text:
## Question types: STACK

qtype_stack
Maintained by Christopher Sangwin, Tim Hunt
The STACK question type adds a sophisticated assessment in mathematics and related disciplines, with emphasis on formative assessment underpinned by computer algebra.
15k
3
Moodle 2.3, 2.4, 2.5

The STACK system is a computer aided assessment package for mathematics, which provides a question type for the Moodle quiz. In computer aided assessment (CAA), there are two classes of question types.

• Selected response questions
In these questions, a student makes a selection from, or interacts with, potential answers which the teacher has selected. Examples include multiple choice, multiple response and so on.
• Student-provided answer question
In these questions the student's answer contains the content. It is not a selection. Examples of these are numeric questions.

STACK concentrates on student-provided answers which are mathematical expressions. For example, a student might respond to a question with a polynomial or matrix. Essentially STACK asks for mathematical expressions and evaluates these using computer algebra. The prototype test is the following pseudo-code.

if
then
mark = 1,
else
mark = 0.

STACK uses a computer algebra system (CAS) to implement these mathematical functions. A CAS provides a library of functions with which to manipulate students' answers and generate outcomes such as providing feedback. Establishing algebraic equivalence with a correct answer is only one kind of manipulation which is possible.

Using CAS can also help generate random yet structured problems, and corresponding worked solutions.

In STACK a lot of attention has been paid to allowing teachers to author and manage their own questions. The following are the key features.

• Question versions are randomly generated within structured templates.
• There are many different kinds of inputs. These are, for example, where the student enters a mathematical expression, or makes a true/false selection.
• Mathematical properties of students' answers are established using answer tests within the CAS Maxima.
• Feedback is assigned on the basis of these properties using a potential response tree. This feedback includes:
1. Textual comments for the student.
2. A numerical mark.
3. Answer notes from which statistics for the teacher are compiled.

These broadly correspond to formative, summative and evaluative functions of assessment. Which of these outcomes is available to the student, and when, is under the control of the teacher.

• Multi-part mathematical questions are possible: each question may have any number of inputs and any number of potential response trees. There need not be a one-to-one correspondence between these.
• Partial credit is possible when an expression only satisfies some of the required properties.
• Plots can be dynamically generated and included within any part of the question, including feedback in the form of a plot of the student's expression.

### Sets

This plugin is part of set STACK.

### Contributors

Christopher Sangwin (Lead maintainer)
Please login to view contributors details and/or to contact them

• Thu, Mar 21, 2013, 1:50 AM
Hi,
I tried to instal the plugin several times. I think I followed step by steps the instructions but when running STACK healthcheck I get the message:
The version of the STACK-Maxima libraries being used (So old the version is unknown!) does not match what is expected (2012122800) by this version the the STACK question type. It is not really clear how that happened. You will need to debug this problem yourself.

And CAS result
Warning, empty result!
unpack_raw_result: no results were returned by the CAS.

Unpacked result as
Array
(
)
Has anybody similar problema?
• Thu, Mar 21, 2013, 2:19 AM
Emilio,

It looks to me like PHP could not start a CAS process at all. Hence, there is no version number being returned. If you email me the HTML page you get back from the healthcheck, and let me know a little about your system I will be able to help more.

If you are on unix you need to ensure maxima is installed and working. Try the command line "maxima".

Chris
• Wed, Jul 17, 2013, 3:00 PM
Hi:
Getting this error with latest release:
Coding error detected, it must be fixed by a programmer: Cannot require a CSS file after has been printed.

Was able to test with Moodle 2.5 and previous release. Any idea what to verify?

Thanks
• Wed, Jul 17, 2013, 3:55 PM
I have not seen that error from STACK.

Can you try again with Debugging set to DEVELOPER level, and send us the full error message, including the stack trace.
• Fri, Jul 19, 2013, 12:37 AM
Hi Tim,
I hope you don't mind that STACK is now listed in http://docs.moodle.org/25/en/Question_types#Third-party_question_types

and there is a Moodle Docs page at http://docs.moodle.org/25/en/STACK_question_type , whish is also translated into spanish.

The STACK plugin English language strings are 90% translated into spanish and the translation should be finished in a month.
• Fri, Jul 19, 2013, 12:40 AM
Thanks German. Very helpful.
• Mon, Jan 13, 2014, 3:17 AM
Wonderfull, plugin. Thank you so much.
• Sun, Jan 19, 2014, 5:04 AM
Is there a french translation for Stack's feed back. Or a guide on how I could translate them?
• Sun, Jan 19, 2014, 5:48 AM
Malik, STACK conforms exactly to the normal Moodle system of language strings for plugins. E.g. http://docs.moodle.org/24/en/Translation Please do get involved in translating STACK. This would be very welcome. If you need STACK specific changes, e.g. splitting up strings for any reason, please let me know. Chris
• Tue, Mar 18, 2014, 8:53 AM
Loving STACK! --I entered a bug on your github page. Is that the best way to alert you of a problem?
• Tue, Mar 18, 2014, 3:38 PM
Yes. That is the preferred way to report bugs.

If you are not sure whether something is a bug or not, or if you just want advice, then the Moodle Maths forum (linked to as 'Discussion' above) is the best place to ask.
• Sat, Apr 5, 2014, 2:10 AM
Is it possible to have a subscript in a student answer? For example, I'd like to have as an answer ta: 1/(4*pi*epsilon[0])*q/r^2 (Coulomb's law for a point charge). When I preview the question, the epsilon[0] renders correctly as the Greek letter epsilon with a subscript zero, but STACK rejects this when entered as the answer with the message "map: improper argument: epsilon[0]". Is there any way to allow a student to enter a subscript?

Kevin
• Mon, Apr 7, 2014, 1:32 PM
HI Chris, sorry to keep bothering you. Getting ready for the new quarter by writing new STACK questions! I have a question about plotting:

I'm using STACK 2014032800 (installed via git). None of my plots are showing grid lines; they do show zero lines. ("The grid has been changed by adding the gnuplot commands set zeroaxis, set grid. This adds more grid lines than Maxima's default...") I am using Gnuplot 4.2.6 and Maxima 5.32.1 (I compiled Maxima on my OSX Lion machine, but not gnuplot):

STACK_SETUP(ex):=block(
MAXIMA_VERSION_NUM:31.3,
TMP_IMAGE_DIR:"/Library/WebServer/moodledata/stack/tmp/",
IMAGE_DIR:"/Library/WebServer/moodledata/stack/plots/",
PLOT_TERMINAL:"png",
PLOT_TERM_OPT:"large transparent size 450,300",
DEL_CMD:"rm",
GNUPLOT_CMD:"/Applications/Gnuplot.app/Contents/Resources/bin/gnuplot",
MAXIMA_VERSION:"5.31.3",
URL_BASE:"!ploturl!",
true)$/* We are using an optimised lisp image with maxima and the stack libraries pre-loaded. That is why you don't see the familiar load("stackmaxima.mac")$ here. */

Any insight as to why grid lines are not showing up? In case it's relevant, changing [nticks] has no effect, either.

Thanks again,

Kevin.

• Tue, Dec 2, 2014, 12:38 AM
It seems that, if you edit the requirements in the version.php files for all the required plugins, specifically
* $plugin->requires change that value to 2013051400 * (qbehaviours only)$plugin->dependencies change all the version numbers there to 2013050100.

then STACK v3.3.3 will work with Moodle 2.5. However, we don't recomment that you use an un-supported version of Moodle. There are security risks. You should upgrade your version of Moodle instead. But, if you can't you may still be able to run the latest STACK.
• Wed, Mar 25, 2015, 9:52 PM
Hi, i'm getting an error "CAS failed to return any evaluated expressions. Please check your connection with the CAS". Anyone can help with this?


text:
7th Class Mathematics Algebraic Expressions Algebraic identities

Algebraic identities

Category : 7th Class

Algebraic identities

Abatement of equality which holds, for all values of the variable is called algebraic identities.

Now we recall some important algebraic identities:

• ${{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
• ${{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
• ${{a}^{2}}-{{b}^{2}}=(a-b)(a+b)$
• ${{(a+b+c)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca$
• ${{a}^{3}}+{{b}^{3}}={{(a+b)}^{3}}-3ab(a+b)$or ${{(a+b)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab(a+b)$
• ${{(a-b)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab(a-b)$
• ${{a}^{3}}+{{b}^{3}}={{(a+b)}^{3}}-3ab(a+b)$or $(a+b)({{a}^{2}}-ab+{{b}^{2}})$
• ${{a}^{3}}-{{b}^{3}}={{(a-b)}^{3}}+3ab(a-b)$or $(a-b)({{a}^{2}}+ab+{{b}^{2}})$
• ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=(a+b+c)$$({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca)$

If $(a+b+c)=0$${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc$

Simplify:  ${{(2p+3q-+4r)}^{2}}+{{(2p-3q-4r)}^{2}}$

(a) $2(4{{p}^{2}}+9{{q}^{2}}+16{{r}^{2}}-16rp)$

(b) $-2(4{{p}^{2}}+9{{q}^{2}}+16{{r}^{2}}-16rp)$

(c) $2(-4{{p}^{2}}+9{{q}^{2}}-16{{r}^{2}}+16rp)$

(d)$~2(5{{p}^{2}}+9{{q}^{2}}-16{{r}^{2}}-98rp)$

(e) None of these

Explanation

Let us first solve,

${{\left[ 2p+3q+(-4r) \right]}^{2}}={{(2p)}^{2}}+{{(3q)}^{2}}$$+{{(-4r)}^{2}}+2(2p)(3q)+2(3q)(-4r)$ $+2(-4r)(2p)=4{{p}^{2}}+4{{p}^{2}}+9{{q}^{2}}+16{{r}^{2}}$$+12pq-24qr-16rp..........(i)$

Now solve, $~{{(2p-3q-4r)}^{2}}={{\left[ 2p+(-3q)+(-4r) \right]}^{2}}$

$={{(2p)}^{2}}+{{(-3q)}^{2}}+{{(-4r)}^{2}}+2(2p)(-3q)+2(-3q)$$(-4r)+2(-4r)(2p)$

$=4{{p}^{2}}+9{{q}^{2}}+16r2-12pq+24qr-16rp\text{ }.......\left( ii \right)$

Adding, (i) & (ii) we get,

${{(2p+3a-4r)}^{2}}+{{(2p-3q-4r)}^{2}}$

$=4{{p}^{2}}+9{{q}^{2}}+16{{r}^{2}}+12pq-24qr-16rp$+ $(4{{p}^{2}}+9{{q}^{2}}+16{{r}^{2}}-12pq+24qr-16rp)$

