If a cylinder with a 4 inch diameter and a 6 inch height holds 1 pound of oatmeal. How much oatmeal will a cylinder (similar) with a 10 inch diameter hold? Volume of 4 in-----24pi volume of 10 in----137pi I get 91 oz choiced are37,44,74, 192
Tuesday, May 10, 2016 by lulu

Math (Trig) lowering powers
If cos^4(3x)-sin^4(3x)=cos(g(x)) then g(x)=? Use a^4-b^4=(a^2-b^2)(a^2+b^2)! thanks in advance for any help!
Monday, May 9, 2016 by Giovanni

trig
sec = 7/2 and is in Q1, then what is double angle of tan?
Saturday, May 7, 2016 by PLEASE HELP ME

Trig
Two rays with common endpoint O form a 30 degree angle. Point A lies on one ray, point B on the other ray, and AB=1. What is the maximum possible length of OB?
Saturday, May 7, 2016 by Pip

Pre-Cal (Trig) Help?
The following relationship is known to be true for two angles A and B: cos(A)cos(B)-sin(A)sin(B)=0.957269 Express A in terms of the angle B. Work in degrees and report numeric values accurate to 2 decimal places. So I'm pretty lost on how to even begin this problem. I do know ...
Friday, May 6, 2016 by Giovanni

trig, precalc, math, algebra, trigonometry
Rewrite: tan(cos^−1 (v)) as an algebraic expression in v. Please help me understand.
Thursday, May 5, 2016 by Veronica

trig
sin 25° cos 35° + cos 25° sin 35°
Wednesday, May 4, 2016 by EBB

Trig
find the equation for the cosine function that has an amplitude of 3/5, a period of 3pi/2 and a y-intercept of 5.
Tuesday, May 3, 2016 by Jon

Math -Trig
Determine the exact value if cot(3pi/4) using special triangles. I don't understand how it could be 1/tan3pi/4 in a triangle....
Tuesday, May 3, 2016 by Jen

Math - Trig
if the diameter of a tire is 3.5 m and makes one rotation every 5 seconds, how far does the tire rotates in one minute?
Tuesday, May 3, 2016 by Jen

Geometry/Trig
In triangle ABC, A is a right angle, and m b=45 degrees. What is the length of BC? If your answer is not an integer, leave it in simplest radical form. a. 16 ft b. 16 sq 2 ft c. 16 sq 3 ft d. 32 ft I'm really confused as to how to do this... Could someone please explain and ...
Tuesday, May 3, 2016 by Gina

trig
tan(theta)/tan(2theta)-tan(theta)= cos(2theta)
Sunday, May 1, 2016 by aaron

Trig
Problem: A cable from the top of a 60 foot high tower is to be attached to the ground (x) feet from the base of the tower. A) if the cable makes an angle of t radians with the ground when attached, express t as a function of x. B) if t = pi/5, how far is the end of the cable ...
Sunday, May 1, 2016 by Chris

Trig
Solve each equation for exact solutions over the interval [0,360)where appropriate. Round approximate solutions to the nearest tenth degree. Sin^2Theta=Cos^2Theta+1 My Work: Using double angle Identity I subtract 1 to other side therefore: 1-Sin^2theta=Cos^2theta= Cos2theta=...
Wednesday, April 27, 2016 by Ronald

Trig
While fishing on a lake, the fisherman looks up at a cliff. His boat is 23 ft from the base of the cliff, and the angle of elevation is 40 degrees. Find the height of the cliff to the nearest tenth of a foot.
Wednesday, April 27, 2016 by Tg

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We want to estimate the probability of success / failure of the dependent variable ($$\Pr(Y=1)$$), given the explanatory variables: $$\Pr(Y=1|X_1,X_2,\cdots, X_n)$$. The explanatory variables can be categorical or continuous - it doesn’t matter.

Logistic regression addresses the following issues:

1. There is a need to bound at $$1$$ since we are estimating a probability, but lines are not naturally bounded. So we transform probability into odds:

$\text{odds}=\frac{\Pr(Y=1)}{1-\Pr(Y=1)}\tag 1$

In addition, this also helps turning a sigmoid curve of a typical cummulative probability distribution (e.g. normal) into a linear relation.

1. Now, when the probability goes down to zero, the odds also tend to zero; yet there is no floor restriction in a line. This problem can be addressed by expressing the linear relation in log scale: $$\log(\text{odds})$$, which will tend to $$-\infty$$ as the odds tend to zero. This is the logit or log-odds function.

The logistic regression model is therefore:

$$\large \color{red}{\log} \left[\color{blue}{\text{odds(p(Y=1))}}\right]=\color{red}{\text{log}}\left(\frac{\hat p\,(Y=1)}{1-\hat p\,(Y=1)}\right) = X\beta = \beta_o + \beta_1 x_1 + \beta_2 x_2 +\cdots+ \beta_p x_p$$

Consequently,

$\large \color{blue}{\text{odds(Y=1)}} = \frac{\Pr\,(Y=1)}{1\,-\,\Pr\,(Y=1)} = e^{\,\beta_0 + \beta_1 x_1 + \cdots + \beta_p x_p} = e^{\,\beta_o}\,e^{\,\beta_1x_1}\,e^{\,\beta_2x_2}\,\cdots\,e^{\,\beta_px_p}$

Therefore a unit increase in $$x_1$$ increases the odds $$\large e^{\,\beta_1}$$ times. The factors $$\large e^{\,\beta_i}$$ are the ODDS RATIO’s. On the other hand, $$\beta_i$$ (the coefficients) are the LOG ODDS-RATIO:

For CATEGORICAL VARIABLES, the ODDS RATIO denote how much the presence or absence of a factor variable increases the odds of a positive dependent variable.

For CONTINOUS VARIABLES the odds ratio tell us how much the odds increase multiplicatively with a one-unit change in the independent variable. To calculate the difference in odds you raise the OR to the power of the difference between the observations.

To get the exponentiated coefficients and their confidence intervals in R, we use cbind to bind the coefficients and confidence intervals column-wise:

exp(cbind(OR = coef(mylogit), confint(mylogit)))

See here.

#### Translation into probabilities:

From equation $$(1):$$

\begin{align} \Pr(Y = 1) &= (1-\Pr(Y=1))\times\,e^{\,\beta_0 + \beta_1 x_1 + \cdots + \beta_p x_p}\\[2ex] e^{\,\beta_0 + \beta_1 x_1 + \cdots + \beta_p x_p}&= \Pr(Y = 1)\,\times (1 \,+\,e^{\,\beta_0 + \beta_1 x_1 + \cdots + \beta_p x_p}) \\[2ex] \color{green}{\Pr(Y = 1)} = \frac{\color{blue}{\text{odds(Y=1)}}}{1\,+\,\color{blue}{\text{odds(Y=1)}}}&=\frac{e^{\,\beta_0 + \beta_1 x_1 + \cdots + \beta_p x_p}}{1 \,+\,e^{\,\beta_0 + \beta_1 x_1 + \cdots + \beta_p x_p}}=\frac{1}{1 \,+\,e^{-(\,\beta_0 + \beta_1 x_1 + \cdots + \beta_p x_p)}} \end{align}

####INTERPRETATION OF LOGISTIC REGRESSION RESULTS:

QUESTION:

(Initially posted here)

I ran a linear regression of acceptance into college against SAT scores and family / ethnic background. The data are fictional. The question focuses in the gathering and interpretation of odds ratios when leaving the SAT scores aside for simplicity.

The variables are Accepted (0 or 1) and Background (“red” or “blue”). I set up the data so that people of “red” background were more likely to get in:

fit <- glm(Accepted ~ Background, data=dat, family="binomial")
exp(cbind(Odds_Ratio_RedvBlue = coef(fit), confint(fit)))

Odds_Ratio_RedvBlue             2.5 %       97.5 %
(Intercept)             0.7088608                     0.5553459   0.9017961
Backgroundred           2.4480042                     1.7397640   3.4595454

Questions:

1. Is 0.7 the odd ratio of a person of “blue” background being accepted? I’m asking this because I also get 0.7 for “Backgroundblue” if instead I run the following code:

 fit <- glm(Accepted~Background - 1, data=dat, family="binomial")
exp(cbind(OR=coef(fit), confint(fit)))
2. Shouldn’t the odds ratio of “red” being accepted ($$\rm Accepted/Red:Accepted/Blue$$) just the reciprocal: ($$\rm OddsBlue = 1 / OddsRed$$)?

I’ve been working on answering my question by calculating manually the odds and odds ratios:

Acceptance   blue            red            Grand Total
0              158           102                260
1              112           177                289
Total          270           279                549

So the Odds Ratio of getting into the school of Red over Blue is:

$\frac{\rm Odds\ Accept\ If\ Red}{\rm Odds\ Acccept\ If\ Blue} = \frac{^{177}/_{102}}{^{112}/_{158}} = \frac {1.7353}{0.7089} = 2.448$

And this is the Backgroundredreturn of:

fit <- glm(Accepted~Background, data=dat, family="binomial")
exp(cbind(Odds_and_OR=coef(fit), confint(fit)))

Odds_and_OR                         2.5 %      97.5 %
(Intercept)             0.7088608                     0.5553459   0.9017961
Backgroundred           2.4480042                     1.7397640   3.4595454

At the same time, the (Intercept)corresponds to the numerator of the odds ratio, which is exactly the odds of getting in being of ‘blue’ family background: $$112/158 = 0.7089$$.

If instead, I run:

fit2 <- glm(Accepted~Background-1, data=dat, family="binomial")
exp(cbind(Odds=coef(fit2), confint(fit2)))

Odds            2.5 %      97.5 %
Backgroundblue     0.7088608        0.5553459   0.9017961
Backgroundred      1.7352941        1.3632702   2.2206569

The returns are precisely the odds of getting in being ‘blue’: Backgroundblue (0.7089) and the odds of being accepted being ‘red’: Backgroundred (1.7353). No Odds Ratio there. Therefore the two return values are not expected to be reciprocal.

