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## reference request – Energy estimation for linear hyperbolic system (without Fourier)

Where can I find proof of an energy estimate (under appropriate assumptions) for the following linear hyperbolic system which is not based on Fourier methods?

$$frac { partial U} { partial t} + sum_ {j = 1} ^ m A_j frac { partial U} { partial x_j} + B cdot U = 0,$$
or $$x in mathbb R ^ m$$ and $$U in mathbb R ^ n$$ and $$A$$ is a symmetric matrix.

## Solving a system and taking inverse Laplace / Fourier transforms

I have a set of linear equations for 4 quantities that have been Fourier and Laplace transforms. The system must be solved for the quantities and then each of the quantities must be transformed into an inverse Laplace, then transformed into an inverse Fourier (where the transforms are all one-dimensional so there is only one frequency).

I tried to solve the system with Maple, then I use the invlaplace function to take the Laplace inverse transforms of the quantities. The code I use is as follows:

e1: = -2 * DIkpi + As = 0

e2: = 2 * AIk * pi + 2 * CIkpi + Bs = a

e3: = s * C + 4/5 * Pw3(2 * ftIk * D – 2 * 1/3 * piIkD) / w2 = -2(C + 2 * K * (2 * piIk * B – 2 * 1/3 * piIk * B)) / (w2 * K)

e4: = s * D + 2 * 5/4 * P * t4 * piIk * C / (t2 * K) = -5 / (2 * P) * D / (t2 * K)

sys: = {s * C + 4/5 * Pw3(2 * ftIk * D – 2 * 1/3 * piIkD) / w2 = -2(C + 2 * K * (2 * piIk * B – 2 * 1/3 * piIk * B)) / (w2 * K), s * D + 2 * 5/4 * P * t4 * piIk * C / (t2 * K) = -5 / (2 * P) * D / (t2 * K), -2 * DIkpi + As = 0.2 * AIk * pi + 2 * CIkpi + Bs = a}

solve (sys, (A, B, C, D)),

where everything is a constant apart from the frequency $$k$$, the variable $$s$$ is from Laplace's transformation and $$i = sqrt {-1}$$. The solution Maple produces for the 4 quantities is a fairly complicated fraction, and when I use invlaplace on each quantity, I get the answer as a sum on the roots of a quartic polynomial.

I was wondering if it would be possible to use Mathematica to solve the system and get the quantities in the simplest form possible, then if there is a function similar to invlaplace which then allows me to Get a relatively simple analytical expression for the inverse 1D Laplace transformation for each of the magnitudes, so that they can be passed to a function that gives the inverse Fourier transform 1D?

## ag.algebraic geometry – Why is the Fourier transform so ubiquitous?

Many operations and equivalences in mathematics appear as a kind of Fourier transform. By Fourier transform, I mean the following:

Let $$X$$ and $$Y$$ be two objects of a certain category with products, and consider correspondence $$X leftarrow X times Y to Y$$. If we have an object (think of the sheaf, the function, the space, etc.) $$mathcal {P}$$ more than $$X times Y$$ and another, let's say $$mathcal {F}$$ more than $$X$$, assuming the existence of proper push and pull operations, we can get another object on $$Y$$ pulling $$mathcal {F}$$ back to product, tensor with $$mathcal {P}$$, then pushing towards $$Y$$.

The standard example is the Fourier transform of functions on a locally compact abelian group $$G$$ (for example. $$mathbb {R}$$). In that case, $$Y$$ is the Pontryagin dual of $$G$$, $$mathcal {P}$$ is the exponential function on the product, and the push and pull are given by integration and precomposition, respectively.

We also have the Fourier-Mukai functors for coherent showers in algebraic geometry which provide the equivalence of coherent showers on double abelian varieties. In fact, almost all of the interesting functors between coherent bundles on fairly pleasant varieties are examples of Fourier-Mukai transforms. A variant of this example also provides the geometric Langlands correspondence

$$D (Bun_T (C)) simeq QCoh (LocSys_1 (C))$$

for a torus $$T$$ and a curve $$C$$. In fact, the Langlands geometric correspondence for general reducing groups also seems to come from such a transformation.

Speak $$SYZ$$ guess, two Calabi-Yau mirror collectors $$X$$ and $$Y$$ are Lagrangian double-core fibrations. As such, conjectured equivalence

$$D (Coh (X)) simeq Fuk (Y)$$

is obtained morally by applying a Fourier-Mukai transform which ignites coherent sheaves $$X$$ in Lagrangiens in $$Y$$.