$=4{{p}^{2}}+9{{q}^{2}}+16{{r}^{2}}+12pq-24qr-16rp+4{{p}^{2}}+9{{q}^{2}}$    $+16{{r}^{2}}-12pq+24qr-16rp$

$=8{{p}^{2}}+18{{q}^{2}}+32{{r}^{2}}-32rp$

$=2(4{{p}^{2}}+9{{q}^{2+}}16r2-16rp)$

The expanded form of ${{(2x+3y-5z)}^{2}}$ is:

(a) $4{{x}^{2}}+9{{y}^{2}}+25{{z}^{2}}+12xy-30yz-20zx$

(b) $5{{x}^{2}}-6{{y}^{3}}+15{{z}^{3}}+12xy-36{{y}^{6}}+21x{{y}^{2}}$

(c) $8{{a}^{2}}+{{a}^{3}}-{{c}^{2}}+7{{x}^{2}}+2{{c}^{2}}+9{{c}^{2}}y$

(d) $1{{z}^{2}}-{{z}^{4}}-{{8}^{c}}-{{75}^{2}}+2{{c}^{2}}-98{{c}^{2}}$

(e) None of these

Explanation

${{(2x+3y-5z)}^{2}}={{(2x)}^{2}}+{{(3y)}^{2}}+{{(-5z)}^{2}}$$+2(2x)(3y)+2(3y)(-5z)+2(5z)(2x)\text{ }$

Expand: ${{(3a-b+4c)}^{2}}$

(a) $~5{{x}^{2}}-6{{y}^{3}}+15{{z}^{3}}+12xy-36y+21xy$

(b) $9{{a}^{2}}+{{b}^{2}}+16{{c}^{2}}-6ab-8bc+24ac$

(c) $2{{x}^{2}}+7{{y}^{2}}+2{{z}^{2}}+xy+3yz+2zx$

(d) $15a2+22{{b}^{2}}+2{{c}^{4}}+ab+4bc+102ca$

(e) $5{{x}^{2}}-6{{y}^{3}}+15{{z}^{3}}+12xy-36y+21xy$

Explanation

${{(3a-b+4c)}^{2}}={{(3a(-b)+4c)}^{2}}={{(3a)}^{2}}$$+{{(-b)}^{2}}+{{(4c)}^{2}}+2(3a)(-b)+2(-b)$ $(4c)+2(4c)(3a)=9{{a}^{2}}+{{b}^{2}}+16{{c}^{2}}-6ab-8bc+24ac$

Find the value of ${{(3a+5b)}^{3}}.$

(a) $27{{a}^{3}}+125{{b}^{3}}+135{{a}^{2}}b+225a{{b}^{2}}$

(b) $27{{a}^{3}}-125{{b}^{3}}-135{{a}^{2}}b+2225a{{b}^{2}}$

(c) $27{{a}^{2}}+155{{b}^{2}}-135{{a}^{2}}b-225{{a}^{2}}b$

(d) $29{{a}^{2}}-156{{b}^{2}}-156{{a}^{2}}b-225{{a}^{4}}c$

(e) None of these

Find the Cube of  $x-2y.$

(a)$~{{x}^{3}}+8{{y}^{2}}+6{{x}^{2}}y-12x{{y}^{3}}$

(b) ${{x}^{3}}-8{{y}^{3}}-6{{x}^{2}}y+12x{{y}^{2}}$v

(c)${{x}^{2}}+87y+7xy-7xy$

(d) $7x{{y}^{3}}-6x-74{{x}^{2}}y+2xy$

(e) None of these

Evaluate: ${{\left( -6p+\frac{1}{3}q-r \right)}^{2}}$

(a) $36{{p}^{2}}\frac{9}{1}{{p}^{2}}+{{5}^{2}}-4pq-\frac{2}{3}gr+12rp$

(b) $36{{p}^{3}}\frac{9}{1}{{q}^{2}}+{{p}^{2}}+4qp-\frac{3}{2}qr+13rp$

(c)$~36{{p}^{2}}+\frac{1}{9}{{q}^{2}}+{{r}^{2}}-4qp-\frac{2}{3}qr+12rp$

(d) $36{{x}^{2}}-\frac{99}{8}{{s}^{2}}-{{s}^{2}}-4pq-\frac{2}{3}qr+14rp$

(e) None of these

format: plain


text:
27
Dec 22

## Final exam in Advanced Statistics ST2133, 2022

Unlike most UoL exams, here I tried to relate the theory to practical issues.

KBTU International School of Economics

Compiled by Kairat Mynbaev

The total for this exam is 41 points. You have two hours.

Everywhere provide detailed explanations. When answering please clearly indicate question numbers. You don’t need a calculator. As long as the formula you provide is correct, the numerical value does not matter.

Question 1. (12 points)

a) (2 points) At a casino, two players are playing on slot machines. Their payoffs $X,Y$ are standard normal and independent. Find the joint density of the payoffs.

b) (4 points) Two other players watch the first two players and start to argue what will be larger: the sum $U = X + Y$ or the difference $V = X - Y$. Find the joint density. Are variables $U,V$ independent? Find their marginal densities.

c) (2 points) Are $U,V$ normal? Why? What are their means and variances?

d) (2 points) Which probability is larger: $P(U > V)$ or $P\left( {U < V} \right)$?

e) (2 points) In this context interpret the conditional expectation $E\left( {U|V = v} \right)$. How much is it?

Reminder. The density of a normal variable $X \sim N\left( {\mu ,{\sigma ^2}} \right)$ is ${f_X}\left( x \right) = \frac{1}{{\sqrt {2\pi {\sigma ^2}} }}{e^{ - \frac{{{{\left( {x - \mu } \right)}^2}}}{{2{\sigma ^2}}}}}$.

Question 2. (9 points) The distribution of a call duration $X$ of one Kcell [largest mobile operator in KZ] customer is exponential: ${f_X}\left( x \right) = \lambda {e^{ - \lambda x}},\,\,x \ge 0,\,\,{f_X}\left( x \right) = 0,\,\,x < 0.$ The number $N$ of customers making calls simultaneously is distributed as Poisson: $P\left( {N = n} \right) = {e^{ - \mu }}\frac{{{\mu ^n}}}{{n!}},\,\,n = 0,1,2,...$ Thus the total call duration for all customers is ${S_N} = {X_1} + ... + {X_N}$ for $N \ge 1$. We put ${S_0} = 0$. Assume that customers make their decisions about calling independently.

a) (3 points) Find the general formula (when ${X_1},...,{X_n}$ are identically distributed and $X,N$ are independent but not necessarily exponential and Poisson, as above) for the moment generating function of $S_N$ explaining all steps.

b) (3 points) Find the moment generating functions of $X$, $N$ and ${S_N}$ for your particular distributions.

c) (3 points) Find the mean and variance of ${S_N}$. Based on the equations you obtained, can you suggest estimators of parameters $\lambda ,\mu$?

Remark. Direct observations on the exponential and Poisson distributions are not available. We have to infer their parameters by observing ${S_N}$. This explains the importance of the technique used in Question 2.

Question 3. (8 points)

a) (2 points) For a non-negative random variable $X$ prove the Markov inequality $P\left( {X > c} \right) \le \frac{1}{c}EX,\,\,\,c > 0.$

b) (2 points) Prove the Chebyshev inequality $P\left( {|X - EX| > c} \right) \le \frac{1}{c^2}Var\left( X \right)$ for an arbitrary random variable $X$.

c) (4 points) We say that the sequence of random variables $\left\{ X_n \right\}$ converges in probability to a random variable $X$ if $P\left( {|{X_n} - X| > \varepsilon } \right) \to 0$ as $n \to \infty$ for any $\varepsilon > 0$.  Suppose that $E{X_n} = \mu$ for all $n$ and that $Var\left(X_n \right) \to 0$ as $n \to \infty$. Prove that then $\left\{X_n\right\}$ converges in probability to $\mu$.

Remark. Question 3 leads to the simplest example of a law of large numbers: if $\left\{ X_n \right\}$ are i.i.d. with finite variance, then their sample mean converges to their population mean in probability.

Question 4. (8 points)

a) (4 points) Define a distribution function. Give its properties, with intuitive explanations.

b) (4 points) Is a sum of two distribution functions a distribution function? Is a product of two distribution functions a distribution function?

Remark. The answer for part a) is here and the one for part b) is based on it.

Question 5. (4 points) The Rakhat factory prepares prizes for kids for the upcoming New Year event. Each prize contains one type of chocolates and one type of candies. The chocolates and candies are chosen randomly from two production lines, the total number of items is always 10 and all selections are equally likely.

a) (2 points) What proportion of prepared prizes contains three or more chocolates?

b) (2 points) 100 prizes have been sent to an orphanage. What is the probability that 50 of those prizes contain no more than two chocolates?

24
Oct 22

## A problem to do once and never come back

There is a problem I gave on the midterm that does not require much imagination. Just know the definitions and do the technical work, so I was hoping we could put this behind us. Turned out we could not and thus you see this post.

Problem. Suppose the joint density of variables $X,Y$ is given by

$f_{X,Y}(x,y)=\left\{ \begin{array}{c}k\left( e^{x}+e^{y}\right) \text{ for }0

I. Find $k$.

II. Find marginal densities of $X,Y$. Are $X,Y$ independent?

III. Find conditional densities $f_{X|Y},\ f_{Y|X}$.

IV. Find $EX,\ EY$.

When solving a problem like this, the first thing to do is to give the theory. You may not be able to finish without errors the long calculations but your grade will be determined by the beginning theoretical remarks.

### I. Finding the normalizing constant

Any density should satisfy the completeness axiom: the area under the density curve (or in this case the volume under the density surface) must be equal to one: $\int \int f_{X,Y}(x,y)dxdy=1.$ The constant $k$ chosen to satisfy this condition is called a normalizing constant. The integration in general is over the whole plain $R^{2}$ and the first task is to express the above integral as an iterated integral. This is where the domain where the density is not zero should be taken into account. There is little you can do without geometry. One example of how to do this is here.

The shape of the area $A=\left\{ (x,y):0 is determined by a) the extreme values of $x,y$ and b) the relationship between them. The extreme values are 0 and 1 for both $x$ and $y$, meaning that $A$ is contained in the square $\left\{ (x,y):0 The inequality $y means that we cut out of this square the triangle below the line $y=x$ (it is really the lower triangle because if from a point on the line $y=x$ we move down vertically, $x$ will stay the same and $y$ will become smaller than $x$).

In the iterated integral:

a) the lower and upper limits of integration for the inner integral are the boundaries for the inner variable; they may depend on the outer variable but not on the inner variable.

b) the lower and upper limits of integration for the outer integral are the extreme values for the outer variable; they must be constant.

This is illustrated in Pane A of Figure 1.