Finally, How to read the results if there are 3 factors in the categorical regressor?

Same manual versus [R] calculation:

I created a different fictitious data set with the same premise, but this time there were three ethnic backgrounds: “red”, “blue” and “orange”, and ran the same sequence:

First, the contingency table:

Acceptance  blue    orange  red   Total
0             86        65  130     281
1             64        42  162     268
Total        150       107  292     549

And calculated the Odds of getting in for each ethnic group:

• Odds Accept If Red = 1.246154;
• Odds Accept If Blue = 0.744186;
• Odds Accept If Orange = 0.646154

As well as the different Odds Ratios:

• OR red v blue = 1.674519;
• OR red v orange = 1.928571;
• OR blue v red = 0.597186;
• OR blue v orange = 1.151717;
• OR orange v red = 0.518519; and
• OR orange v blue = 0.868269

And proceeded with the now routine logistic regression followed by exponentiation of coefficients:

fit <- glm(Accepted~Background, data=dat, family="binomial")
exp(cbind(ODDS=coef(fit), confint(fit)))

ODDS     2.5 %   97.5 %
(Intercept)      0.7441860 0.5367042 1.026588
Backgroundorange 0.8682692 0.5223358 1.437108
Backgroundred    1.6745192 1.1271430 2.497853

Yielding the odds of getting in for “blues” as the (Intercept), and the Odds Ratios of Orange versus Blue in Backgroundorange, and the OR of Red v Blue in Backgroundred .

On the other hand, the regression without intercept predictably returned just the three independent odds:

fit2 <- glm(Accepted~Background-1, data=dat, family="binomial")
exp(cbind(ODDS=coef(fit2), confint(fit2)))

ODDS     2.5 %    97.5 %
Backgroundblue   0.7441860 0.5367042 1.0265875
Backgroundorange 0.6461538 0.4354366 0.9484999
Backgroundred    1.2461538 0.9900426 1.5715814

####CATEGORICAL INDEPENDENT VARIABLES:

Here is an example extracted form this site:

# Define an explanatory variable with two levels:
parentsmoke <- as.factor(c(1,0))
# create response vector:
response <- cbind(yes=c(816,188), no=c(3203,1168))
response
##      yes   no
## [1,] 816 3203
## [2,] 188 1168
smoke.logistic <- glm(response~parentsmoke, family=binomial(link=logit))
smoke.logistic
##
## Call:  glm(formula = response ~ parentsmoke, family = binomial(link = logit))
##
## Coefficients:
##  (Intercept)  parentsmoke1
##      -1.8266        0.4592
##
## Degrees of Freedom: 1 Total (i.e. Null);  0 Residual
## Null Deviance:       29.12
## Residual Deviance: 4.365e-13     AIC: 19.24
summary(smoke.logistic)
##
## Call:
## glm(formula = response ~ parentsmoke, family = binomial(link = logit))
##
## Deviance Residuals:
## [1]  0  0
##
## Coefficients:
##              Estimate Std. Error z value Pr(>|z|)
## (Intercept)  -1.82661    0.07858 -23.244  < 2e-16 ***
## parentsmoke1  0.45918    0.08782   5.228 1.71e-07 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
##     Null deviance: 2.9121e+01  on 1  degrees of freedom
## Residual deviance: 4.3654e-13  on 0  degrees of freedom
## AIC: 19.242
##
## Number of Fisher Scoring iterations: 2

This information gives us the fitted model:

$\text{logit}(\hat\pi_i)=\log\frac{\hat\pi_i}{1-\hat\pi_i}=\hat\beta_0+\hat\beta_1X_i=-1.837+0.459\,(\text{parents smoke = 1})$ where parentsmoke is a dummy variable ((e.g. design variable) that takes value $$1$$ if at least one parents is smoking and $$0$$ if neither is smoking.

parent smoking = 0 is the baseline. We are modeling here probability of “children smoking”. Estimated $$\beta_0=1.827$$ with a standard error of $$0.078$$ is significant and it says that log-odds of a child smoking versus not smoking if neither parents is smoking (the baseline level) is $$-1.827$$ and it’s statistically significant.

Estimated $$\beta_1=0.459$$ with a standard error of $$0.088$$ is significant and it says that log-odds-ratio of a child smoking versus not smoking if at least one parent is smoking versus neither parents is smoking (the baseline level) is $$0.459$$ and it’s statistically significant. $$\exp(0.459)=1.58$$ are the estimated odds-ratios.

Why not just ran a Chi square test:

prop.test(response, correct = F)
##
##  2-sample test for equality of proportions without continuity
##  correction
##
## data:  response
## X-squared = 27.677, df = 1, p-value = 1.434e-07
## alternative hypothesis: two.sided
## 95 percent confidence interval:
##  0.04218944 0.08659558
## sample estimates:
##    prop 1    prop 2
## 0.2030356 0.1386431

It does detect differences in proportions but it doesn’t address:

• Modeling the “risk” of student smoking as a function of parents’ smoking behavior.
• Describing the differences between student smokers and nonsmokers as a function of parents smoking behavior via descriptive discriminate analyses. -Predicting the probabilities that individuals fall into two categories of the binary response as a function of some explanatory variables, e.g. what is the probability that a student is a smoker given that neither of his/her parents smokes.
• Predicting that a student is a smoker given that neither of his/her parents smokes, i.e. probabilities that individuals fall into two categories of the binary response as a function of some explanatory variables, we want to classify new students into “smoking” or “nonsmoking” group depending on parents smoking behavior. we want to develop a social network model, adjust for “bias”, These are just some of the possibilities of logistic regression, which cannot be handled within a framework of goodness-of-fit only.

Here is the same model splitting mothers and fathers as predictors:

kids.smoke <- matrix(c(188,599,816-205,1168,1456,3203-1456), nrow = 3)
dimnames(kids.smoke) = list(Parents = c("None", "Mother", "Father"), Kids = c("Smokers", "Non-smokers"))
kids.smoke
##         Kids
## Parents  Smokers Non-smokers
##   None       188        1168
##   Mother     599        1456
##   Father     611        1747
bad.example <- as.factor(c(0,"1Mother","2Father"))
smoke.logistic
##
## Call:  glm(formula = kids.smoke ~ bad.example, family = binomial(link = logit))
##
## Coefficients:
##            -1.8266              0.9384              0.7760
##
## Degrees of Freedom: 2 Total (i.e. Null);  0 Residual
## Null Deviance:       119.2
## Residual Deviance: -1.06e-12     AIC: 28.77
summary(smoke.logistic)
##
## Call:
## glm(formula = kids.smoke ~ bad.example, family = binomial(link = logit))
##
## Deviance Residuals:
## [1]  0  0  0
##
## Coefficients:
##                    Estimate Std. Error z value Pr(>|z|)
## (Intercept)        -1.82661    0.07858 -23.244   <2e-16 ***
## bad.example1Mother  0.93842    0.09237  10.160   <2e-16 ***
## bad.example2Father  0.77605    0.09157   8.475   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
##     Null deviance:  1.1915e+02  on 2  degrees of freedom
## Residual deviance: -1.0603e-12  on 0  degrees of freedom
## AIC: 28.768
##
## Number of Fisher Scoring iterations: 2

An example with credit defaults with some calculations extracted from here:

set.seed(123)

gender <- factor(round(runif(100,0,1)),
labels=c("female", "male"))
age <- round(runif(100,20,80))
profession <- factor(round(runif(100,1,4)),
labels=c("student", "worker", "teacher", "self-employed"))

# Let's make males, teachers and self-employed more likely to default:

credit.default <-  round(((as.numeric(gender)/2 + as.numeric(profession)/4)/2) * runif(100,0.5,1))