To make things more mysterious, many of these examples are the result of the existence of a perfect match. For example, the bundle of Poincaré lines which provides the equivalence of coherent sheaves on double abelian varieties $$A$$ and $$A ^ *$$ results from the perfect combination

$$A times A ^ * à B mathbb {G} _m.$$

Likewise, the Langlands geometric correspondence for the toroids, as well as the GLC for the Hitchin system, arise in a way from the self-duality of the Picard stack of the underlying curve. These examples seem to show that the non-degenerate quadratic forms seem to be fundamental in a very deep sense (for example perhaps even the Poincaré duality could be considered as a Fourier transform).

I don't have a specific question, but I would like to know why we should expect Fourier transformations to be so fundamental. These transformations are also found in physics as well as in many other "real world" situations of which I am even less qualified than my examples above. Nevertheless, I have the feeling that something deep is going on here and I would like an explanation, even a philosophical one, on why this model seems to appear everywhere.

## Fourier analysis – Reference request: explicit construction of Kakeya sets using Perron's tree

I have found many excellent notes online that illustrate how to build a set of Kakeya needles (with measure $$< varepsilon$$.) Yet none of them gives a complete argument on the construction of a set of Kakeya (with zero measure). The closest is given on page 6 of

https://web.stanford.edu/~yuvalwig/math/teaching/KakeyaNotes.pdf,

which unfortunately leaves the detail to argue for the existence of a unitary line segment in $$cap_ {i = 1} ^ infty U_n$$. He says this is proven subtly using a compactness argument.

What is this argument?

## Help verifying a Fourier transformation identity

I don't know why I'm getting stuck, but I'm having trouble verifying the following identity in one of my textbooks:

$$chi (T ( lambda – lambda_j)) = frac {1} {T pi} int _ {- infty} ^ { infty} hat { chi} (t / T) e ^ {it lambda} cos (t lambda_j) , dt + chi (T ( lambda + lambda_j))$$

I know I have to separate the complex exponential and use the Euler identity, but I'm having trouble getting the $$chi (T ( lambda + lambda_j))$$ term to work. My calculations keep coming with $$chi (T ( lambda_j – lambda))$$.

I don't think it's important for the calculation, but $$T> 1$$, $$lambda, lambda_j geq 0$$, and $$chi$$ is a function of the Schwartz class.

## derived categories – Fourier Mukai kernel which gives equivalence in one direction

Yes $$X$$ and $$Y$$ are two schemes and $$F in Perf (X times Y)$$, we can then define a functor $$Perf (X)$$ at $$Perf (Y)$$ like the Fourier Mukai transformation $$Phi ^ {X rightarrow Y} = q _ { ast} (F otimes p ^ { ast} -)$$, or $$p$$ and $$q$$ are the projections of $$X times Y$$ at $$X$$ and $$Y$$ respectively. Yes $$X$$ and $$Y$$ are smooth we can change $$Perf (-)$$ at $$D ^ b (-)$$, while if you choose $$F$$ to be a complex with almost coherent cohomologies, we change $$Perf (-)$$ at $$D_ {qc} (-)$$, the unbounded derived category of complexes with almost coherent cohomologies. Obviously, we could exchange the role of $$p$$ and $$q$$ and get a functor of $$Y$$ at $$X$$. My question is whether the fact that $$Phi ^ {X rightarrow Y}$$ is an equivalence implies that $$Phi ^ {Y rightarrow X}$$ is an equivalence, and if not, I would like a counterexample.

The reason why I think this might not be the case that the implication holds is that there is no apparent relationship between the two functors that I have described above. on it, even if they share the same core. However, I couldn't find a counterexample.

## What is the physical meaning of this property of the Fourier transform?

What is the physical meaning of this property of the Fourier transform?

$$int _ {- ∞} ^ {+ ∞} （f (x)） ^ 2 dx = frac1 {2π} int _ {- ∞} ^ {+ ∞} | F (ω) | ^ 2 dω,$$ ($$F (ω)$$ is the Fourier transform of $$f (x)$$)

The Fourier transform is widely used in electricity. Explanations in this or other areas are all welcome. A reference in the relevant articles is appreciated!

By the way, how do you type a formula in the text box when you ask a question? Thank you!