Figure 1. Integration order

Always take the inner integral in parentheses to show that you are dealing with an iterated integral.

a) In the inner integral integrating over $x$ means moving along blue arrows from the boundary $x=y$ to the boundary $x=1.$ The boundaries may depend on $y$ but not on $x$ because the outer integral is over $y.$

b) In the outer integral put the extreme values for the outer variable. Thus,

$\underset{A}{\int \int }f_{X,Y}(x,y)dxdy=\int_{0}^{1}\left(\int_{y}^{1}f_{X,Y}(x,y)dx\right) dy.$

Check that if we first integrate over $y$ (vertically along red arrows, see Pane B in Figure 1) then the equation

$\underset{A}{\int \int }f_{X,Y}(x,y)dxdy=\int_{0}^{1}\left(\int_{0}^{x}f_{X,Y}(x,y)dy\right) dx$

results.

In fact, from the definition $A=\left\{ (x,y):0 one can see that the inner interval for $x$ is $\left[ y,1\right]$ and for $y$ it is $\left[ 0,x\right] .$

### II. Marginal densities

The condition for independence of $X,Y$ is $f_{X,Y}\left( x,y\right) =f_{X}\left( x\right) f_{Y}\left( y\right)$ (this is a direct analog of the independence condition for events $P\left( A\cap B\right) =P\left( A\right) P\left( B\right)$). In words: the joint density decomposes into a product of individual densities.

### III. Conditional densities

In this case the easiest is to recall the definition of conditional probability $P\left( A|B\right) =\frac{P\left( A\cap B\right) }{P\left(B\right) }.$ The definition of conditional densities $f_{X|Y},\ f_{Y|X}$ is quite similar:

(2) $f_{X|Y}\left( x|y\right) =\frac{f_{X,Y}\left( x,y\right) }{f_{Y}\left( y\right) },\ f_{Y|X}\left( y|x\right) =\frac{f_{X,Y}\left( x,y\right) }{f_{X}\left( x\right) }$.

Of course, $f_{Y}\left( y\right) ,f_{X}\left( x\right)$ here can be replaced by their marginal equivalents.

### IV. Finding expected values of $X,Y$$X,Y$

The usual definition $EX=\int xf_{X}\left( x\right) dx$ takes an equivalent form using the marginal density:

$EX=\int x\left( \int f_{X,Y}\left( x,y\right) dy\right) dx=\int \int xf_{X,Y}\left( x,y\right) dydx.$

Which equation to use is a matter of convenience.

Another replacement in the usual definition gives the definition of conditional expectations:

$E\left( X|Y\right) =\int xf_{X|Y}\left( x|y\right) dx,$ $E\left( Y|X\right) =\int yf_{Y|X}\left( y|x\right) dx.$

Note that these are random variables: $E\left( X|Y=y\right)$ depends in $y$ and $E\left( Y|X=x\right)$ depends on $x.$

### Solution to the problem

Being a lazy guy, for the problem this post is about I provide answers found in Mathematica:

I. $k=0.581977$

II. $f_{X}\left( x\right) =-1+e^{x}\left( 1+x\right) ,$ for $x\in[ 0,1],$ $f_{Y}\left( y\right) =e-e^{y}y,$ for $y\in \left[ 0,1\right] .$

It is readily seen that the independence condition is not satisfied.

III. $f_{X|Y}\left( x|y\right) =\frac{k\left( e^{x}+e^{y}\right) }{e-e^{y}y}$ for $0

$f_{Y|X}\left(y|x\right) =\frac{k\left(e^x+e^y\right) }{-1+e^x\left( 1+x\right) }$ for $0

IV. $EX=0.709012,$ $EY=0.372965.$

24
Oct 22

## Marginal probabilities and densities

This is to help everybody, from those who study Basic Statistics up to Advanced Statistics ST2133.

### Discrete case

Suppose in a box we have coins and banknotes of only two denominations: $1 and$5 (see Figure 1).

Figure 1. Illustration of two variables

We pull one out randomly. The division of cash by type (coin or banknote) divides the sample space (shown as a square, lower left picture) with probabilities $p_{c}$ and $p_{b}$ (they sum to one). The division by denomination ($1 or$5) divides the same sample space differently, see the lower right picture, with the probabilities to pull out $1 and$5 equal to $p_{1}$ and $p_{5}$, resp. (they also sum to one). This is summarized in the tables

 Variable 1: Cash type Prob coin $p_{c}$$p_{c}$ banknote $p_{b}$$p_{b}$
 Variable 2: Denomination Prob $1 $p_{1}$$p_{1}$$5 $p_{5}$$p_{5}$

Now we can consider joint events and probabilities (see Figure 2, where the two divisions are combined).

Figure 2. Joint probabilities

For example, if we pull out a random $item$ it can be a $coin$ and \$1 and the corresponding probability is $P\left(item=coin,\ item\ value=\1\right) =p_{c1}.$ The two divisions of the sample space generate a new division into four parts. Then geometrically it is obvious that we have four identities:

Adding over denominations: $p_{c1}+p_{c5}=p_{c},$ $p_{b1}+p_{b5}=p_{b},$

Adding over cash types: $p_{c1}+p_{b1}=p_{1},$ $p_{c5}+p_{b5}=p_{5}.$

Formally, here we use additivity of probability for disjoint events

$P\left( A\cup B\right) =P\left( A\right) +P\left( B\right) .$

In words: we can recover own probabilities of variables 1,2 from joint probabilities.

### Generalization

Suppose we have two discrete random variables $X,Y$ taking values $x_{1},...,x_{n}$ and $y_{1},...,y_{m},$ resp., and their own probabilities are $P\left( X=x_{i}\right) =p_{i}^{X},$ $P\left(Y=y_{j}\right) =p_{j}^{Y}.$ Denote the joint probabilities $P\left(X=x_{i},Y=y_{j}\right) =p_{ij}.$ Then we have the identities

(1) $\sum_{j=1}^mp_{ij}=p_{i}^{X},$ $\sum_{i=1}^np_{ij}=p_{j}^{Y}$ ($n+m$ equations).

In words: to obtain the marginal probability of one variable (say, $Y$) sum over the values of the other variable (in this case, $X$).

The name marginal probabilities is used for $p_{i}^{X},p_{j}^{Y}$ because in the two-dimensional table they arise as a result of summing table entries along columns or rows and are displayed in the margins.

### Analogs for continuous variables with densities

Suppose we have two continuous random variables $X,Y$ and their own densities are $f_{X}$ and $f_{Y}.$ Denote the joint density $f_{X,Y}$. Then replacing in (1) sums by integrals and probabilities by densities we get

(2) $\int_R f_{X,Y}\left( x,y\right) dy=f_{X}\left( x\right) ,\ \int_R f_{X,Y}\left( x,y\right) dx=f_{Y}\left( y\right) .$

In words: to obtain one marginal density (say, $f_{Y}$) integrate out the other variable (in this case, $x$).

27
Jan 21

## AP Stats and Business Stats

Its content, organization and level justify its adoption as a textbook for introductory statistics for Econometrics in most American or European universities. The book's table of contents is somewhat standard, the innovation comes in a presentation that is crisp, concise, precise and directly relevant to the Econometrics course that will follow. I think instructors and students will appreciate the absence of unnecessary verbiage that permeates many existing textbooks.

Having read Professor Mynbaev's previous books and research articles I was not surprised with his clear writing and precision. However, I was surprised with an informal and almost conversational one-on-one style of writing which should please most students. The informality belies a careful presentation where great care has been taken to present the material in a pedagogical manner.

Carlos Martins-Filho
Professor of Economics
Boulder, USA

26
May 20

2
Mar 20

## Statistical calculator

In my book I explained how one can use Excel to do statistical simulations and replace statistical tables commonly used in statistics courses. Here I go one step further by providing a free statistical calculator that replaces the following tables from the book by Newbold et al.:

Table 1 Cumulative Distribution Function, F(z), of the Standard Normal Distribution Table

Table 2 Probability Function of the Binomial Distribution

Table 5 Individual Poisson Probabilities

Table 7a Upper Critical Values of Chi-Square Distribution with $\nu$ Degrees of Freedom

Table 8 Upper Critical Values of Student’s t Distribution with $\nu$ Degrees of Freedom

Tables 9a, 9b Upper Critical Values of the F Distribution

The calculator is just a Google sheet with statistical functions, see Picture 1:

Picture 1. Calculator using Google sheet

## How to use Calculator

1. Open an account at gmail.com, if you haven't already. Open Google Drive.
3. Find the sheet on my Google drive and copy it to your Google drive (File/Make a copy). An icon of my calculator will appear in your drive. That's not the file, it's just a link to my file. To the right of it there are three dots indicating options. One of them is "Make a copy", so use that one. The copy will be in your drive. After that you can delete the link to my file. You might want to rename "Copy of Calculator" as "Calculator".
5. When you click a cell, you can enter what you need either in the formula bar at the bottom or directly in the cell. You can also see the functions I embedded in the sheet.
6. In cell A1, for example, you can enter any legitimate formula with numbers, arithmetic signs, and Google sheet functions. Be sure to start it with =,+ or - and to press the checkmark on the right of the formula bar after you finish.
7. The cells below A1 replace the tables listed above. Beside each function there is a verbal description and further to the right - a graphical illustration (which is not in Picture 1).
8. On the tab named Regression you can calculate the slope and intercept. The sample size must be 10.
9. Keep in mind that tables for continuous distributions need two functions. For example, in case of the standard normal distribution one function allows you to go from probability (area of the left tail) to the cutting value on the horizontal axis. The other function goes from the cutting value on the horizontal axis to probability.
10. Feel free to add new sheets or functions as you may need. You will have to do this on a tablet or computer.
17
Mar 19

## AP Statistics the Genghis Khan way 2

Last semester I tried to explain theory through numerical examples. The results were terrible. Even the best students didn't stand up to my expectations. The midterm grades were so low that I did something I had never done before: I allowed my students to write an analysis of the midterm at home. Those who were able to verbally articulate the answers to me received a bonus that allowed them to pass the semester.

This semester I made a U-turn. I announced that in the first half of the semester we will concentrate on theory and we followed this methodology. Out of 35 students, 20 significantly improved their performance and 15 remained where they were.

### Midterm exam, version 1

#### 1. General density definition (6 points)

a. Define the density $p_X$ of a random variable $X.$ Draw the density of heights of adults, making simplifying assumptions if necessary. Don't forget to label the axes.

b. According to your plot, how much is the integral $\int_{-\infty}^0p_X(t)dt?$ Explain.

c. Why the density cannot be negative?

d. Why the total area under the density curve should be 1?

e. Where are basketball players on your graph? Write down the corresponding expression for probability.

f. Where are dwarfs on your graph? Write down the corresponding expression for probability.