df <- data.frame(credit.default,gender,age,profession)
head(df)
##   credit.default gender age    profession
## 1              0 female  56        worker
## 2              1   male  40 self-employed
## 3              1 female  49       teacher
## 4              1   male  77       teacher
## 5              1   male  49        worker
## 6              1 female  73 self-employed
catlogis <- glm(credit.default ~ gender+age+profession, df, family=binomial(link="logit"))
summary(catlogis)
##
## Call:
## glm(formula = credit.default ~ gender + age + profession, family = binomial(link = "logit"),
##     data = df)
##
## Deviance Residuals:
##     Min       1Q   Median       3Q      Max
## -2.2580  -0.9295  -0.2042   0.5361   1.5200
##
## Coefficients:
##                           Estimate Std. Error z value Pr(>|z|)
## (Intercept)             -3.860e+00  1.327e+00  -2.910 0.003616 **
## gendermale               3.084e+00  7.844e-01   3.931 8.46e-05 ***
## age                      9.465e-07  1.692e-02   0.000 0.999955
## professionworker         6.325e-01  8.211e-01   0.770 0.441075
## professionteacher        3.245e+00  1.003e+00   3.236 0.001214 **
## professionself-employed  4.464e+00  1.193e+00   3.740 0.000184 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
##     Null deviance: 136.663  on 99  degrees of freedom
## Residual deviance:  88.237  on 94  degrees of freedom
## AIC: 100.24
##
## Number of Fisher Scoring iterations: 5
require(pROC)
## Loading required package: pROC
## Type 'citation("pROC")' for a citation.
##
## Attaching package: 'pROC'
## The following objects are masked from 'package:stats':
##
##     cov, smooth, var
auc(df$credit.default, predict(catlogis, type="response")) ## Area under the curve: 0.8517 We can generate an ANOVA to analyze the table of variance, paying attention at the difference between the NULL variance and the Resid. Deviance to see which variable reduce this residual variance in comparison to the model with only an intecept. The p-values also help select variables. We want a drop in the deviance and AIC. catlogis <- glm(credit.default ~ gender+profession, df, family=binomial(link="logit")) summary(catlogis) ## ## Call: ## glm(formula = credit.default ~ gender + profession, family = binomial(link = "logit"), ## data = df) ## ## Deviance Residuals: ## Min 1Q Median 3Q Max ## -2.2580 -0.9295 -0.2042 0.5361 1.5200 ## ## Coefficients: ## Estimate Std. Error z value Pr(>|z|) ## (Intercept) -3.8604 0.9996 -3.862 0.000113 *** ## gendermale 3.0836 0.7841 3.933 8.39e-05 *** ## professionworker 0.6325 0.8202 0.771 0.440570 ## professionteacher 3.2448 1.0028 3.236 0.001213 ** ## professionself-employed 4.4636 1.1873 3.759 0.000170 *** ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ## ## (Dispersion parameter for binomial family taken to be 1) ## ## Null deviance: 136.663 on 99 degrees of freedom ## Residual deviance: 88.237 on 95 degrees of freedom ## AIC: 98.237 ## ## Number of Fisher Scoring iterations: 5 auc(df$credit.default, predict(catlogis, type="response"))
## Area under the curve: 0.8527
genlogis <- glm(credit.default ~ gender, df, family=binomial(link="logit"))
summary(genlogis)
##
## Call:
## glm(formula = credit.default ~ gender, family = binomial(link = "logit"),
##     data = df)
##
## Deviance Residuals:
##     Min       1Q   Median       3Q      Max
## -1.4261  -0.7502  -0.7502   0.9476   1.6765
##
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)
## (Intercept)  -1.1239     0.3193  -3.520 0.000431 ***
## gendermale    1.6919     0.4405   3.841 0.000123 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
##     Null deviance: 136.66  on 99  degrees of freedom
## Residual deviance: 120.56  on 98  degrees of freedom
## AIC: 124.56
##
## Number of Fisher Scoring iterations: 4
auc(df$credit.default, predict(genlogis, type="response")) ## Area under the curve: 0.6997 fitted.results <- predict(genlogis, type="response") fitted.results <- ifelse(fitted.results > 0.5,1,0) misClasificError <- mean(fitted.results != df$credit.default)
print(paste('Accuracy', 1-misClasificError))
## [1] "Accuracy 0.7"
library(ROCR)
## Loading required package: gplots
##
## Attaching package: 'gplots'
## The following object is masked from 'package:stats':
##
##     lowess
pr <- prediction(predict(genlogis, type="response"), df$credit.default) prf <- performance(pr, measure = "tpr", x.measure = "fpr") plot(prf) auc <- performance(pr, measure = "auc") auc <- auc@y.values[[1]] auc ## [1] 0.6997144 proflogis <- glm(credit.default ~ profession, df, family=binomial(link="logit")) summary(proflogis) ## ## Call: ## glm(formula = credit.default ~ profession, family = binomial(link = "logit"), ## data = df) ## ## Deviance Residuals: ## Min 1Q Median 3Q Max ## -1.7941 -0.7710 -0.6231 1.0508 1.8626 ## ## Coefficients: ## Estimate Std. Error z value Pr(>|z|) ## (Intercept) -1.5404 0.6362 -2.421 0.01547 * ## professionworker 0.4796 0.7445 0.644 0.51950 ## professionteacher 1.8458 0.7272 2.538 0.01114 * ## professionself-employed 2.9267 0.9063 3.229 0.00124 ** ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ## ## (Dispersion parameter for binomial family taken to be 1) ## ## Null deviance: 136.66 on 99 degrees of freedom ## Residual deviance: 115.75 on 96 degrees of freedom ## AIC: 123.75 ## ## Number of Fisher Scoring iterations: 4 auc(df$credit.default, predict(proflogis, type="response"))
## Area under the curve: 0.7438
pr <- prediction(predict(proflogis, type="response"), df$credit.default) prf <- performance(pr, measure = "tpr", x.measure = "fpr") plot(prf) auc <- performance(pr, measure = "auc") auc <- auc@y.values[[1]] auc ## [1] 0.743778 Another example from this video: set.seed(420) n.samples <- 100 weight <- sort(rnorm(n.samples,172,29)) obese <- ifelse(runif(n.samples) < (rank(weight)/100), yes=1, no=0) plot(weight,obese, pch=19, col='tan4', cex=.7) glm.fit <- glm(obese ~ weight, family="binomial") lines(weight, glm.fit$fitted.values)

par(pty="s")
roc(obese, glm.fit$fitted.values, plot=T) ## ## Call: ## roc.default(response = obese, predictor = glm.fit$fitted.values,     plot = T)
##
## Data: glm.fit$fitted.values in 45 controls (obese 0) < 55 cases (obese 1). ## Area under the curve: 0.8291 roc.info <- roc(obese, glm.fit$fitted.values)
roc.df <- data.frame(ttp=roc.info$sensitivities*100, fpp=(1-roc.info$specificities) *100, thresholds=roc.info$thresholds) head(roc.df) ## ttp fpp thresholds ## 1 100 100.00000 -Inf ## 2 100 97.77778 0.01349011 ## 3 100 95.55556 0.03245008 ## 4 100 93.33333 0.05250145 ## 5 100 91.11111 0.07017225 ## 6 100 88.88889 0.08798755 tail(roc.df) ## ttp fpp thresholds ## 96 9.090909 0 0.9275222 ## 97 7.272727 0 0.9371857 ## 98 5.454545 0 0.9480358 ## 99 3.636364 0 0.9648800 ## 100 1.818182 0 0.9735257 ## 101 0.000000 0 Inf Notice that the first row of head(roc.df) with a threshold for labeling “obese” of $$-\inf$$ the ttp (true positive percentage) will be of $$100\%,$$ and so will be the fpp (false positive percentage), and it corresponds to the upper-right hand part of the ROC curve. The bottom-left part of the curve corresponds to the last row of tail(roc.df). We can isolate part of the data, for example with ttp between $$60$$ and $$80$$, to select an optimal threshold value: roc.df[roc.df$ttp > 60 & roc.df$ttp < 80,] ## ttp fpp thresholds ## 42 78.18182 35.55556 0.5049310 ## 43 78.18182 33.33333 0.5067116 ## 44 78.18182 31.11111 0.5166680 ## 45 76.36364 31.11111 0.5287933 ## 46 76.36364 28.88889 0.5429351 ## 47 76.36364 26.66667 0.5589494 ## 48 74.54545 26.66667 0.5676342 ## 49 74.54545 24.44444 0.5776086 ## 50 74.54545 22.22222 0.5946054 ## 51 72.72727 22.22222 0.6227449 ## 52 70.90909 22.22222 0.6398136 ## 53 69.09091 22.22222 0.6441654 ## 54 67.27273 22.22222 0.6556705 ## 55 67.27273 20.00000 0.6683618 ## 56 67.27273 17.77778 0.6767661 ## 57 65.45455 17.77778 0.6802060 ## 58 65.45455 15.55556 0.6831936 ## 59 65.45455 13.33333 0.6917225 ## 60 63.63636 13.33333 0.6975300 ## 61 61.81818 13.33333 0.6982807 We can plot the ROC differently: par(pty="s") roc(obese, glm.fit$fitted.values, plot=T, legacy.axes=T, percent=T,
xlab='False positive percentage', ylab='True positive percentage',
col="darkslateblue",lwd=4, print.auc=T)

##
## Call:
## roc.default(response = obese, predictor = glm.fit$fitted.values, percent = T, plot = T, legacy.axes = T, xlab = "False positive percentage", ylab = "True positive percentage", col = "darkslateblue", lwd = 4, print.auc = T) ## ## Data: glm.fit$fitted.values in 45 controls (obese 0) < 55 cases (obese 1).
## Area under the curve: 82.91%

To calculate the partial AUC of the beginning, where only a small number of false positives are allowed…

par(pty="s")
roc(obese, glm.fit$fitted.values, plot=T, legacy.axes=T, percent=T, xlab='False positive percentage', ylab='True positive percentage', col='darkslateblue',lwd=4, print.auc=T, print.auc.x=45, partial.auc=c(100,90), auc.polygon=T, auc.polygon.col='midnightblue') ## ## Call: ## roc.default(response = obese, predictor = glm.fit$fitted.values,     percent = T, plot = T, legacy.axes = T, xlab = "False positive percentage",     ylab = "True positive percentage", col = "darkslateblue",     lwd = 4, print.auc = T, print.auc.x = 45, partial.auc = c(100,         90), auc.polygon = T, auc.polygon.col = "midnightblue")
##
## Data: glm.fit$fitted.values in 45 controls (obese 0) < 55 cases (obese 1). ## Partial area under the curve (specificity 100%-90%): 4.727% Can we bootstrap? ci.auc(roc(obese, glm.fit$fitted.values), conf.level=0.95, method=
"bootstrap", boot.n = 2000, boot.stratified = TRUE, reuse.auc=TRUE,
progress = getOption("pROCProgress")\$name, parallel=FALSE)
## 95% CI: 0.7471-0.9006 (2000 stratified bootstrap replicates)

####TESTING A LOGISTIC REGRESSION MODEL:

1. GOODNESS-OF-FIT: Model calibration

The Hosmer-Lemeshow goodness of fit test divides up in boxes the predicted probabilities (in R the function is fitted as opposed to predict), and runs a chi-square test comparing to the percentage of cases that have $$Y=1$$ among those with predicted probability within a certain interval. See here:

The Hosmer-Lemeshow goodness of fit test is based on dividing the sample up according to their predicted probabilities, or risks. Specifically, based on the estimated parameter values $$\hat{\beta}_{0},\hat{\beta}_{1},..,\hat{\beta}_{p},$$ for each observation in the sample the probability that $$Y=1$$ is calculated, based on each observation’s covariate values:

$\hat{\pi} = \frac{\exp(\hat{\beta}_{0}+\hat{\beta}_{1}X_{1}+..+\hat{\beta}_{p}X_{p})}{1+\exp(\hat{\beta}_{0}+\hat{\beta}_{1}X_{1}+..+\hat{\beta}_{p}X_{p})}$

The observations in the sample are then split into $$g$$ groups (we come back to choice of g later) according to their predicted probabilities. Suppose (as is commonly done) that $$g=10.$$ Then the first group consists of the observations with the lowest $$10\%$$ predicted probabilities. The second group consists of the $$10\%$$ of the sample whose predicted probabilities are next smallest, etc etc.