## Laplace and Fourier transform to derive an unstable fundamental solution

I consider the equations of the unsteady Stokes flux

$$nabla cdot u = 0,$$

$$rho frac { partial u} { partial t} = – nabla p + mu nabla ^ 2 u.$$

In an article (which I can provide if necessary), it is stated that one can use Fourier and Laplace transforms on these equations to obtain the unsteady fundamental solution.

I haven't used a lot of transformation theory before, so I was wondering how it would work, do we separate the variables to have an ODE in the time variable which is solved with the Laplace transform and the Fourier transform is preserved for the spatial part?

## Fourier analysis – Scaling of a Mexican hat wavelet into continuous wavelet transform

My general question is related to the concept of scaling a wavelet whose analytical form exists. An example is that of a Mexican hat.

The Mexican hat wavelet, obtained from the second derivative of a Gaussian, has a functional form of 2 * (1- (t / s). ^ 2). * Exp (-t. ^ 2 * 0.5 / s ^ 2) divided by (pi) ^ (1/4) * sqrt (3 * s), where "s" is the standard deviation of ; a Gaussian and $$t$$ is the independent variable.

If we want to graphically see its compressions and dilations graphically, let's say on a scale $$a$$ = 1 to 5, the typical form of a wavelet function $$psi$$ has a scale setting $$a$$ and a translation parameter $$b$$. The functional form of the Mexican hat given in MATLAB does not explicitly have them both.

Standard deviation $$s$$ is the only variable here that controls the width, but this is not strictly equal to the scale $$a$$. How to generate scale versions of the Mexican hat?

Thank you.

P.S. I posted this digital signal processing, but received no response.

## Fourier analysis – Reference (fundamental soil and grad estimation, etc.): a particular elliptical PDE

in $$mathbb {R} ^ d$$, consider the following equation
$$Delta u -x cdot nabla u = f$$
or $$f$$ may be $$C ^ infty$$ and exponentially rapid decay.

I would like to know the fundamental soil. to this equation or to a gradient estimate, preferably punctually. In particular I am interested in the following quantity
begin {align *} int _ { mathbb {R} ^ d} | nabla u | ^ 2e ^ {- frac {| x | ^ 2} {2}} dx. end {align *}

Thank you!


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# Bag of tricks in Advanced Calculus/ Real Analysis/Complex Analysis

I am studying for an exam and I have been studying my butt off
during the winter break for it. During the course of my study I have written
down quite a number of tricks, which in my opinion were ‘outrageous’ :-). Meaning there
was no way I would come up with that during an exam if I hadn’t seen that before.

Couple of examples.

1. Sometimes, when you want to prove something about $\max$, $\min$, you write ( I got this from Baby Rudin)

1. To prove Hölder’s inequality (in its simplest case)
You write $\int (f+tg)^2 \geq 0$ and since this stays positive
you get that the discriminant of this must be negative, and magically

2. When you want to show something about distinct zeroes of complex functions you kind of eliminate the zeroes of f by dividing them with the appropriate Möbius transforms and you still get an analytic functions which has nice properties.

The value of these is that they can be used in other contexts to write neat proofs.

That’s what I mean by “tricks”. This might be difficult to answer,
but what are some of the tricks you wise folks have up your sleeve when it comes to
Advanced Calculus (Both single variable, multivariable) and complex Analysis.

Anything you have to share will be greatly appreciated. Thanks so much for all your help.

Here are a couple of tricks and general plans of approach I know:

• If $x\in\Bbb R$ and for all $\epsilon>0$ we have that $|x|\leq\epsilon$, then $x=0$. I think of this fact as being Real Analysis in a nutshell.
• Never forget that if $A\subseteq\Bbb R$ is bounded above and $M=\sup A$, then for all $\epsilon>0$ there is a $y\in A$ such that $M< y+\epsilon$. Likewise if $A$ is bounded below and $m=\inf A$, then for all $\epsilon>0$ there is a $y\in A$ such that $y-\epsilon< m$. This is the most important tie between $\Bbb R$'s algebraic and ordering properties.
• Never underestimate the binomial theorem even if you just want an inequality. As an example, look at theorem 3.20(c) in Rudin and how he uses it.
• For all $x,y\in\Bbb R$ and any $\epsilon> 0$, we have the following inequality:
This can be derived from the observiation that $(\epsilon\,|x|-|y|)^2\geq 0$. This inequality allows us to decide how much 'weight' we want to give to a particular term in a product. This can be used to show that the product of Riemann integrable functions is still Riemann integrable.
• A simple inequality to remember is for $a,b,p\geq 0$. This can be derived from the even simpler inequality $(a+b)\leq 2\max(a,b)$, again for positive values. This inequality can be used to show that the $L^p$ spaces are vector spaces.
• The Weierstrass M-test is the first friend you call when dealing with series of functions.
• Ask yourself whether the problem you're working on can be generalized to topology first. Think about compactness and connectedness and the abstract theorems about them you already know.
• Perhaps the greatest topological property that $\Bbb R$ has is second-countability. This means that $\Bbb R$ is hereditarily-separable, sequential, Frechet-Urysohn, and c.c.c. This property of $\Bbb R$ allows us to consider sequences and sequential continuity in place of neighborhoods and continuity. As an adage, if you are working with $\epsilon$, wonder to yourself if you can instead work with $1/n$ with $n\in\Bbb N$.