This question is about the interval formula. In each case students have to write the equation for the probability and the corresponding integral of the density. At this level, I don't talk about the distribution function and introduce the density by the interval formula.

#### 2. Properties of means (8 points)

a. Define a discrete random variable and its mean.

b. Define linear operations with random variables.

c. Prove linearity of means.

d. Prove additivity and homogeneity of means.

e. How much is the mean of a constant?

f. Using induction, derive the linearity of means for the case of $n$ variables from the case of two variables (3 points).

#### 3. Covariance properties (6 points)

a. Derive linearity of covariance in the first argument when the second is fixed.

b. How much is covariance if one of its arguments is a constant?

c. What is the link between variance and covariance? If you know one of these functions, can you find the other (there should be two answers)? (4 points)

#### 4. Standard normal variable (6 points)

a. Define the density $p_z(t)$ of a standard normal.

b. Why is the function $p_z(t)$ even? Illustrate this fact on the plot.

c. Why is the function $f(t)=tp_z(t)$ odd? Illustrate this fact on the plot.

d. Justify the equation $Ez=0.$

e. Why is $V(z)=1?$

f. Let $t>0.$ Show on the same plot areas corresponding to the probabilities $A_1=P(0 $A_2=P(z>t),$ $A_3=P(z<-t),$ $A_4=P(-t Write down the relationships between $A_1,...,A_4.$

#### 5. General normal variable (3 points)

a. Define a general normal variable $X.$

b. Use this definition to find the mean and variance of $X.$

c. Using part b, on the same plot graph the density of the standard normal and of a general normal with parameters $\sigma =2,$ $\mu =3.$

### Midterm exam, version 2

#### 1. General density definition (6 points)

a. Define the density $p_X$ of a random variable $X.$ Draw the density of work experience of adults, making simplifying assumptions if necessary. Don't forget to label the axes.

b. According to your plot, how much is the integral $\int_{-\infty}^0p_X(t)dt?$ Explain.

c. Why the density cannot be negative?

d. Why the total area under the density curve should be 1?

e. Where are retired people on your graph? Write down the corresponding expression for probability.

f. Where are young people (up to 25 years old) on your graph? Write down the corresponding expression for probability.

#### 2. Variance properties (8 points)

a. Define variance of a random variable. Why is it non-negative?

b. Define the formula for variance of a linear combination of two variables.

c. How much is variance of a constant?

d. What is the formula for variance of a sum? What do we call homogeneity of variance?

e. What is larger: $V(X+Y)$ or $V(X-Y)$? (2 points)

f. One investor has 100 shares of Apple, another - 200 shares. Which investor's portfolio has larger variability? (2 points)

#### 3. Poisson distribution (6 points)

a. Write down the Taylor expansion and explain the idea. How are the Taylor coefficients found?

b. Use the Taylor series for the exponential function to define the Poisson distribution.

c. Find the mean of the Poisson distribution. What is the interpretation of the parameter $\lambda$ in practice?

#### 4. Standard normal variable (6 points)

a. Define the density $p_z(t)$ of a standard normal.

b. Why is the function $p_z(t)$ even? Illustrate this fact on the plot.

c. Why is the function $f(t)=tp_z(t)$ odd? Illustrate this fact on the plot.

d. Justify the equation $Ez=0.$

e. Why is $V(z)=1?$

f. Let $t>0.$ Show on the same plot areas corresponding to the probabilities $A_1=P(0 $A_2=P(z>t),$ $A_{3}=P(z<-t),$ $A_4=P(-t Write down the relationships between $A_{1},...,A_{4}.$

#### 5. General normal variable (3 points)

a. Define a general normal variable $X.$

b. Use this definition to find the mean and variance of $X.$

c. Using part b, on the same plot graph the density of the standard normal and of a general normal with parameters $\sigma =2,$ $\mu =3.$

4
Nov 18

## Little tricks for AP Statistics

This year I am teaching AP Statistics. If the things continue the way they are, about half of the class will fail. Here is my diagnosis and how I am handling the problem.

On the surface, the students lack algebra training but I think the problem is deeper: many of them have underdeveloped cognitive abilities. Their perception is slow, memory is limited, analytical abilities are rudimentary and they are not used to work at home. Limited resources require  careful allocation.

### Terminology

Short and intuitive names are better than two-word professional names.

Instead of "sample space" or "probability space" say "universe". The universe is the widest possible event, and nothing exists outside it.

Instead of "elementary event" say "atom". Simplest possible events are called atoms. This corresponds to the theoretical notion of an atom in measure theory (an atom is a measurable set which has positive measure and contains no set of smaller positive measure).

Then the formulation of classical probability becomes short. Let $n$ denote the number of atoms in the universe and let $n_A$ be the number of atoms in event $A.$ If all atoms are equally likely (have equal probabilities), then $P(A)=n_A/n.$

The clumsy "mutually exclusive events" are better replaced by more visual "disjoint sets". Likewise, instead of "collectively exhaustive events" say "events that cover the universe".

The combination "mutually exclusive" and "collectively exhaustive" events is beyond comprehension for many. I say: if events are disjoint and cover the universe, we call them tiles. To support this definition, play onscreen one of jigsaw puzzles (Video 1) and produce the picture from Figure 1.

Video 1. Tiles (disjoint events that cover the universe)

Figure 1. Tiles (disjoint events that cover the universe)

### The philosophy of team work

We are in the same boat. I mean the big boat. Not the class. Not the university. It's the whole country. We depend on each other. Failure of one may jeopardize the well-being of everybody else.

You work in teams. You help each other to learn. My lectures and your presentations are just the beginning of the journey of knowledge into your heads. I cannot control how it settles there. Be my teaching assistants, share your big and little discoveries with your classmates.

I don't just preach about you helping each other. I force you to work in teams. 30% of the final grade is allocated to team work. Team work means joint responsibility. You work on assignments together. I randomly select a team member for reporting. His or her grade is what each team member gets.

This kind of team work is incompatible with the Western obsession with grades privacy. If I say my grade is nobody's business, by extension I consider the level of my knowledge a private issue. This will prevent me from asking for help and admitting my errors. The situation when students hide their errors and weaknesses from others also goes against the ethics of many workplaces. In my class all grades are public knowledge.

In some situations, keeping the grade private is technically impossible. Conducting a competition without announcing the points won is impossible. If I catch a student cheating, I announce the failing grade immediately, as a warning to others.

To those of you who think team-based learning is unfair to better students I repeat: 30% of the final grade is given for team work, not for personal achievements. The other 70% is where you can shine personally.

### Breaking the wall of silence

Team work serves several purposes.

Firstly, joint responsibility helps breaking communication barriers. See in Video 2 my students working in teams on classroom assignments. The situation when a weaker student is too proud to ask for help and a stronger student doesn't want to offend by offering help is not acceptable. One can ask for help or offer help without losing respect for each other.

Video 2. Teams working on assignments

Secondly, it turns on resources that are otherwise idle. Explaining something to somebody is the best way to improve your own understanding. The better students master a kind of leadership that is especially valuable in a modern society. For the weaker students, feeling responsible for a team improves motivation.

Thirdly, I save time by having to grade less student papers.

On exams and quizzes I mercilessly punish the students for Yes/No answers without explanations. There are no half-points for half-understanding. This, in combination with the team work and open grades policy allows me to achieve my main objective: students are eager to talk to me about their problems.

### Set operations and probability

After studying the basics of set operations and probabilities we had a midterm exam. It revealed that about one-third of students didn't understand this material and some of that misunderstanding came from high school. During the review session I wanted to see if they were ready for a frank discussion and told them: "Those who don't understand probabilities, please raise your hands", and about one-third raised their hands. I invited two of them to work at the board.

Video 3. Translating verbal statements to sets, with accompanying probabilities

Many teachers think that the Venn diagrams explain everything about sets because they are visual. No, for some students they are not visual enough. That's why I prepared a simple teaching aid (see Video 3) and explained the task to the two students as follows:

I am shooting at the target. The target is a square with two circles on it, one red and the other blue. The target is the universe (the bullet cannot hit points outside it). The probability of a set is its area. I am going to tell you one statement after another. You write that statement in the first column of the table. In the second column write the mathematical expression for the set. In the third column write the probability of that set, together with any accompanying formulas that you can come up with. The formulas should reflect the relationships between relevant areas.

Table 1. Set operations and probabilities

 Statement Set Probability 1. The bullet hit the universe $S$$S$ $P(S)=1$$P(S)=1$ 2. The bullet didn't hit the universe $\emptyset$$\emptyset$ $P(\emptyset )=0$$P(\emptyset )=0$ 3. The bullet hit the red circle $A$$A$ $P(A)$$P(A)$ 4. The bullet didn't hit the red circle $\bar{A}=S\backslash A$$\bar{A}=S\backslash A$ $P(\bar{A})=P(S)-P(A)=1-P(A)$$P(\bar{A})=P(S)-P(A)=1-P(A)$ 5. The bullet hit both the red and blue circles $A\cap B$$A\cap B$ $P(A\cap B)$$P(A\cap B)$ (in general, this is not equal to $P(A)P(B)$$P(A)P(B)$) 6. The bullet hit $A$$A$ or $B$$B$ (or both) $A\cup B$$A\cup B$ $P(A\cup B)=P(A)+P(B)-P(A\cap B)$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$ (additivity rule) 7. The bullet hit $A$$A$ but not $B$$B$ $A\backslash B$$A\backslash B$ $P(A\backslash B)=P(A)-P(A\cap B)$$P(A\backslash B)=P(A)-P(A\cap B)$ 8. The bullet hit $B$$B$ but not $A$$A$ $B\backslash A$$B\backslash A$ $P(B\backslash A)=P(B)-P(A\cap B)$$P(B\backslash A)=P(B)-P(A\cap B)$ 9. The bullet hit either $A$$A$ or $B$$B$ (but not both) $(A\backslash B)\cup(B\backslash A)$$(A\backslash B)\cup(B\backslash A)$ $P\left( (A\backslash B)\cup (B\backslash A)\right)$$P\left( (A\backslash B)\cup (B\backslash A)\right)$ $=P(A)+P(B)-2P(A\cap B)$$=P(A)+P(B)-2P(A\cap B)$

During the process, I was illustrating everything on my teaching aid. This exercise allows the students to relate verbal statements to sets and further to their areas. The main point is that people need to see the logic, and that logic should be repeated several times through similar exercises.

8
Oct 17

## Reevaluating probabilities based on piece of evidence

This actually has to do with the Bayes' theorem. However, in simple problems one can use a dead simple approach: just find probabilities of all elementary events. This post builds upon the post on Significance level and power of test, including the notation. Be sure to review that post.