Suppose for the moment that all of the observations in the first group had a predicted probability of $$0.1.$$ Then, if our model is correctly specified, we would expect the proportion of these observations who have $$Y=1$$ to be $$10\%.$$ Of course, even if the model is correctly specified, the observed proportion will deviate to some extent from 10%, but not by too much. If the proportion of observations with $$Y=1$$ in the group were instead $$90\%,$$ this is suggestive that our model is not accurately predicting probability (risk), i.e. an indication that our model is not fitting the data well.

In practice, as soon as some of our model covariates are continuous, each observation will have a different predicted probability, and so the predicted probabilities will vary in each of the groups we have formed. To calculate how many Y=1 observations we would expect, the Hosmer-Lemeshow test takes the average of the predicted probabilities in the group, and multiplies this by the number of observations in the group. The test also performs the same calculation for $$Y=0,$$ and then calculates a Pearson goodness of fit statistic

$\sum^{1}_{k=0} \sum^{g}_{l=1} \frac{(o_{kl}-e_{kl})^{2}}{e_{kl}}$

where $$o_{0l}$$ denotes the number of observed $$Y=0$$ observations in the $$l$$-th group, $$o_{1l}$$ denotes the number of observed $$Y=1$$ observations in the $$l$$-th group, and $$e_{0l}$$ and $$e_{1l}$$ similarly denote the expected number of zeros.

In a 1980 paper Hosmer-Lemeshow showed by simulation that (provided $$p+1<g$$) their test statistic approximately followed a chi-squared distribution on $$g-2$$ degrees of freedom, when the model is correctly specified. This means that given our fitted model, the p-value can be calculated as the right hand tail probability of the corresponding chi-squared distribution using the calculated test statistic. If the p-value is small, this is indicative of poor fit.

It should be emphasized that a large p-value does not mean the model fits well, since lack of evidence against a null hypothesis is not equivalent to evidence in favor of the alternative hypothesis. In particular, if our sample size is small, a high p-value from the test may simply be a consequence of the test having lower power to detect mis-specification, rather than being indicative of good fit.

1. C-STATISTIC or AUC (area under the curve of the ROC): Discriminatory power of the model

It is based on selecting different probabilities from $$0$$ to $$1$$ and calculating sensitivity and specificity in an ROC curve, and then measuring the AUC. See here:

Our model or prediction rule is perfect at classifying observations if it has 100% sensitivity and 100% specificity. Unfortunately in practice this is (usually) not attainable. So how can we summarize the discrimination ability of our logistic regression model? For each observation, our fitted model can be used to calculate the fitted probabilities $$P(Y=1|X_{1},..,X_{p}).$$ On their own, these don’t tell us how to classify observations as positive or negative. One way to create such a classification rule is to choose a cut-point $$c,$$ and classify those observations with a fitted probability above $$c$$ as positive and those at or below it as negative. For this particular cut-off, we can estimate the sensitivity by the proportion of observations with $$Y=1$$ which have a predicted probability above $$c,$$ and similarly we can estimate specificity by the proportion of $$Y=0$$ observations with a predicted probability at or below $$c.$$ If we increase the cut-point $$c,$$ fewer observations will be predicted as positive. This will mean that fewer of the $$Y=1$$ observations will be predicted as positive (reduced sensitivity), but more of the $$Y=0$$ observations will be predicted as negative (increased specificity). In picking the cut-point, there is thus an intrinsic trade off between sensitivity and specificity. Now we come to the ROC curve, which is simply a plot of the values of sensitivity against one minus specificity, as the value of the cut-point $$c$$ is increased from $$0$$ through to $$1.$$ Example:

set.seed(63126)
n <- 1000
x <- rnorm(n)
pr <- exp(x)/(1+exp(x))
y <- 1*(runif(n) < pr)
mod <- glm(y~x, family="binomial")
predpr <- predict(mod,type=c("response"))
library(pROC)
roccurve <- roc(y ~ predpr)
plot(roccurve)

A popular way of summarizing the discrimination ability of a model is to report the area under the ROC curve. We have seen that a model with discrimination ability has an ROC curve which goes closer to the top left hand corner of the plot, whereas a model with no discrimination ability has an ROC curve close to a $$45$$ degree line. Thus the area under the curve ranges from $$1,$$ corresponding to perfect discrimination, to $$0.5,$$ corresponding to a model with no discrimination ability. The area under the ROC curve is also sometimes referred to as the c-statistic (c for concordance).

auc(roccurve)
## Area under the curve: 0.7463

The ROC curve (AUC) has a somewhat appealing interpretation. It turns out that the AUC is the probability that if you were to take a random pair of observations, one with $$Y=1$$ and one with $$Y=0,$$ the observation with $$Y=1$$ has a higher predicted probability than the other. The AUC thus gives the probability that the model correctly ranks such pairs of observations.

In the biomedical context of risk prediction modelling, the AUC has been criticized by some. In the risk prediction context, individuals have their risk of developing (for example) coronary heart disease over the next 10 years predicted. Thus a measure of discrimination which examines the predicted probability of pairs of individuals, one with $$Y=1$$ and one with $$Y=0,$$ does not really match the prospective risk prediction setting, where we do not have such pairs.


text:
# Question on universailty classes

Hi

When a particular model (say Ising) is solved on a particular lattice (say 2D triangular), do the critical exponents of the same model fall within the same universality class (have same critical exponents) as when solved on a different lattice (say 3D cubic)?

Thanks,
Mavi

I found out one part of the answer: changing the dimensionality of the lattice will change the critical exponents for the same model (say Ising). So the modified question would be whether the exponents change while changing the lattice structure (say square to triangle) for the same model in a given dimension?

Mavi

Guessing another part of the answer: it seems that since the continuum limit or the lattice limit must yield the same critical exponents (due to scaling invariance after graining), the lattice structure itself must not matter for the exponents. I wonder if this reasoning is correct.

Mavi

For the ferromagnetic Ising model case, outside of the dimensionality, it would be fair to assume that the lattice does not matter. There is no guarantee on this though. For a dramatic example, it should be fairly clear that the physical properties of an antiferromagnetic Ising model with be strongly dependent on the lattice geometry.

The problem with your argument is that you are assuming that given a model on an arbitrary lattice (with fixed dimensionality) you'll always arrive at the same long distance limit. It is difficult to show such a thing directly from a given model, mostly since the coarse graining procedure must done within some approximation (which may or may not be justified).


text:
60 views

$S_{2}$ and $S_{3}$?

+1 vote

s2  and s3  are true  ,  s1 false

can u plz explain .....about S2

if f(n) $\leq$ g(n) then 2f(n) $\leq$ 2g(n)

sumit goyal 1 explain S1 please

@Mk Utkarsh   only way to do is to solve more no. of examples as possible i follow this method only and with help of past experience i solve it ,  let f(n) = $\frac{1}{n}$   , $(f(n))^{2} = \frac{1}{n^2}$

$\frac{1}{n} \neq O( \frac{1}{n^2})$  take very large values of n its failing , since we find atleast one function which dont satisy  option a false

oh thanks  :D


text:
# nLab powering

Contents

This entry is about the formal dual to tensoring in the generality of category theory. For the different concept of cotensor product of comodules see there.

### Context

#### Enriched category theory

enriched category theory

## Extra stuff, structure, property

### Homotopical enrichment

#### Limits and colimits

limits and colimits

# Contents

## Idea

In a closed symmetric monoidal category $V$ the internal hom $[-,-] : V^{op} \times V \to V$ satisfies the natural isomorphism

$[v_1,[v_2,v_3]] \simeq [v_2,[v_1,v_3]]$

for all objects $v_i \in V$ (prop.). If we regard $V$ as a $V$-enriched category we write $V(v_1,v_2) := [v_1,v_2]$ and this reads

$V(v_1,V(v_2,v_3)) \simeq V(v_2,V(v_1,v_3)) \,.$

If we now pass more generally to any $V$-enriched category $C$ then we still have the enriched hom object functor $C(-,-) : C^{op} \times C \to V$. One says that $C$ is powered over $V$ if it is in addition equipped also with a mixed operation $\pitchfork : V^{op} \times C \to C$ such that $\pitchfork(v,c)$ behaves as if it were a hom of the object $v \in V$ into the object $c \in C$ in that it satisfies the natural isomorphism

$C(c_1,\pitchfork(v,c_2)) \simeq V(v,C(c_1,c_2)) \,.$

## Definition

###### Definition

Let $V$ be a closed monoidal category. In a $V$-enriched category $C$, the power of an object $y\in C$ by an object $v\in V$ is an object $\pitchfork(v,y) \in C$ with a natural isomorphism

$C(x, \pitchfork(v,y)) \cong V(v, C(x,y))$

where $C(-,-)$ is the $V$-valued hom of $C$ and $V(-,-)$ is the internal hom of $V$.

We say that $C$ is powered or cotensored over $V$ if all such power objects exist.

###### Remark

Powers are frequently called cotensors and a $V$-category having all powers is called cotensored, while the word “power” is reserved for the case $V=$ Set. However, there seems to be no good reason for making this distinction. Moreover, the word “tensor” is fairly overused, and unfortunate since a tensor (= a copower) is a colimit, while a cotensor (= power) is a limit.

## Properties

• Powers are a special sort of weighted limits. Conversely, all weighted limits can be constructed from powers together with conical limits. The dual colimit notion of a power is a copower.