format: plain


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## Down the rabbit hole

The following has been a rather interesting journey – from a test question which seemed fine, to a subtlety which seemed easy, to a discussion with a number of different mathematicians about the nature of distributions, measure theory and regularisation. I will try and make it as clear as possible in the post below. Note, as mentioned in the comments, we have actually only found the solution to this problem for a constrained range of x, and not x $\in \mathbb{R}$. I didn’t want to complicate things any more than necessary here for first year students, but the comments are very important too.

In a recent class test, there was a question, written by me, which was not quite the question that I wanted to ask. It turns out that it does have an answer, but it’s not an answer that can yet be found by the means at the class’s disposal.…

## On Convergent Sequences and Prime Numbers

Ever since Euclid first proved that there are infinitely many prime numbers, mathematicians have found ever more creative ways to prove the same result, and also various stronger theorems that imply it. Dirichlet’s Theorem, for example, states that if$m$ and $n$ are relatively prime integers, then there are infinitely many prime numbers of the form $mk + n$ for some integer $k$. It is also known that the sum of the reciprocals of the prime numbers diverges, that the sum

$\displaystyle \sum_{\substack{p \leq n \\ p \text{ prime}}} \frac{1}{p} \sim \log(\log(n))$

and that the number of prime numbers less than $n$ is asymptotically equal to $\displaystyle \frac{n}{\log(n)}$. In this blog post, we will continue this proud tradition by proving that there are infinitely many prime numbers which have your phone number somewhere in their digits, and which simultaneously have a prime number of digits.

To do so, we will look at the convergence of two different sums: that of the reciprocals of the primes with a prime number of digits, and that of the reciprocals of the natural numbers which do not contain your phone number amongst their digits.…

## Nowhere Differentiable Functions

By: Jan Wuzyk

In this article I am going to show that nowhere differentiable functions do in fact exists and give a few examples, some of which are relatively modern. But first I’m going to try to answer a question that is, in my opinion, too rarely discussed in mathematics classes, ”Why do we care?”.

# Why we care

To answer this question we have to look into the history of mathematics. In 1821 Augustin-Louis Cauchy published his seminal book, Cours d’Analyse, this is generally recognized as the first serious attempt to put calculus on a rigours footing[Com][1] , mainly through introducing rigorous definitions of limits,continuity and differentiability among others, and the definitions that go with them2. This was also time the integral was defined as an area instead of simply as the antiderivative.

It should be noted that Cauchy by no means closed the issue of rigour in analysis but he provided a starting point.…

## The Newton-Raphson Method

How would you go about finding the value of $\sqrt{3}$ if you didn’t have a square root button on your calculator? Well, the most obvious thing might be to try some values, based on your knowledge of the square root function. You are being asked to find that x for which:

$x=\sqrt{3}$

or, in other words, that x, which, when squared  gives 3. We have to be a little careful here because we know that there will actually be two numbers which satisfy this (one positive, one negative), and we are interested in the positive one only.

So, we try some values, but we don’t do it randomly, we can see that because $1^2=1$ and $2^2=4$ that whatever number squared gives 3 must be between 1 and 2. We can try something called a binary intersection. This just means taking the values which we know bound the right answer (ie. we know that 1<x<2), and trying the number in the middle.…


text:
# All Questions

9,980 questions
26 views

### Reference on generalization of plane graph duality between bonds and simple cycles

I've also asked this question on Mathoverflow, but it hasn't gotten an answer after several months: https://mathoverflow.net/questions/316132/reference-on-generalization-of-plane-graph-duality-between-...
78 views

### Shortest s-t path when is allowed to ignore k weights

Given an undirected graph $G$ with $n$ vertices and $m$ edges, with non-negative weights on the edges, what's the best algorithm that computes the shortest path from $s$ to $t$, where you are allowed ...
157 views