Here is an example from the guide for Quantitative Finance by A. Patton (University of London course code FN3142).

Activity 7.2 Consider a test that has a Type I error rate of 5%, and power of 50%.

Suppose that, before running the test, the researcher thinks that both the null and the alternative are equally likely.

1. If the test indicates a rejection of the null hypothesis, what is the probability that the null is false?

2. If the test indicates a failure to reject the null hypothesis, what is the probability that the null is true?

Denote events R = {Reject null}, A = {fAil to reject null}; T = {null is True}; F = {null is False}. Then we are given:

(1) $P(F)=0.5;\ P(T)=0.5;$

(2) $P(R|T)=\frac{P(R\cap T)}{P(T)}=0.05;\ P(R|F)=\frac{P(R\cap F)}{P(F)}=0.5;$

(1) and (2) show that we can find $P(R\cap T)$ and $P(R\cap F)$ and therefore also $P(A\cap T)$ and $P(A\cap F).$ Once we know probabilities of elementary events, we can find everything about everything.

Figure 1. Elementary events

Answering the first question: just plug probabilities in $P(F|R)=\frac{P(R\cap F)}{P(R)}=\frac{P(R\cap F)}{P(R\cap T)+P(A\cap T)}.$

Answering the second question: just plug probabilities in $P(T|A)=\frac{P(A\cap T)}{P(A)}=\frac{P(A\cap T)}{P(A\cap T)+P(A\cap F)}.$

Patton uses the Bayes' theorem and the law of total probability. The solution suggested above uses only additivity of probability.

6
Oct 17

## Significance level and power of test

In this post we discuss several interrelated concepts: null and alternative hypotheses, type I and type II errors and their probabilities. Review the definitions of a sample space and elementary events and that of a conditional probability.

## Type I and Type II errors

Regarding the true state of nature we assume two mutually exclusive possibilities: the null hypothesis (like the suspect is guilty) and alternative hypothesis (the suspect is innocent). It's up to us what to call the null and what to call the alternative. However, the statistical procedures are not symmetric: it's easier to measure the probability of rejecting the null when it is true than other involved probabilities. This is why what is desirable to prove is usually designated as the alternative.

Usually in books you can see the following table.

 Decision taken Fail to reject null Reject null State of nature Null is true Correct decision Type I error Null is false Type II error Correct decision

This table is not good enough because there is no link to probabilities. The next video does fill in the blanks.

Video. Significance level and power of test

## Significance level and power of test

The conclusion from the video is that

$\frac{P(T\bigcap R)}{P(T)}=P(R|T)=P\text{(Type I error)=significance level}$ $\frac{P(F\bigcap R)}{P(F)}=P(R|F)=P\text{(Correctly rejecting false null)=Power}$


text:
Power supply rejection ratio

In electronic systems, power supply rejection ratio (PSRR), also supply-voltage rejection ratio[1] (kSVR; SVR), is a term widely used to describe the capability of an electronic circuit to suppress any power supply variations to its output signal.

In the specifications of operational amplifiers, the PSRR is defined as the ratio of the change in supply voltage to the equivalent (differential) output voltage it produces,[2] often expressed in decibels.[3][4][5] An ideal op-amp would have infinite PSRR. The output voltage will depend on the feedback circuit, as is the case of regular input offset voltages. But testing is not confined to DC (zero frequency); often an operational amplifier will also have its PSRR given at various frequencies (in which case the ratio is one of RMS amplitudes of sinewaves present at a power supply compared with the output, with gain taken into account). Unwanted oscillation, including motorboating, can occur when an amplifying stage is too sensitive to signals fed via the power supply from a later power amplifier stage.

Some manufacturers specify PSRR in terms of the offset voltage it causes at the amplifiers inputs; others specify it in terms of the output; there is no industry standard for this issue.[6] The following formula assumes it is specified in terms of output:

${\displaystyle {\text{PSRR}}[{\text{dB}}]=10\log _{10}\left({\frac {\Delta {V_{\text{supply}}}^{2}{A_{v}}^{2}}{\Delta {V_{\text{out}}}^{2}}}\right){\text{dB}}}$

where ${\textstyle A_{v}}$ is the voltage gain.

For example: an amplifier with a PSRR of 100 dB in a circuit to give 40 dB closed-loop gain would allow about 1 millivolt of power supply ripple to be superimposed on the output for every 1 volt of ripple in the supply. This is because

${\displaystyle 100\ {\text{dB}}-40\ {\text{dB}}=60\ {\text{dB}}}$.

And since that's 60 dB of rejection, the sign is negative so:

${\displaystyle 1\ {\text{V}}\cdot 10^{\frac {-60}{20}}=0.001\ {\text{V}}=1\ {\text{mV}}}$

Note:

• The PSRR doesn't necessarily have the same poles as A(s), the open-loop gain of the op-amp, but generally tends to also worsen with increasing frequency (e.g. http://focus.ti.com/lit/ds/symlink/opa2277.pdf).[7]
• For amplifiers with both positive and negative power supplies (with respect to earth, as op-amps often have), the PSRR for each supply voltage may be separately specified (sometimes written: PSRR+ and PSRR−), but normally the PSRR is tested with opposite polarity signals applied to both supply rails at the same time (otherwise the common-mode rejection ratio (CMRR) will affect the measurement of the PSRR).
• For voltage regulators the PSRR is occasionally quoted (confusingly; to refer to output voltage change ratios), but often the concept is transferred to other terms relating changes in output voltage to input: Ripple rejection (RR) for low frequencies, line transient response for high frequencies, and line regulation for DC.


text:
# Into Math Grade 4 Module 12 Lesson 3 Answer Key Identify Equivalent Fractions and Decimals

We included HMH Into Math Grade 4 Answer Key PDF Module 12 Lesson 3 Identify Equivalent Fractions and Decimals to make students experts in learning maths.

## HMH Into Math Grade 4 Module 12 Lesson 3 Answer Key Identify Equivalent Fractions and Decimals

I Can write a fraction in tenths as a fraction in hundredths and as a decimal in tenths and hundredths.

Kelly buys $$\frac{4}{10}$$ kilogram of American cheese at the deli counter.

Write the amount of cheese that Kelly buys as a fraction and as a decimal in hundredths. Justify your answer with a visual model.
Explanation:
Kelly buys $$\frac{4}{10}$$ kilogram
$$\frac{4}{10}$$ = 0.4

Turn and Talk How does a numerator of a fraction in tenths relate to the numerator of an equivalent fraction in hundredths? Describe a visual model you could use to support your answer.
Both are equivalent to each other.
$$\frac{4}{10}$$ = $$\frac{40}{100}$$

Explanation:
If we observe the tenths and hundredths model, each are divided into 10 equal sections.
$$\frac{1}{10}$$ = $$\frac{10}{100}$$
$$\frac{4}{10}$$ = $$\frac{4×10}{10×10}$$
$$\frac{4}{10}$$ = $$\frac{40}{100}$$

Build Understanding

Question 1.
Joel spends 1$$\frac{3}{10}$$ hours shopping for groceries at the supermarket.

How can you represent the time spent as a mixed number in hundredths and as a decimal in hundredths?
Show your thinking and then write the decimal in the place-value chart.

Explanation:
Joel spends 1$$\frac{3}{10}$$ hours shopping
= $$\frac{13}{10}$$
decimal in hundredths = $$\frac{13×10}{10×10}$$ = 1.30

A. What is the relationship between tenths and hundredths?
Tenth is the place value that represents the position of digit after decimal place equal to 1/10th
and hundredth represents the position of digit in a number after decimal place equal to 1/100th.
Explanation:
For example, nine tenths is equal to 4/10 = 0.4
and 5 hundredth is equal to $$\frac{5}{100}$$ = 0.05
decimal pace changes between tenths and hundredths

Connect to Vocabulary
Equivalent decimals are two or more decimals that name the same number.

B. How can you use that relationship to rename tenths as an equivalent fraction in hundredths?
Decimal in tenths and decimal in hundredths also divided evenly and both are equivalent.
$$\frac{1}{10}$$ ÷ $$\frac{1}{100}$$= 10

C. How can you write the fraction in hundredths as a decimal in hundredths?
hundredth is equal to $$\frac{5}{100}$$
= 0.05
Explanation:
To write fraction in hundredths as a decimal in hundredths,
divide the numerator with denominator.

Step It Out

Question 2.
Pam walks with her mother from the market to her home. Write the distance in fraction form and decimal form in tenths.

A. Write the distance as a fraction in hundredths.
$$\frac{60}{100}$$
Explanation:
Pam walks with her mother from the market to her home = 0.6 miles
Distance fraction form in hundredths = $$\frac{60}{100}$$

B. Use division to rename the fraction in hundredths as a fraction in tenths. Show your work.
By dividing both the denominator by 10.
and when we go from hundredths to tenths and you would also divide the numerator by.
10 going from hundredths to tenths

C. Write the fraction in tenths as a decimal in tenths.
$$\frac{6}{10}$$ = 0.6
Explanation:
Pam walks with her mother from the market to her home = 0.6 miles
Distance fraction form = $$\frac{6}{10}$$
Distance in decimal tenths = 0.6

Turn and Talk How can you rename a decimal in tenths as a decimal in hundredths without multiplying?
Decimal in tenths and decimal in hundredths also divided evenly
$$\frac{1}{10}$$ ÷ $$\frac{1}{100}$$= 10
Explanation:
one tenth = one hundredth
0.1 = 0.01
100 ÷ 10 = 10

Check Understanding

Question 1.
Cara drinks $$\frac{2}{10}$$ liter of apple juice. Write the amount of juice as a fraction and a decimal in hundredths.
fraction: ________
decimal: _________
fraction : $$\frac{1}{5}$$
decimal : 0.2
Explanation:
Cara drinks $$\frac{2}{10}$$ liter of apple juice.
Amount of juice as a fraction and a decimal in hundredths = $$\frac{1}{5}$$ = 0.2

Question 2.
A store clerk unpacks 2$$\frac{4}{10}$$ cases of milk cartons. Rename the number of cases as a mixed number and a decimal in hundredths.
2.4 cartons
Explanation:
2$$\frac{4}{10}$$ cases of milk cartons
Rename mixed number into fraction = $$\frac{20 + 4}{10}$$
decimal in hundredths = $$\frac{24}{10}$$ = 2.4

Question 3.
Jennifer buys $$\frac{70}{100}$$ kilogram of mixed nuts. Rename the number of kilograms as a fraction and a decimal in tenths.
0.7 kilograms
Explanation:
Jennifer buys $$\frac{70}{100}$$ kilogram of mixed nuts
To convert decimal in tenths divided both numerator and denominator with 10.
$$\frac{70÷10 }{100÷10}$$ = $$\frac{7}{10}$$ = 0.7