## Examples

### In 1-category theory

• $V$ itself is always powered over itself, with $\pitchfork(v_1,v_2) := [v_1,v_2]$.

• Every locally small category $C$ ($V = (Set,\times)$ ) with all products is powered over Set: the powering operation

$\pitchfork(S,c) := \prod_{s\in S} c$

of an object $c$ by a set $S$ forms the $|S|$-fold cartesian product of $c$ with itself, where $|S|$ is the cardinality of $S$.

The defining natural isomorphism

$Hom_C(c_1,\pitchfork(S,c_2))\simeq Hom_{Set}(S,Hom_C(c_1,c_2))$

is effectively the definition of the product (see limit).

### Powering of $\infty$-toposes over $\infty$-groupoids

We discuss how the powering of $\infty$-toposes over $Grpd_\infty$ is given by forming mapping stacks out of locally constant $\infty$-stacks. All of the following formulas and their proogfs hold verbatim also for Grothendieck toposes, as they just use general abstract properties.

Let $\mathbf{H}$ be an $\infty$-topos

$\mathbf{H} \big( (-) ,\, Maps(X,A) \big) \;\simeq\; \mathbf{H} \big( (-) \times X ,\, A \big) \xrightarrow{ \mathbf{H} \big( (-) \times f ,\, A \big) } \mathbf{H} \big( (-) \times Y ,\, A \big) \;\simeq\; \mathbf{H} \big( (-) ,\, Maps(X,A) \big) \,.$

By definition, for any $S \in Grpd_\infty$ and $X \in \mathbf{H}$ the powering] is the (∞,1)-limit over the diagram constant on $X$

$X^K \,=\, {\lim_\leftarrow}_K X$

while the tensoring is the (∞,1)-colimit over the diagram constant on $X$

$K \cdot X \,=\, {\lim_{\to}}_K X \,.$

###### Remark

Under Isbell duality, the powering operations on homotopy types $X$ corresponds to higher order Hochschild cohomology of suitable algebras of functions on $X$, as discussed there.

###### Proposition

The powering of $\mathbf{H}$ over $Grpd_\infty$ is given by the mapping stack out of the locally constant $\infty$-stacks:

$\array{ Grpd_\infty^{op} \times \mathbf{H} & \overset{ LConst^{op} \times \mathrm{id} }{\longrightarrow} & \mathbf{H}^{op} \times \mathbf{H} & \overset{Maps(-,-)}{\longrightarrow} & \mathbf{H} }$

in that this operation has the following properties:

1. For all $X,\,A \,\in\, \mathbf{H}$ and $S \,\in\, Grpd_\infty$ we have a natural equivalence

$\mathbf{H} \Big( X ,\, Maps \big( LConst(S) ,\, A \big) \Big) \;\; \simeq \;\; Grpd_\infty \Big( S ,\, \mathbf{H} \big( X ,\, A \big) \Big)$
2. In its first argument the operation

1. sends the terminal object (the point) to the identity:

(3)$Maps \big( LConst(\ast) ,\, X \big) \;\; \simeq \;\; X$
2. (4)$Maps \Big( \underset{ \longrightarrow }{\lim} \, LConst\big(S_\bullet\big) ,\, X \Big) \;\; \simeq \;\; \underset{ \longleftarrow }{\lim} \, Maps \Big( LConst\big(S_\bullet\big) ,\, X \Big) \,,$

where all equivalences shown are natural.

###### Proof

For the first statement to be proven, consider the following sequence of natural equivalences:

$\begin{array}{lll} \mathbf{H} \Big( X ,\, Maps \big( LConst(S) ,\, A \big) \Big) & \;\simeq\; \mathbf{H} \big( X \times LConst(S) ,\, A \big) & \text{(2)} \\ & \;\simeq\; \mathbf{H} \Big( LConst(S) ,\, Maps \big( X ,\, A \big) \Big) & \text{(2)} \\ & \;\simeq\; Grpd_\infty \Big( S ,\, \Gamma \, Maps \big( X ,\, A \big) \Big) & \text{ (1) } \\ & \;\simeq\; Grpd_\infty \bigg( S ,\, \mathbf{H} \Big( \ast_{\mathbf{H}} ,\, Maps \big( X ,\, A \big) \Big) \bigg) & \text{by}\;\text{this Prop.} \\ & \;\simeq\; Grpd_\infty \Big( S ,\, \mathbf{H} \big( \ast_{\mathbf{H}} \times X ,\, A \big) \Big) & \text{(2)} \\ & \;\simeq\; Grpd_\infty \Big( S ,\, \mathbf{H} \big( X ,\, A \big) \Big) \end{array}$

For the second statement, recall that hom-functors preserve limits in that there are natural equivalences of the form

(5)$\mathbf{H} \Big( \underset{\underset{i}{\longrightarrow}}{\lim} \,, X_i ,\, \underset{\underset{j}{\longleftarrow}}{\lim} \,, A_j \Big) \;\; \simeq \;\; \underset{\underset{i}{\longleftarrow}}{\lim} \, \underset{\underset{j}{\longleftarrow}}{\lim} \, \mathbf{H} \Big( X_i ,\, A_j \Big) \,,$

and that $\infty$-toposes have universal colimits, in particular that the product operation is a left adjoint (2) and hence preserves colimits:

(6)$(-) \,\times\, \underset{{\longrightarrow}}{\lim} \, S_\bullet \;\; \simeq \;\; \underset{{\longrightarrow}}{\lim} \, \big( (-) \,\times\, S_\bullet \big) \,.$

With this, we get the following sequences of natural equivalences:

$\begin{array}{lll} & \mathbf{H} \bigg( (-) ,\, Maps \Big( \underset{\longrightarrow}{\lim} \, LConst(S_\bullet) ,\, X \Big) \bigg) \\ & \;\simeq\; \mathbf{H} \Big( (-) \times \underset{\longrightarrow}{\lim} \, LConst(S_\bullet) ,\, X \Big) & \text{ (2) } \\ & \;\simeq\; \mathbf{H} \Big( \underset{\longrightarrow}{\lim} \big( (-) \times LConst(S_\bullet) \big) ,\, X \Big) & \text{ (6) } \\ & \;\simeq\; \underset{\longleftarrow}{\lim} \, \mathbf{H} \big( (-) \times LConst(S_\bullet) ,\, X \big) & \text{ (5) } \\ & \;\simeq\; \underset{\longleftarrow}{\lim} \, \mathbf{H} \Big( (-) ,\, Maps \big( LConst(S_\bullet) ,\, X \big) \Big) & \text{ (2) } \\ & \;\simeq\; \mathbf{H} \Big( (-) ,\, \underset{\longleftarrow}{\lim} \, Maps \big( LConst(S_\bullet) ,\, X \big) \Big) & \text{ (5) } \,. \end{array}$

This implies (4) by the $\infty$-Yoneda lemma over $\mathbf{H}$ (which is large but locally small, so that the lemma does apply).

Finally (3) is immediate from the fact that $LConst$ preserves the terminal object, by definition:

$Maps \big( LConst(\ast) ,\, X \big) \;\simeq\; Maps \big( \ast_{\mathbf{H}} ,\, X \big) \;\simeq\; X \,.$

## References

Textbook accounts:

Last revised on October 18, 2021 at 06:07:51. See the history of this page for a list of all contributions to it.


text:
# SoundCloud Adds $50 Million And Meeker For U.S. Expansion Almost a year to the day after it last took funding, audio-hosting service SoundCloud is taking a further round from new investors. The round led by Kleiner Perkins Caufield & Byers is also joined by GGV Capital and is for$50 million and a \$200 million valuation, TechCrunch reports, adding SoundCloud “will use the funding to expand more rapidly, especially in the US”.

SoundCloud CEO Alexander Ljung, who just fractured his foot in Miami, seems pretty pleased with the additions of KPCB and its partner Mary Meeker.

“With Kleiner Perkins’ impeccable track record for recognizing and investing in companies with a long-term view, and their understanding and support for the potential of sound, we couldn’t be more excited about working together,” says Ljung in SoundCloud’s announcement.

“Mary Meeker showed very clearly in her recent State of the Internet 2011 presentation that sound is the next frontier on the web. That vision, along with spotlighting mobile and international markets, made it clear that Kleiner Perkins aligns with SoundCloud’s mission.”

SoundCloud, a freemium service, raised a €2.5 million http://paidcontent.co.uk/article/419-soundcloud-gets-2.5-million-for1/round, led by Doughty Hanson Technology Ventures, in April 2009, followed by a round from Union Square and Index Ventures in January 2011.