Question 4.
Janelle saves $$\frac{1}{10}$$ of her paycheck each week. Shade the hundredths model to show that amount.
What is $$\frac{1}{10}$$ written as a decimal in hundredths?

$$\frac{10}{100}$$ = 0.1
Explanation:
Janelle saves $$\frac{1}{10}$$ of her pay check each week.
To write decimal in hundredths multiply both numerator and denominator with 10.
$$\frac{1×10}{10×10}$$ = $$\frac{10}{100}$$= 0.1

Question 5.
Use Structure Mr. Vidal serves 0.40 liter of turkey gravy. Explain how you can rename 0.40 in tenths in fraction form and in decimal form.

$$\frac{40}{100}$$ = 0.4 liter
Explanation:
Mr. Vidal serves 0.40 liter of turkey gravy.
Rename 0.40 in tenths in fraction form is $$\frac{40}{10}$$

Question 6.
A roll of foil has 3$$\frac{9}{10}$$ meters left on the roll. Rename the length of foil as a mixed number and as a decimal in hundredths.
3.90 meters
Explanation:
A roll of foil has 3$$\frac{9}{10}$$ meters left on the roll.
3$$\frac{9}{10}$$ = $$\frac{39}{10}$$
To write decimal in hundredths multiply numerator and denominator with 10
= $$\frac{39×10}{10×100}$$ = $$\frac{390}{100}$$ = 3.90

Question 7.
Mason waits for 4$$\frac{50}{100}$$ minutes. Rename the amount of time as a mixed number and as a decimal in tenths.
4.5
Explanation:
Mason waits 4$$\frac{50}{100}$$
= $$\frac{450}{100}$$
To write decimal in tenths divided the numerator and denominator with 10
= $$\frac{450÷10 }{100÷10}$$ = $$\frac{45 }{10}$$
= 4.50

Question 8.
Critique Reasoning Chloe says that 0.90 is equivalent to $$\frac{9}{100}$$. Is Chloe correct? Explain.
No, Chloe is wrong.
Explanation:
0.90 is equivalent to $$\frac{9}{100}$$
$$\frac{9}{100}$$ = 0.09
$$\frac{90}{100}$$ = 0.90

I’m in a Learning Mindset!

What strategies for writing a fraction and an equivalent decimal did I share with a partner?
$$\frac{2×2}{3×2}$$ = $$\frac{4}{6}$$


text:
## Algebra 2 (1st Edition)

$$3x^2+7x+4$$
We start with the factored form of the equation that would make these solutions, and then we use the FOIL method to get the equation into the form $ax^2+bx+c$. Thus, we obtain: $$(3x+4)(x+1) \\ 3x^2+3x+4x+4 \\ 3x^2+7x+4$$


text:
# zbMATH — the first resource for mathematics

Well-solved special cases. (English) Zbl 0631.90081
The traveling salesman problem, a guided tour of combinatorial optimization, 87-143 (1985).
[For the entire collection see Zbl 0562.00014.]
The authors give an excellent survey on special travelling salesman problems which can be solved by polynomial algorithms. There are two kinds of conditions possible: algebraic conditions for the cost matrix or graph theoretic conditions on the underlying network. In particular the following problem classes are treated: the “constant” TSP yields the same objective function value for all tours. Its cost elements have the form $$c_{ij}=ai+b_ j$$ for given vectors $$a=(a_ i)$$ and $$b=(b_ j)$$. Also “small TSP’s” with $$c_{ij}=\min \{a_ i,b_ j\}$$ can be solved polynomially, whereas Sarvanov showed that the TSP with product matrices $$c_{ij}=a_ ib_ j$$ is NP-hard. A proof of this famous theorem is given. Moreover, an extensive survey on the general Demidenko- conditions, the theory of pyramidal tours and subtour patching is given (inclusive proofs). Here also the Gilmore-Gomory case is treated.
Moreover it is mentioned that circulants as cost-matrices allow to find polynomially an open Hamiltonian path whereas it is an open question whether the Hamiltonian tour problem is NP-hard. TSPs with graded matrices $$(c_{ij}\leq c_{ij+1}$$ for all i,j) are NP-hard for sum objectives, but polynomially tractible for bottleneck objectives. TSPs with upper triangular cost-matrices can be solved via assignment problems. This leads to graphtheoretic conditions: TSPs on networks with limited bound widths and on Halin graphs can be solved by efficient algorithms.
Reviewer: R.E.Burkard

##### MSC:
 90C35 Programming involving graphs or networks 90-02 Research exposition (monographs, survey articles) pertaining to operations research and mathematical programming 68Q25 Analysis of algorithms and problem complexity 05C35 Extremal problems in graph theory


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Avisynth And Its Plugins

A question that people ask me on a pretty regular basis is "What the heck is Avisynth?" Well, I'll tell you. But be warned, this post can get a bit..."wordy".

A Little History...

Avisynth is a program that started as a small software project by Berkley student Ben Rudiak-Gould. Since it's creation, Avisynth has seen a explosion in development from the open source community and has been released in numerous versions (at the time of this writing, it is at version 2.5.8 stable, and 2.6.0 alpha 2).

Additional development has been contributed by authors Edwin van Eggelen, Klaus Post, Richard Berg, Ian Brabham, and many others. There have even been sub-versions created to add support for multithreading and fix several bugs (the current plan is to incorporate multithreading support into version 2.6.0).

But I digress. So what exactly isAvisynth? Avisynth is a scripting language of sorts. More specifically, it is a scripting language targeted at video post-production. When I say "video" I actually (kind of) mean "film" in the essence of both video and audio, as Avisynth can manipulate both video and audio streams.

Anyways, Avisynth is essentially a way to use a simple text file to decode and transform a video file.

Ex.

Avisource("C:\\A\\path\\to\\my\\video.avi")

The example above can be all that is necessary in some instances. For example, someone could create a file with the above contents and then load that ".avs" file into a program like MeGUI or VirtualDub and start encoding.

But frankly that is a little boring.

What you will see more often is something along the lines of the following:

Ex.

DirectshowSource("C:\\A\\path\\to\\my\\video.mp4")
Crop(4,0,-4,-0)
Spline36Resize(720,480)

Okay, so let's go over what I just did. First off, notice how I used DirectShowSource instead of Avisource? The difference is that Avisource should only be used to open AVI files (files that end with the ".avi" file extension) while DirectshowSource can be used to open pretty much anything (with some caveats). One caveat is opening DVD VOB files. When opening VOB files, you should use Donald Graft's DGDecode plugin, but we will get to that later.

Note: I wrote a tutorial a little while back about properly configuring your DirectShow settings. You can read it here.

Secondly, notice that Crop line? I'll give you one guess as to what it does. That's right! It crops the video! To be more specific, it crops 4 pixels from the left side of the video and 4 pixels from the right side of the video. The top and the bottom are left alone. This can be used to remove dirty edges from videos or to crop off black bars. For more information on the crop filter, refer to the Avisynth documentation or the Avisynth Crop wiki page.

Thirdly, look at the Spline36Resize line. I don't expect you to know right away what this does, but the hint is at the very end of the name. "Resize". That's right, this little function resizes my video to 720 pixels wide by 480 pixels high. Does that make sense?

To be clear, when working with Avisynth scripts everything you do will be through the use of functions calls. For those of you who are not programmers, a function call looks like this:

Ex.

FunctionName(parameter 1, parameter 2, parameter 3, etc...)

Not all functions have more than one parameter, or even any parameters. Often, you can call a function just like this:

Ex.

DirectShowSource("myvideo.mp4")
FunctionName()

Now, some functions do require parameters, like our Spline36Resize function. Remember how I wanted to resize to 720pixels by 480 pixels? Well, all I needed to do is enter those numbers into the parameters of the Spline36Resize function and viola!

Ex.

Spline36Resize(720,480)

This will now produce an video stream that is 720 pixels wide by 480 pixels high!

Internal Filters

All of the function calls that I have made in the above examples have been to Internal Filters. What this means is that these filters come with Avisynth when you install it. You can see all of the available filters in the documentation or on the Avisynth Wiki. Don't be overwhelmed because even I don't know all of the internal filters. Just get familiar with some of them and the rest will come naturally.

External Filters

This is where the power of Avisynth really starts to shine. An external filter is an Avisynth filter that you can download from the internet to expand Avisynth's functionality. All filters take the form of ".dll" files and are placed in Avisynth's "Plugins" folder, which can be found in the Avisynth install folder (usually in your Program Files folder in Windows).

In order to call a filter, all you have to do is make sure that the filter is in your Plugins directory, and then use the same function calling syntax that I detailed earlier.

Ex.

DirectshowSource("myvideo.mp4")
AnExternalFilter(param 1, param 2)

There is a whole host of external filters available. More and more are debuted on the Doom9 forums practically every day. No body actually knows how many external plugins exist, but this Avisynth Wiki page is a good place to start when looking for a new plugin.

External Scripts

One last section that I want to talk about is using External Scripts in your Avisynth files. The cool thing about external scripts is that they are just text files. They can be downloaded and placed in the Plugins folder just like external filters, but need a special file extension if you want them to load automatically. If you want a script to be automatically loaded by Avisynth every time you use it, give the script a ".avsi" extension, instead of the standard ".avs" Avisynth file extension. Now, you can use an external script just like a filter when you are ready to call it.

Ex.

DirectshowSource("myvideo.mp4")
MyExternalScript(param 1, param 2, param 3)

Conclusion

I hope that this little introduction into Avisynth has helped you in some way, and if it hasn't, let me know in the comments and I'll add on to this post to make it more clear in the future.

Again, this post is just an introduction into Avisynth to make it a little less scary. If you want to get into Avisynth, I suggest you start reading the documentation that is installed with Avisynth. In addition, the Doom9 forums and the Avisynth Wiki are fantastic resources.

All in all, I wish you good luck!

References

Anime Music Videos Guide - http://www.animemusicvideos.org/guides/avtech/amvappavisynth.html

Avisynth Wiki - http://avisynth.org/mediawiki/Main_Page

Doom9 Forums - http://forum.doom9.org/forumdisplay.php?f=69


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# Madular multiplicative inverse

Find the largest natural number $$n$$ such that every natural number less than $$n$$ and coprime to $$n$$ is its own modular multiplicative inverse mod $$n$$.

×

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# Need some help with arcus functions

• Jan 2nd 2010, 08:16 AM
Vey
Need some help with arcus functions
Hey, I'm not sure wether this is univerity or pre-university stuff cuz I'm in university now, but I did it in high school too, so I'm sorry if this is at the wrong place.

My problem is this:
I need to calculate arcsin( 3/sqrt(10) ) + arccos( 1/sqrt(5) ) and the answer may not contain any cyklometric functions.