The notion of a “YouTube (NSDQ: GOOG) for audio” is as old as the hills though, perhaps strangely, not tried by many players. With its new resources, SoundCloud is giving rival Audioboo a run for its money.


text:
REPORTS > KEYWORD > ALGORITHM:
Reports tagged with algorithm:
TR95-033 | 29th June 1995
Richard Beigel, David Eppstein

#### 3-Coloring in time O(1.3446^n): a no-MIS algorithm

We consider worst case time bounds for NP-complete problems
including 3-SAT, 3-coloring, 3-edge-coloring, and 3-list-coloring.
Our algorithms are based on a common generalization of these problems,
called symbol-system satisfiability or, briefly, SSS [R. Floyd &
R. Beigel, The Language of Machines]. 3-SAT is equivalent to
(2,3)-SSS while the other problems ... more >>>

TR06-025 | 19th January 2006
Leonid Gurvits

#### Hyperbolic Polynomials Approach to Van der Waerden/Schrijver-Valiant like Conjectures :\\ Sharper Bounds , Simpler Proofs and Algorithmic Applications

Let $p(x_1,...,x_n) = p(X) , X \in R^{n}$ be a homogeneous polynomial of degree $n$ in $n$ real variables ,
$e = (1,1,..,1) \in R^n$ be a vector of all ones . Such polynomial $p$ is
called $e$-hyperbolic if for all real vectors $X \in R^{n}$ the univariate polynomial
equation ... more >>>

TR08-074 | 19th July 2008
Neeraj Kayal, Timur Nezhmetdinov

#### Factoring groups efficiently

We give a polynomial time algorithm that computes a
decomposition of a finite group G given in the form of its
multiplication table. That is, given G, the algorithm outputs two
subgroups A and B of G such that G is the direct product
of A ... more >>>

TR13-123 | 6th September 2013
Joshua Grochow, Youming Qiao

#### Algorithms for group isomorphism via group extensions and cohomology

The isomorphism problem for groups given by multiplication tables (GpI) is well-known to be solvable in n^O(log n) time, but only recently has there been significant progress towards polynomial time. For example, in 2012 Babai et al. (ICALP 2012) gave a polynomial-time algorithm for groups with no abelian normal subgroups. ... more >>>

TR13-141 | 8th October 2013
Leonid Gurvits

#### Boolean matrices with prescribed row/column sums and stable homogeneous polynomials: combinatorial and algorithmic applications

Revisions: 1

We prove a new efficiently computable lower bound on the coefficients of stable homogeneous polynomials and present its algorthmic and combinatorial applications. Our main application is the first poly-time deterministic algorithm which approximates the partition functions associated with
boolean matrices with prescribed row and (uniformly bounded) column sums within simply ... more >>>

ISSN 1433-8092 | Imprint


text:
## Taiwanese Journal of Mathematics

### A Nekhoroshev Type Theorem of Higher Dimensional Nonlinear Schrödinger Equations

#### Abstract

In this paper, we prove a Nekhoroshev type theorem for high dimensional NLS (nonlinear Schrödinger equations):$\mathrm{i} \partial_{t} u - \Delta u + V * u + \partial_{\overline{u}} g(x,u,\overline{u}) = 0, \quad x \in \mathbb{T}^d, \; t \in \mathbb{R}$ where real-valued function $V$ is sufficiently smooth and $g$ is an analytic function. We prove that, for any given $M \in \mathbb{N}$, there exists an $\varepsilon_0 \gt 0$, such that for any solution $u = u(t,x)$ with initial data $u_0 = u_0(x)$ whose Sobolev norm $\|u_{0}\|_{s} = \varepsilon \lt \varepsilon_0$, during the time $|t| \leq \varepsilon^{-M}$, its Sobolev norm $\|u(t)\|_s$ remains bounded by $C_s \varepsilon$.

#### Article information

Source
Taiwanese J. Math. Volume 21, Number 5 (2017), 1115-1132.

Dates
Revised: 2 January 2017
Accepted: 8 January 2017
First available in Project Euclid: 1 August 2017

https://projecteuclid.org/euclid.twjm/1501599185

Digital Object Identifier
doi:10.11650/tjm/7951

#### Citation

Zhou, Shidi; Geng, Jiansheng. A Nekhoroshev Type Theorem of Higher Dimensional Nonlinear Schrödinger Equations. Taiwanese J. Math. 21 (2017), no. 5, 1115--1132. doi:10.11650/tjm/7951. https://projecteuclid.org/euclid.twjm/1501599185

#### References

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• ––––, An averaging theorem for quasilinear Hamiltonian PDEs, Ann. Henri Poincaré, 4 (2003), no. 4, 685–712.
• ––––, Birkhoff normal form for some nonlinear PDEs, Comm. Math. Phys. 234 (2003), no. 2, 253–285.
• D. Bambusi and B. Grébert, Birkhoff normal form for partial differential equations with tame modulus, Duke Math. J. 135 (2006), no. 3, 507–567.
• G. Benettin, L. Galgani and A. Giorgilli, A proof of Nekhoroshev's theorem for the stability times in nearly integrable Hamiltonian systems, Celestial Mech. 37 (1985), no. 1, 1–25.
• J. Bourgain, On diffusion in high-dimensional Hamiltonian systems and PDE, J. Anal. Math. 80 (2000), 1–35.
• J. Colliander, M. Keel, G. Staffilani, H. Takaoka and T. Tao, Transfer of energy to high frequencies in the cubic defocusing nonlinear Schrödinger equation, Invent. Math. 181 (2010), no. 1, 39–113.
• L. H. Eliasson, B. Grébert and S. B. Kuksin, KAM for the nonlinear beam equation, Geom. Funct. Anal. 26 (2016), no. 6, 1588–1715.
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• M. Guardia, E. Haus and M. Procesi, Growth of Sobolev norms for the analytic NLS on $\mb{T}^2$, Adv. Math. 301 (2016), 615–692.
• M. Guardia and V. Kaloshin, Growth of Sobolev norms in the cubic defocusing nonlinear Schrödinger equation, J. Eur. Math. Soc. (JEMS) 17 (2015), no. 1, 71–149.
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text:
# If |X|=5, how many non-ordered pairs {A,B} consisting of disjoint subsets exist?

• March 8th 2011, 05:38 PM
jmgilbert
If |X|=5, how many non-ordered pairs {A,B} consisting of disjoint subsets exist?
Hi, I have the following exercise, but I don't know how to solve it so any help you can give me would be great... Thanks!

It says

Suppose that |X|=5. Find how many non-ordered pairs {A,B} consisting of non-empty disjoint subsets A and B of the set X exist.
• March 9th 2011, 12:18 PM
tom@ballooncalculus
How many pairs of singletons could you draw from the power set? <single, single>

How many pairs of any one pair and the triple which is disjoint to it could you draw? <pair, triple>

How many distinct pairs of one pair and one of the 3 pairs disjoint to it could you draw? <pair, pair>

How many pairs of any one pair and one of the 3 singletons disjoint to it could you draw? <pair, single>

How many pairs of any triple and one of the 2 singletons disjoint to it? <triple, single>

How many pairs of any quadruple and the singleton disjoint to it? <quadruple, single>
• March 9th 2011, 02:52 PM
Plato
@tom@ballooncalculus
Where have you considered the $\emptyset$ in any of your counting?
• March 9th 2011, 05:25 PM
topspin1617
Quote:

Originally Posted by Plato
@tom@ballooncalculus
Where have you considered the $\emptyset$ in any of your counting?

The question says not to.

Don't you need to do a little more though, considering different possibilities for, e.g., the particular 4-subset?
• March 9th 2011, 10:51 PM
tom@ballooncalculus
Well, for any of the 5 particular quadruples (each giving a pairing of that quadruple with its disjoint single), no. What different possibilities do you see?
• March 10th 2011, 08:22 AM
topspin1617
Like you said, 5 possibilities.

Maybe it's just the way you interpret what your wrote... I'm reading it as counting for one fixed quadruple instead of all possible ones.


text:
# Introduction

Exact Inference is computationally expensive. If we are given a bayes net, calculating the marginal distribution of a variable is a lot of work.

Consider the following Bayesian network:

To compute p(D), we calculate the joint probability and sum out everything but D:

$p(D) = \sum_a \sum_b \sum_c p(A,B,C,D) = \sum_a \sum_b \sum_c p(A)p(B|A)p(C|B)p(D|C)$

It seems like this is more work than it is worth. If each of the variables have k values, then we end up having to generate $k^n$ probabilities in the joint distribution that we have to sum over. There must be a more efficient way to compute these marginal probabilities.

Enter Variable Elimination.

# The Basic Ideas

Variable Elimination fundamentally contains two main ideas that help address this exponential blowup of the joint distribution.

The first idea is about exploiting properties of arithmetic. For example, if you have an expression:

$yx_1 + yx_2 + yx_3 + yx_4$

This expression has four multiplications and three additions. However, using the well known properties of arithmetic we can rearrange the expression:

$y(x_1 + x_2 + x_3 + x_4)$

Now we have ONE multiplication and three additions.

The second idea is about exploiting the structure of the Bayesian network. Many of the sub expressions in the joint only depend on a small number of variables.

For example, lets take our Bayes net from above (which is structured as a chain), again the marginal probability of p(D) is:

$p(D) = \sum_a \sum_b \sum_c p(a)p(b|a)p(c|b)p(D|c)$

Now if we apply what we know about arithmetic and the structure of the network we can rearrange:

$p(B) = \sum_a p(a)p(B|a)$

$p(C) = \sum_b p(b)p(C|b)$

$p(D) = \sum_c p(c)p(D|c)$

Essentially what we did here was perform the computation from the inside out. If we stick the above expressions back into the equation for p(D), we get:

$p(D) = \sum_c p(D|c) \Bigg( \sum_b p(C|b) \Bigg( \sum_a p(a)p(B|a) \Bigg)\Bigg)$

The inner expression (or p(B) ) is computed first for all values of B and stored so we only compute them once (thus eliminating A). Then p(C) is computed with the values of p(B) and stored. Finally p(D) is computed with the stored values of p(C).

In general terms, lets assume a Bayesian network that has the structure of a chain with n variables $X_1 \rightarrow \ldots \rightarrow X_n$ ,where each variable has $k$ possible values. Computing $p(X_{i+1})$ can be defined recursively.

$p(X_{i+1}) = \sum_{x_{i}} p(X_{i+1} | x_i)p(x_i)$

Each recursive step is $O(k^2)$ and we recurse through all n variables in the worst case scenerio, so the total bigO is $O(nk^2)$. This is much better than generating the full $k^n$ probabilities to sum over in the joint distribution.

# The Algorithm

To define the basic variable elimination algorithm, we must first define a few terms.

• $\phi$ is called a factor,or a conditional probability from a Bayesian network. For example, from the above network, p(B|A) represents one factor.
• $Scope[\phi]$ is the set of variables that are used by the factor $\phi$.

VariableElimination(
$\Phi$  // Set of factors for a given Bayesian network
Q  // Query variable
E=e  // Set of observed variables (and their values)
Z  // Ordered set of variables not in Q$\cup$E
)

* For each $\phi \in \Phi$ set all observed variables E $\in Scope[\phi]$ to E=e

* For each $Z_i \in Z$
$\Phi \leftarrow Eliminate(\Phi,Z_i)$

* $\phi^{*} \leftarrow \prod_{\phi \in \Phi} \phi$

* $\alpha \leftarrow \sum_{q \in Val(Q)} \phi^{*}(q)$

* return $\phi^{*} / \alpha$
Eliminate(
$\Phi$ \\ Set of factors
Z \\ Variable to eliminate

* $\Phi' \leftarrow \{$ Every $\phi \in \Phi$ where $Z \in Scope[\phi]\}$
* $\Phi'' \leftarrow \Phi - \Phi'$
* $\psi \leftarrow \prod_{\phi \in \Phi'} \phi$
* $\tau \leftarrow \sum_{z} \psi$
* return $\Phi'' \cup \{\tau\}$

Note that this algorithm is based off the version presented in (Koller, Friedman, 2009, see references).

# Example walk through

For an example, lets take the following Bayesian network that you have seen before (Russel,Norvig, 2005):

For convenience let B = Burglary, E = Earthquake, A = Alarm, M = Marycalls, and J = Johncalls.

p(B) = .001
p(E) = .002

p(A|B,E):                p(M|A):             p(J|A):
----------------         ------------       ------------
| B | E | p(A) |         | A | p(M) |       | A | p(J) |
----------------         ------------       ------------
| t | t | .95  |         | t | .70  |       | t | .90  |
| t | f | .94  |         | f | .01  |       | f | .05  |
| f | t | .29  |         ------------       ------------
| f | f | .001 |
----------------

Lets say we would like to compute p(B|J,M), which is the probability that there was a Burglary given John and Mary called.

$p(B|M,J) \propto \sum_E \sum_A p(B,E,A,M,J) \propto \sum_E \sum_A p(B)p(E)p(A|B,E)p(M|A)p(J|A)$

We will now apply the Variable Elimination Algorithm:

• M and J are observed and B is the query variable, so we need to eliminate E and A

Eliminate E:

$\tau_1(B,A) = \sum_E p(E)*p(A|B,E)$

• Note that $\tau_1(B,A)$ is a vector of values for each combination of B and A. Specifically:

$\tau_1(B,A) = p(E)*p(A|B,E) + p(\bar{E})*p(A|B,\bar{E}) = (.002 * .95) + (.998 * .94) =$.94002

$\tau_1(B,\bar{A}) = p(E)*p(\bar{A}|B,E) + p(\bar{E})*p(\bar{A}|B,\bar{E}) = (.002 * .05) + (.998 * .06) =$.05998

$\tau_1(\bar{B},A) = p(E)*p(A|\bar{B},E) + p(\bar{E})*p(A|\bar{B},\bar{E}) = (.002 * .29) + (.998 * .001) =$.00158

$\tau_1(\bar{B},\bar{A}) = p(E)*p(\bar{A}|\bar{B},E) + p(\bar{E})*p(\bar{A}|\bar{B},\bar{E}) = (.002 * .71) + (.998 * .999) =$.99842

So we now have $p(B)\tau_1(B,A)p(M|A)p(J|A)$

Eliminate A:

$\tau_2(B,M,J) = \sum_A \tau_1(B,A)*p(M|A)*p(J|A)$

• Note that M and J are observed as true and so $\tau_2(B,M,J)$ is only a vector over the values of B and can be written as $\tau_2'(B)$. Specifically:

$\tau_2'(B) = \tau_1(B,A)*p(J|A)*p(M|A) + \tau_1(B,\bar{A})*p(J|\bar{A})*p(M|\bar{A}) = (.94002 * .90 * .70) + (.05998 * .05 * .01) =$.59224

$\tau_2'(\bar{B}) = \tau_1(\bar{B},A)*p(J|A)*p(M|A) + \tau_1(\bar{B},\bar{A})*p(J|\bar{A})*p(M|\bar{A}) = (.00158 * .90 * .70) + (.99842 * .05 * .01) =$.00149

Finishing up

So we now have $p(B)\tau_2'(B)$ and can easily compute p(B|M,J):

$p(B|M,J) = \frac{p(B)\tau_2'(B)}{\sum_B p(B)\tau_2'(B)} = \frac{p(B)\tau_2'(B)}{p(B)\tau_2'(B) + p(\bar{B})\tau_2'(\bar{B})} = \frac{(.001 * .59224)}{(.001 * .59224)+(.999 * .00149)} =$.2846

Note that this whole algorithm performed only 18 multiplication, 7 additions and 1 division for a total of 26 arithmetic operations. We never had to calculate the full joint distribution which would have required 128 multiplications alone.

# Other Considerations

## Graph Structure

It should be apparent that the structure of the network has a lot to do with how much advantage is gained by the Variable Elimination algorithm. More complicated networks will result in factors that depend on a large number of variables that cannot be easily eliminated. A simple example is shown here:

The resulting factors are: p(B)p(C)p(D)p(E)p(A|B,C,D,E). Trying to eliminate even one variable will result in having to enumerate almost every combination of values for each variable, not much different than computing the whole joint distribution.

## Elimination Ordering

The order in which you eliminate variables also has an impact on the efficiency of the Variable Elimination Algorithm. For example, in the previous section, had we eliminated the Alarm variable first, we would have had to perform four additional multiplications. Not a big deal for a simple example, but for a more complicated network, the right ordering will make a significant difference. Unfortunately, finding the optimal elimination ordering is an NP-Hard problem. However, there are a few techniques out there that can find good elimination orderings. A couple examples are:

• Chordal Graphs - Uses, as a heuristic, the definition of a Chordal graph and how it relates to the resultant graph after a variable is eliminated to find a good elimination ordering.
• Minimal Fill/Size/Weight Search - Basically does a greedy search that orders the variables one at a time by determining which one will result in the smallest graph after that variable has been eliminated.

# Candidate Question

Given the following Bayesian Network (binary values), calculate p(B|A,E) using variable elimination (show steps).

Probabilities:   p(variable|condition) = {condition=false, condition=true}

p(A|B) = {.25, .33}   p(B|C) = {.10, .99}

p(C|D) = {.40, .70}   p(E|C) = {.20, .04}

p(F|B) = {.08, .85}   p(G|B) = {.50, .60}

p(D) = .115

# References

• Probabilistic Graphical Models Principles and Techniques, by Daphne Koller and Nir Friedman, 2009 MIT Press
• Artificial Intelligence A Modern Approach, by Stuart Russell and Peter Norvig, 2003 Pearson Education, Inc
• A Simple approach to Bayesian Network Computations, by N. Zhang and David Poole, 1994
• Efficient Reasoning in Graphical Models, by Rina Dechter, 1999


text:
# Equation not satisfied in Policy Iteration Algorithm

In equation 4.9 of Sutton and Barto's book on page 79, we have(for policy iteration algo):

$$\pi ^{'}(s) = arg \max_{a}\sum_{s',r}p(s',r|s,a)[r+\gamma v_{\pi}(s')]$$

where $$\pi$$ is the previous policy and $$\pi ^{'}$$ is the new policy. Hence in iterations $$k$$ it must mean

$$\pi _{k+1}(s) = arg \max_{a}\sum_{s',r}p(s',r|s,a)[r+\gamma v_{\pi_{k}}(s')]$$

But in the example given in the same book on page 77 we have:

Now for the concerned state marked in red -

So $$v_{\pi_{1}}$$ = -1 for all four surrounding states

r = -1 for all four surrounding states

p(s',r|s,a) = 1 for all four surrounding states

$$\pi _{2}(s) = arg \max_{a}[1*[-1+1*-1],1*[-1+1*-1],1*[-1+1*-1],1*[-1+1*-1]]$$ $$\pi _{2}(s) = arg \max_{a}(-2,-2,-2,-2)$$

Hence this should give us a criss-cross symbol(4 directional arrow) in $$\pi_{2}$$(s) but here a left arrow symbol is given.

What's wrong with my calculations.

Your calculations are correct, but you have misinterpreted the equations and the diagram. The index $$k$$ in $$v_k$$ for the diagram refers to the policy evaluation update iteration only, and is not related to the policy update step (which uses the notation $$\pi'$$ and does not mention $$k$$).
Policy improvement consists of multiple sweeps through states to fully evaluate the current policy and estimate the value function for it. After that, it updates the policy in a separate policy improvement step. There are two loops - an inner loop indexed by $$k$$ in the equations and diagram, plus an outer loop which is not given an index notation.
The diagram is not showing incremental $$\pi'$$ policies from outer loops over policy iteration. Instead it is showing "Greedy Policy w.r.t. $$v_k$$" steps in the inner loop - you can think of that as the policy $$\pi'$$ that you would get inside the first outer loop, if you terminated the policy evaluation stage after that iteration $$k$$ of the inner loop.
• In the case of this very simple environment, if you ran a single outer loop with long enough policy evaluation stage ($$k \ge 3$$) you would find the optimal policy.
• Even before the value function estimate is close to convergence (with a high $$k$$), the policy that could be derived from the new estimates could be used to improve the policy. That leads to the idea of the value iteration method.