• Jan 2nd 2010, 08:24 AM
skeeter
Quote:

Originally Posted by Vey
Hey, I'm not sure wether this is univerity or pre-university stuff cuz I'm in university now, but I did it in high school too, so I'm sorry if this is at the wrong place.

My problem is this:
I need to calculate arcsin( 3/sqrt(10) ) + arccos( 1/sqrt(5) ) and the answer may not contain any cyklometric functions.

let $a = \arcsin\left(\frac{3}{\sqrt{10}}\right)$

$b = \arccos\left(\frac{1}{\sqrt{5}}\right)$

$\sin(a+b) = \sin{a}\cos{b} + \cos{a}\sin{b}$

$\sin(a+b) = \frac{3}{\sqrt{10}} \cdot \frac{1}{\sqrt{5}} + \frac{1}{\sqrt{10}} \cdot \frac{2}{\sqrt{5}} = \frac{1}{\sqrt{2}}
$

$a+b = \frac{3\pi}{4}$
• Jan 2nd 2010, 08:41 AM
Hello Vey

Welcome to Math Help Forum!
Quote:

Originally Posted by Vey
Hey, I'm not sure wether this is univerity or pre-university stuff cuz I'm in university now, but I did it in high school too, so I'm sorry if this is at the wrong place.

My problem is this:
I need to calculate arcsin( 3/sqrt(10) ) + arccos( 1/sqrt(5) ) and the answer may not contain any cyklometric functions.

The best thing to do with inverse trig functions is to get rid of them as soon as you can! So:

Let:
$x = \arcsin \left(\frac{3}{\sqrt{10}}\right)$
and
$y = \arccos \left(\frac{1}{\sqrt5}\right)$
Then get rid of the inverse functions:
$\sin x = \frac{3}{\sqrt{10}}$ and $\cos y = \frac{1}{\sqrt5}$
So, using Pythagoras' Theorem:
$\cos x = \frac{1}{\sqrt{10}}$ and $\sin y = \frac{2}{\sqrt5}$
Now we need $x + y$, so, using $\cos(x+y) = \cos x \cos y - \sin x \sin y$, we get:
$\cos(x+y) = \frac{1}{\sqrt{10}}\frac{1}{\sqrt{5}}- \frac{3}{\sqrt{10}}\frac{2}{\sqrt{5}}$
$=\frac{1}{5\sqrt2}-\frac{6}{5\sqrt2}$

$= -\frac{1}{\sqrt2}$
Of course, there's an infinite number of angles whose cosine is $= -\frac{1}{\sqrt2}$, but, noting that $0 and $0 < y < \frac{\pi}{2}$, it's clear that:
$x+y = \frac{3\pi}{4}$


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# Extended Rolle's Theorem

## Theorem

Let $f: D \to \R$ be differentiable on a closed interval $I \subseteq \R$.

Let $x_0 < x_1 < \dots < x_n \in I$.

Let $f \left({x_i}\right) = 0$ for $i = 0, \ldots, n$.

Then for all $i = 0, \ldots, n-1$:

$\exists \xi_i \in \left({x_i \,.\,.\, x_{i+1}}\right): f' \left({\xi_i}\right) = 0$

## Proof

Since $f$ is differentiable on $I$, f is differentiable on $\left[{x_i \,.\,.\, x_{i+1}}\right]$ for $i = 0, \ldots, n-1$.

Thus a fortiori, $f$ is also continuous on the closed interval $\left[{x_i \,.\,.\, x_{i+1}}\right]$ and differentiable on the open interval $\left({x_i \,.\,.\, x_{i+1}}\right)$.

For $i = 0, \ldots, n$ we have $f \left({x_i}\right) = 0$.

Hence the conditions of Rolle's Theorem are fulfilled on each interval which yields the result.

## Comment

If $f$ is differentiable multiple times, this theorem can be applied multiple times recursively until one runs out of zeroes, each interval pair $\left({x_i \,.\,.\, x_{i+1}}\right)$ $\left({x_{i+1} \,.\,.\, x_{i+2}}\right)$ becoming $\left({\xi_i \,.\,.\, \xi_{i+1}}\right)$ with $f \left({\xi_i}\right) = 0$, $f \left({\xi_{i+1}}\right) = 0$ and all $\xi \in I$.

## Source of Name

This entry was named for Michel Rolle.


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# In a linear regression model with both categorical and continuous predictors, what is the interpretation of a categorical predictor coefficient?

Suppose we have a linear regression model where the response $Y$ is continuous and have $X_1, X_2$ as continuous predictor variables and $I_1, I_2, I_3, I_4$ as categorical predictors that refer to the seasons, i.e., $I_1$ is fall, $I_2$ is winter, etc.

Suppose we have:

$$Y = \beta_0 + \beta_1X_1 + \beta_2X_2 + \beta_3I_1 + \beta_4I_2 + \beta_5I_3 + \epsilon$$

where $I_4$ was dropped to prevent multicollinearity.

Normally, the interpretation of $\beta_3$ in a model where there are ONLY categorical predictors is that it is the difference in effects between $I_4$ and $I_1$, and that the intercept is the case where $\beta_0$ is the effect of $I_4$.

However, in this model, how do we interpret $\beta_0$ and $\beta_3$? Do we set the continuous predictors to zero? Thanks.

Effect of Categorical Variable
In the absence of an interaction* with a continuous predictor, the effect of the categorical predictor is the difference with the intercept ($\beta_0$) at any given level of the continuous predictor. After all, if there is no interaction, they are presumed to affect the outcome independently of each other.

Do we set the continuous predictors to zero?

No, at least not necessarily.

*If there would be an interaction with a continuous predictor, you can only consider the effect of the interaction (i.e. the slope per category), due to the principle of marginality.

Edit
To put it more simple, $\beta_3$ is already the difference between $I_4$ and $I_1$. There is no need to take the continuous predictor into consideration. To whom downvoted this, care to elaborate?

• I am not sure that a variable season would be any different from four categories. It would still be treated the exact same way in the regression (i.e. a 4x4 dummy variable matrix with a column dropped due to multicolinearity) Oct 10, 2017 at 12:41
• So why then opt to include separate dummy variables? A single factor seems simpler to me. Oct 10, 2017 at 14:25
• Depends on the audience? Some reviewers want the model spelled out, some are happy with just having a single variable represent levels. All in all, its mathematically equivalent. Oct 10, 2017 at 15:50
• Very well, I'll remove that part Oct 10, 2017 at 21:24


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With all due respect to tan's inverse

Calculus Level 3

Find the derivative of $$\displaystyle \tan ^{ -1 }{ \left( \cfrac { \sqrt { 1+{ x }^{ 2 } } -1 }{ x } \right) }$$ with respect to $$\displaystyle \tan ^{ -1 }{ x }$$, when $$x\neq 0$$.

Notation: $$\displaystyle \tan ^{ -1 }{ y } =\arctan { y }$$.

×


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Question regarding the continuity of functions

Homework Statement

so a function was only continuous if and only if lim x ---> a = f(a)

but I read some other books saying a functions continuity can only be judged based on their allowed domains. Does that mean the first definition fails if the domains of where the rule fails are explicitly stated or when using a "familiar" parent graph?

refer to above

The Attempt at a Solution

the book I read,the humongus of calculus, said but I read some other books saying a functions continuity can only be judged based on their allowed domains. The book doesnt seem to serious with math. I dont know how to approach this. I think the book meant to say that this only applies to continuity over an interval rather than the meaning a whole function was continuous.

Dick
Homework Helper

Homework Statement

so a function was only continuous if and only if lim x ---> a = f(a)

but I read some other books saying a functions continuity can only be judged based on their allowed domains. Does that mean the first definition fails if the domains of where the rule fails are explicitly stated or when using a "familiar" parent graph?

refer to above

The Attempt at a Solution

the book I read,the humongus of calculus, said but I read some other books saying a functions continuity can only be judged based on their allowed domains. The book doesnt seem to serious with math. I dont know how to approach this. I think the book meant to say that this only applies to continuity over an interval rather than the meaning a whole function was continuous.

Your first statement defines continuity at x=a. f needs to be defined in an open interval around x=a for that to work. Other than that I don't know what the 'allowed domains' thing is about. Can you quote something? Being continuous at a point is certainly different from being continuous everywhere. Is that what you are getting at?

Mark44
Mentor

Homework Statement

so a function was only continuous if and only if lim x ---> a = f(a)
As written, this doesn't mean what you're trying to say - you left part out. It should be like so:
$$\lim_{x \to a}f(x) = f(a)$$
but I read some other books saying a functions continuity can only be judged based on their allowed domains. Does that mean the first definition fails if the domains of where the rule fails are explicitly stated or when using a "familiar" parent graph?

refer to above

The Attempt at a Solution

the book I read,the humongus of calculus, said but I read some other books saying a functions continuity can only be judged based on their allowed domains. The book doesnt seem to serious with math. I dont know how to approach this. I think the book meant to say that this only applies to continuity over an interval rather than the meaning a whole function was continuous.

eumyang
Homework Helper
so a function was only continuous if and only if lim x ---> a = f(a)
If this is supposed to define continuity of a function at a, you're missing some details. If I remember correctly, for a function to be continuous at a, three conditions have to be satisfied:
1. $f(a)$ is defined.
2. $\lim_{x \to a}f(x)$ exists.
3. $\lim_{x \to a}f(x) = f(a)$

For a function to be continuous in some open interval (a, b), the function has to be continuous at every point in the interval.

pasmith
Homework Helper

Homework Statement

so a function was only continuous if and only if lim x ---> a = f(a)

$$\lim_{x \to a} f(x) = f(a)$$

but I read some other books saying a functions continuity can only be judged based on their allowed domains. Does that mean the first definition fails if the domains of where the rule fails are explicitly stated or when using a "familiar" parent graph?

The domain of a function is part of its definition. A function is defined at all points of its domain and is undefined everywhere else. A function can't be continuous at a point outside its domain because its value is not defined there.

The formal definition of continuity of a real function is that $f: U \subset \mathbb{R} \to \mathbb{R}$ is continuous at $a \in U$ if and only if for all $\epsilon > 0$ there exists $\delta > 0$ such that for all $x \in U$, if $|x - a| < \delta$ then $|f(x) - f(a)| < \epsilon$.

The definition does not require that $(a - \delta, a + \delta) \subset U$. Indeed it may happen that $U \cap (a - \delta, a + \delta) = \{a\}$, in which case $f$ being continuous as $a$ tells you nothing about its values at any other points in $U$.

Thus, for example, any $f: U = (-\infty, 1] \cup \{2\} \cup [3, \infty) \to \mathbb{R}$ is continuous at 2, because for all $x \in U$, if $|x - 2| < \frac12$ then $x = 2$. Thus for all $x \in U$, if $|x - 2| < \frac12$ then $|f(x) - f(2)| = 0$.


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# zbMATH — the first resource for mathematics

Existence of $$n$$-factors in powers of connected graphs. (English) Zbl 0735.05069
The author proves the following: If $$G$$ is a connected graph of order $$p\geq n+1$$ $$(n\geq 1)$$, with $$p$$ odd if $$n$$ is odd, then for any vertex $$v$$ of $$G$$, $$G^{n+1}-v$$ has an $$n$$-factor. This supplements an earlier result of the author and Nebeský: If $$G$$ is a connected graph of order $$p\geq n+1$$ $$(n\geq 1)$$ with $$p$$ even when $$n$$ is odd, then $$G^{n+1}$$ has an $$n$$-factor.
Actually the author derives the result by providing it for any spanning tree of $$G$$. Besides the usual induction and case-studies, the proof technique involves a systematic ordering of the vertices in the different branches of the tree $$T$$ at $$v$$, going ’upwards’ from $$v$$ to the leaves. This simplifies the distance computations.
Reviewer: K.R.Parthasarathy
##### MSC:
 05C70 Edge subsets with special properties (factorization, matching, partitioning, covering and packing, etc.) 05C12 Distance in graphs 05C05 Trees
##### Keywords:
$$n$$-factor
Full Text:


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## Introductory Algebra for College Students (7th Edition)

14x$^2$-11
Use the distributive property to get rid of the parentheses and bracket and combine like terms. 4(6x$^2$-3)-[2(5x$^2$-1)+1] 4(6x$^2$-3)-[10x$^2$-2+1] 4(6x$^2$-3)-[10x$^2$-1] 4(6x$^2$-3)-10x$^2$+1 4(6x$^2$)+4(-3)-10x$^2$+1 24x$^2$+(-12)-10x$^2$+1 24$^2$-10x$^2$+(-12)+1 14x$^2$-11


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# Subset simulation

Subset simulation[1] is a method used in reliability engineering to compute small (i.e., rare event) failure probabilities encountered in engineering systems. The basic idea is to express a small failure probability as a product of larger conditional probabilities by introducing intermediate failure events. This conceptually converts the original rare event problem into a series of frequent event problems that are easier to solve. In the actual implementation, samples conditional on intermediate failure events are adaptively generated to gradually populate from the frequent to rare event region. These 'conditional samples' provide information for estimating the complementary cumulative distribution function (CCDF) of the quantity of interest (that governs failure), covering the high as well as the low probability regions. They can also be used for investigating the cause and consequence of failure events. The generation of conditional samples is not trivial but can be performed efficiently using Markov chain Monte Carlo (MCMC).

Subset Simulation takes the relationship between the (input) random variables and the (output) response quantity of interest as a 'black box'. This can be attractive for complex systems where it is difficult to use other variance reduction or rare event sampling techniques that require prior information about the system behaviour. For problems where it is possible to incorporate prior information into the reliability algorithm, it is often more efficient to use other variance reduction techniques such as importance sampling. It has been shown that subset simulation is more efficient than traditional Monte Carlo simulation, but less efficient than line sampling, when applied to a fracture mechanics test problem.[2]

## Basic idea

Let X be a vector of random variables and Y = h(X) be a scalar (output) response quantity of interest for which the failure probability ${\displaystyle P(F)=P(Y>b)}$  is to be determined. Each evaluation of h(·) is expensive and so it should be avoided if possible. Using direct Monte Carlo methods one can generate i.i.d. (independent and identically distributed) samples of X and then estimate P(F) simply as the fraction of samples with Y > b. However this is not efficient when P(F) is small because most samples will not fail (i.e., with Y ≤ b) and in many cases an estimate of 0 results. As a rule of thumb for small P(F) one requires 10 failed samples to estimate P(F) with a coefficient of variation of 30% (a moderate requirement). For example, 10000 i.i.d. samples, and hence evaluations of h(·), would be required for such an estimate if P(F) = 0.001.

Subset simulation attempts to convert a rare event problem into more frequent ones. Let ${\displaystyle b_{1}  be an increasing sequence of intermediate threshold levels. From the basic property of conditional probability,

{\displaystyle {\begin{aligned}P(Y>b)&=P(Y>b_{m}\mid Y>b_{m-1})P(Y>b_{m-1})\\&=P(Y>b_{m}\mid Y>b_{m-1})P(Y>b_{m-1}\mid Y>b_{m-2})P(Y>b_{m-2})\\&=\cdots \\&=P(Y>b_{m}\mid Y>b_{m-1})P(Y>b_{m-1}\mid Y>b_{m-2})\cdots P(Y>b_{2}\mid Y>b_{1})P(Y>b_{1})\end{aligned}}}

The 'raw idea' of subset simulation is to estimate P(F) by estimating ${\displaystyle P(Y>b_{1})}$  and the conditional probabilities ${\displaystyle P(Y>b_{i}\mid Y>b_{i-1})}$  for ${\displaystyle i=2,\ldots ,m}$ , anticipating efficiency gain when these probabilities are not small. To implement this idea there are two basic issues:

1. Estimating the conditional probabilities by means of simulation requires the efficient generation of samples of X conditional on the intermediate failure events, i.e., the conditional samples. This is generally non-trivial.
2. The intermediate threshold levels ${\displaystyle b_{i}}$  should be chosen so that the intermediate probabilities are not too small (otherwise ending up with rare event problem again) but not too large (otherwise requiring too many levels to reach the target event). However, this requires information of the CCDF, which is the target to be estimated.

In the standard algorithm of subset simulation the first issue is resolved by using Markov chain Monte Carlo.[3] More generic and flexible version of the simulation algorithms not based on Markov chain Monte Carlo have been recently developed.[4] The second issue is resolved by choosing the intermediate threshold levels {bi} adaptively using samples from the last simulation level. As a result, subset simulation in fact produces a set of estimates for b that corresponds to different fixed values of pP(Y > b), rather than estimates of probabilities for fixed threshold values.

There are a number of variations of subset simulation used in different contexts in applied probability and stochastic operations research[5][6] For example, in some variations the simulation effort to estimate each conditional probability P(Y > bi | Y > bi−1) (i = 2, ..., m) may not be fixed prior to the simulation, but may be random, similar to the splitting method in rare-event probability estimation. [7] These versions of subset simulation can also be used to approximately sample from the distribution of X given the failure of the system (that is, conditional on the event ${\displaystyle \{Y>b\}}$ ). In that case, the relative variance of the (random) number of particles in the final level ${\displaystyle m}$  can be used to bound the sampling error as measured by the total variation distance of probability measures.[8]

## Notes

• See Au & Wang[9] for an introductory coverage of subset simulation and its application to engineering risk analysis.
• Schuëller & Pradlwarter[10] reports the performance of Subset Simulation (and other variance reduction techniques) in a set of stochastic mechanics benchmark problems.
• Chapter 4 of Phoon [11] discusses the application of subset simulation (and other Monte Carlo methods) to geotechnical engineering problems.
• Zio & Pedroni[12] discusses the application of subset simulation (and other methods) to a problem in nuclear engineering.

## References

1. ^ Au, S.K.; Beck, James L. (October 2001). "Estimation of small failure probabilities in high dimensions by subset simulation". Probabilistic Engineering Mechanics. 16 (4): 263–277. CiteSeerX 10.1.1.131.1941. doi:10.1016/S0266-8920(01)00019-4.
2. ^ Zio, E; Pedroni, N (2009). "Subset simulation and line sampling for advanced Monte Carlo reliability analysis". Reliability, Risk, and Safety (PDF). doi:10.1201/9780203859759.ch94. ISBN 978-0-415-55509-8. S2CID 9845287.
3. ^ Au, Siu-Kui (2016). "On MCMC algorithm for Subset Simulation". Probabilistic Engineering Mechanics. 43: 117–120. doi:10.1016/j.probengmech.2015.12.003.
4. ^ Au, Siu-Kui; Patelli, Edoardo (2016). "Rare event simulation in finite-infinite dimensional space" (PDF). Reliability Engineering & System Safety. 148: 67–77. doi:10.1016/j.ress.2015.11.012.
5. ^ Villén-Altamirano, Manuel; Villén-Altamirano, José (1994). "Restart: a straightforward method for fast simulation of rare events". Written at San Diego, CA, USA. Proceedings of the 26th Winter simulation conference. WSC '94. Orlando, Florida, United States: Society for Computer Simulation International. pp. 282–289. ISBN 0-7803-2109-X. acmid 194044.
6. ^ Botev, Z. I.; Kroese, D. P. (2008). "An Efficient Algorithm for Rare-event Probability Estimation, Combinatorial Optimization, and Counting". Methodology and Computing in Applied Probability. 10 (4): 471–505. CiteSeerX 10.1.1.399.7912. doi:10.1007/s11009-008-9073-7. S2CID 1147040.
7. ^ Botev, Z. I.; Kroese, D. P. (2012). "Efficient Monte Carlo simulation via the generalized splitting method". Statistics and Computing. 22 (1): 1–16. doi:10.1007/s11222-010-9201-4. S2CID 14970946.
8. ^ Botev, Z. I.; L’Ecuyer, P. (2020). "Sampling Conditionally on a Rare Event via Generalized Splitting". INFORMS Journal on Computing. arXiv:1909.03566. doi:10.1287/ijoc.2019.0936. S2CID 202540190.
9. ^ Au, S.K.; Wang, Y. (2014). Engineering Risk Assessment with Subset Simulation. Singapore: John Wiley & Sons. ISBN 978-1-118-39804-3.
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# Prove that $\neg p \to (q \to r)$ and $q \to (p \vee r)$ are logically equivalent using the laws of logical equivalences

by Nuvali   Last Updated September 12, 2019 04:20 AM

Please help, I cannot figure out how $\neg p \to (q \to r)$ and $q \to (p \vee r)$ are logically equivalent using the laws of logical equivalences.

Here's what I came up, please help to explain how to show that the equation is logically equivalent using the laws of logical equivalences

$\neg p \to (q \to r) = q \to (p \vee r)$

$\neg p (\neg q \vee r) = q (p \vee r)$

$p (q \vee r) = q \to p$

Thank you!

Tags :

~p --> (q --> r)
~~p v (q --> r)
p v (~q v r)
p v (~q) v r
~q v (p v r)
q --> (p v r)

Юрій Ярош
May 07, 2018 16:44 PM

We have $(\neg p\to (q\to r))\equiv (p\lor (q\to r))\equiv (p\lor\neg q\lor r)\equiv (\neg q \lor p\lor r)\equiv (q\to (p\lor r))$.

J.G.
May 07, 2018 17:00 PM