text:
## Near-Optimal Sample Complexity Bounds for Constrained MDPs

Abstract: In contrast to the advances in characterizing the sample complexity for solving Markov decision processes (MDPs), the optimal statistical complexity for solving constrained MDPs (CMDPs) remains unknown. We resolve this question by providing minimax upper and lower bounds on the sample complexity for learning near-optimal policies in a discounted CMDP with access to a generative model (simulator). In particular, we design a model-based algorithm that addresses two settings: (i) relaxed feasibility, where small constraint violations are allowed, and (ii) strict feasibility, where the output policy is required to satisfy the constraint. For (i), we prove that our algorithm returns an $\epsilon$-optimal policy with probability $1 - \delta$, by making $\tilde{O}\left(\frac{S A \log(1/\delta)}{(1 - \gamma)^3 \epsilon^2}\right)$ queries to the generative model, thus matching the sample-complexity for unconstrained MDPs. For (ii), we show that the algorithm's sample complexity is upper-bounded by $\tilde{O} \left(\frac{S A \, \log(1/\delta)}{(1 - \gamma)^5 \, \epsilon^2 \zeta^2} \right)$ where $\zeta$ is the problem-dependent Slater constant that characterizes the size of the feasible region. Finally, we prove a matching lower-bound for the strict feasibility setting, thus obtaining the first near minimax optimal bounds for discounted CMDPs. Our results show that learning CMDPs is as easy as MDPs when small constraint violations are allowed, but inherently more difficult when we demand zero constraint violation.


text:
# B Rocket problem

Tags:
1. Apr 26, 2015

### Stephanus

Dear Physics Forum,
I really don't understand the concept of watt and newton.
I have two questions, perhaps someone can explain it to me.
=======================================================
Case 1:
If a one newton force is applied to a one kg rocket in space with no or very little effect of gravity, for ....
A. 10 seconds, it will travel (1/2 x a x t^2 = 1/2 x 1 x10^2 = 50) 50 metres, it has consumed 50 joules, so its power is 5 watt.
B. 20 seconds, it will travel 200 metres, it consumes 200 joules, power is 10 watt.
C. 30 seconds, 450 metres, 450 joules, 15 watt
It seems that the power is increasing with each passing time(second). Its energy consume is not constant, but accelerating with each passing time.
Question1: Is that true?
======================================================
Case 2:
Before I ask, I would like to insert some equations here.
E = P.t (energy is power x time, second)
E = F.d (energy is force x distance, metre)
F = M.a (force is mass x accel)
d = 1/2 a.tt (dist is accel x time squared)
P.t = F.d
E = (F) x (d)
E = (M x a) x (1/2 a x tt)
E = 1/2 M x aa x tt
aa = 2E/M/tt
a = sqrt(2E/M)/t
F =
M x sqrt(2E/M)/t = sqrt(2EM)/t
d =
1/2 a.tt = sqrt(1/2 x E/M)/t x tt = sqrt(1/x2 x E/M) x t
Note:
F.d = sqrt(2EM)/t x sqrt(1/2 x E/M) x t = sqrt(2 x 1/2 x EE) x t / t = sqrt(EE) = E
F.d is of course (E)nergy

--------------------------------------------------------------------------------
A one kg rocket in space (no gravity) having a one watt power. Assuming it has a very, very efficient fuel, so the fuel mass loss is negligible.
A. After 2 seconds, its energy consume is 1 watt x 2 seconds = 2 Joule.
Its Force will be 1 newton, its distance has to be 2 metres, 1 newton x 2 metres = 2 joule
B. After 8 seconds, its energy consume is 8 joules
Its Force is 0.5 newton x its distance is 16 metres = 8 joules
C. After 32 seconds , 32 joules
Force: 0.25 Newton, distance: 128 metres.
It seems with each passing time the force is diminishing.
Is that true?

Note: It will take 5.6 billion years to push it near the speed of light
11 billion years = twice the speed of light??? But that's another question for relavity I think.

======================================================

Conclusion:
1. If the force is constant, does the power increase?
2. If the power is constant, does the force decrease?

Peace on earth

Steven

2. Apr 26, 2015

### jbriggs444

Yes. The power produced from applying a fixed force increases with the velocity of its target.

One watt power applied to what?

Rocket motors are normally rated based on their total power output -- power going to the craft plus power going to the exhaust stream. That power level turns out to be the same no matter how fast the rocket is moving.

The faster the craft is moving, the more power is going into accelerating the craft. And the less power is going into the exhaust stream. Go fast enough and the exhaust stream is actually less energetic than the unburned fuel (as viewed from the perspective of the ground that you are whizzing past).

3. Apr 26, 2015

### Stephanus

Thank you for the answer Sir, (I guess you are male judging from you avatar)

Yes. The power produced from applying a fixed force increases with the velocity of its target.

"Power increases, applying a fixed force with the velocity"
I have a little "trouble" with the word "velocity" here.
- Supposed the target is in intergalactic void, with really no or miniscule gravity. How in the hell, (hell might be right if you're in Bootes void) that the rocket knows its velocity?. Supposed there is an astronout in that rocket and he when goes to sleep, the rocket accelerates for some hours and stops before he wakes up. After he wakes up, he would feel no acceleration at all, altough its speed might be hundreds of kms per hour. Does it require more power than before he slept, to thrust or add velocity to the rocket by 1 m/s?
-----------------------------------------------------------------------------------------------------------------

One watt power applied to what?

1. Rocket motors are normally rated based on their total power output...
2. The faster the craft is moving, the more power is going into accelerating the craft...
----------------
1. Supposed it does not use motor, but some very advanced ion thruster as in Startrek. That it ejects 1 molecule, or 1 polimer for that matter, near the speed of light to accel that 1 kg rocket by 1 m/s. Does the principle still apply?
2. The faster the craft...,
Perhaps you are refering air or ground friction or gravitation bound? Supposed the rocket is in intergalactic void where there's no friction or gravitation effect. Does it still require more power going into accelerationg the craft?

Again, thank you very much for you attention, sir.

Last edited: Apr 26, 2015
4. Apr 26, 2015

### jbriggs444

Pick a reference frame, any reference frame. The principle applies regardless.

Yes. Do the math. Take a 1.000001 kg object at rest and eject one milligram of exhaust at 1000 meters per second. Compute the total energy of the remaining 1 kg object and of the 1 milligram exhaust. Repeat the exercise but assume that the object starts with a velocity of one meter per second.

Compute the amount by which total energy increases in the first case.
Compute the amount by which total energy increases in the second case.
Compare the two.

5. Aug 7, 2015

### Stephanus

Dear PF Forum,
Can I refresh my understanding of kinetic energy?
Supposed a rocket floating in space. With relative velocity 0 wrt to a planet. Let's say that the rocket is some distance away, so that the effect of that planet gravity is very small.
The rocket is 11 kg.
Then, the rocket propels its material 1 kg to the "west".
It only produces 220 joules for that process.
In momentum conservation, we'll have...
$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$
Here:
m1: 10 kg
m2: 1 kg
n: The ratio m1/m2 = 10
Because u1 and u2 is zero, the rocket is at rest first wrt planet, so
$0 = m_1v_1 + m_2v_2$
The rocket produces 220 joules for the ejecting process, so...
$\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = 220$
$m_1v_1 = -m_2v_2$
$m_2 = \frac{1}{n}m_1$
$v_2 = -nv_1$
combine those equations, we'll have
$\frac{1}{2}m_1v_1^2 + \frac{1}{2}nm_1v_1^2 = 220$
$\frac{1}{2}m_1v_1^2(1+n) = 220$
$v_1 = 2$
So, M1 will travel to the 'east' 2 m /s and m2 will travel to the 'west' 20 m/s is that so?

6. Aug 8, 2015

### Qwertywerty

Yes , I believe it's correct .

7. Aug 16, 2015

### Stephanus

Dear PF Forum,
I have a question about rocket problem. That I asked in previous thread, but I didn't get enough feed back.
Actually I want to know the relation between watt and newton/power and force. But before that I'd like to know about this answers first.
A rocket is floating in space.
The mass of the rocket: 1,000 kg.
The rocket spends energy for this action for 7,200,000 joules
Let's say n is 10
So the rocket eject 1/10th of it's mass backward to propel the rocket forward.
So $M_{rocket} = 1000 kg$
$M_{body} = M_1; M_{ejecta} = M_2$

Based on momentum equation
$M_1 * V_1 = M_2 * V_2$
Based on kinetic energy equation
$0.5 M_1 * {V_1}^2 + 0.5 M_2 * {V_2}^2 = 7,200,000 joules$

So, trough a lengthy calculation I come up with this:
$M_1 = 900 kg$
$M_2 = 100 kg$, that is easy enough.
$V_1 = 40m/s$
$V_2 = 360m/s$

I calculate the energy consumed by each objects
$Ek_{body} = 0.5 * 900 kg * 40 * 40 = 720,000 joules$
$Ek_{ejecta} = 0.5 * 100 kg * 360 * 360 = 6,480,000 joules$

So the rocket consumes energy much smaller then its ejecta?

I play around with n. Now I plug 100 into n
It means that the rocket throw 1/100th of its mass as the ejecta, to propel forward according to Newton third law:

$M_1 = 990kg$
$M_2 = 10kg$
$V_1 = 12.06m/s$
$V_2 = 1,193.98m/s$
$Ek_1 = 72,000 joules$
$Ek_2 = 7,128,000 joules$
That low?

n=1000
$Ek_1 = 7,200 joules$
$Ek_2 = 7,192,800 joules$

Is my calculation wrong? So a rocket really wastes its energy enormously?

8. Aug 16, 2015

### paisiello2

I think your calculation might be right but your assumption for n is not realistic. Most real life rockets have n=1.11 to 1.25.

Also, all rockets that have achieved orbit are multiple stage rockets so there is an additional loss of mass there.

9. Aug 16, 2015

### Staff: Mentor

In terms of energy that goes to rocket and exhaust, rockets are horribly inefficient at low speeds, that is correct. There is no more efficient way in space, however. In an atmosphere, you can accelerate the air around the rocket, which is much more effective.

The ratio is better in a frame where the rocket has some initial speed. That shows that the energy ratio is not a very useful quantity. Rockets are all about velocity changes ("delta v").

10. Aug 16, 2015

### Staff: Mentor

Now you have one thread with plenty of feedback.

As was told to you each and every time you asked, yes, most of a rocket's energy goes into the exhaust at low speeds where they are very inefficient.

Thread closed. And don't reopen another one, the answer is still the same as the first three times and there is no need to say it again.