### Intuition Behind Strict Positivity?

I'm wondering if someone can give me the intuition behind why strict positivity of inductive data types guarantees strong normalization. To be clear, I see how having negative occurrences leads to ...
171 views
+100

120 views

### How small can extension complexity be?

In this article on extension complexity of regular polygons https://arxiv.org/pdf/1505.08031.pdf it is mentioned that extension complexity of $n$ regular polygons should be $\theta(\log n)$. This is ...
135 views

### Are Turing machines still useful as model of computation?

Often when I hear "Turing machine," my mind's eye pictures a quaint infinite ticker-tape with a small little machine writing and erasing $0$'s and $1$'s. But when I'm forced to think about a Turing ...
105 views

### Canonical complete problem for $\mathrm{FP}^{\Sigma^p_2}$

Given a $\Sigma^p_2$-complete oracle (i.e., $\Sigma_2 \mathrm{SAT}$), I have a problem that requires to call this oracle polynomially many times and returns an integer. Essentially, this is a function ...
66 views

### Entropy bounds on solutions to problems in BPP and other complexity classes based on entropy demands

Has anyone studied the asymptotics of problems in complexity classes like $BPP$? The thought came to me that if a problem in $BPP$ only requires $O(log(n))$ bits of entropy to solve then, intuitively, ...
94 views

249 views

### Naive definition of treewidth

Treewidth has arguably pretty involved definition. Recently I was thinking about a problem and turns out it easy to solve it for graphs with small naive treewidth''. Naive treewidth is defined as ...
158 views

### Extending Hindley-Milner to type mutable references

I have been trying to implement a programming language from scratch, and have gotten reasonably far. It reads just like Python, other than the fact that let is used ...
126 views


text:
#### Larvel – Eloquent way to connect three tables using two pivots

New to laravel,

I have three tables users, roles, and permissions. Those are connected with the following two pivot tables role_user, and permission_role

Here I want to connect three tables so I can get a result like many-to-many and one-to-many relationship. For instance, I can get all permissions and roles for the given user (one-to-many) and vise-versa for roles and permissions to each other. Also, I want to get all users and respective roles and permissions and vise-versa for roles and permissions.

I have just solved the issue getting roles for the given user with the help of the community but now I stuck to get the above mention result. Can anyone help me to make it happen?

## User Model

<?php

namespace App;

use IlluminateFoundationAuthUser as Authenticatable;

class User extends Authenticatable
{
use Notifiable;

/**
* The attributes that are mass assignable.
*
* @var array
*/
protected $fillable = [ 'name', 'email', 'password', ]; /** * The attributes that should be hidden for arrays. * * @var array */ protected$hidden = [
'remember_token',
];

/**
* The attributes that should be cast to native types.
*
* @var array
*/
protected $casts = [ 'email_verified_at' => 'datetime', ]; public function roles() { return$this->belongsToMany('AppAdminRole');
}
}

## Role Model

<?php

use IlluminateDatabaseEloquentModel;

class Role extends Model
{
protected $fillable = ['name', 'display']; public function permissions() { return$this->belongsToMany('AppAdminPermission');
}

{
return $this->belongsToMany('AppAdminAdmin'); } public function users() { return$this->belongsToMany('AppUser');
}

}

# Permission Model

<?php

protected $fillable = ['model', 'can']; public function roles() { return$this->belongsToMany('AppAdminRole');


text:
# Tag Info

4

Package ctable loads package transparent. Both package transparent and tikz implements transparency. This needs access to a page resource, a dictionary ExtGState for the transparent values. Unhappily, the LaTeX format has overslept the PDF development quite entirely. Managing global resources is the prime task for an OS, format in TeX speek. Because of the ...

0

Just add handout as an option for the beamer document class. \documentclass[handout]{beamer}

3

I do not know which is your system --- in TeXLive pdftex should embed all fonts by default since ages. In the linked article you can see it. To check, I used this MWE: \documentclass{article} \usepackage{mathpazo} \begin{document} This is just an example $a^b-\sqrt{c}$. \end{document} and with the utility pdffonts: [romano:~/tmp] pdffonts example.pdf ...

4

I will explain the preamble you should use if you want to follow the French canon des ateliers for B5 paper. The text width is a fraction of paperwidth – 3/4 for ordinary printing, 2/3 for neater printing and even 5/8 for luxury books. What remains is white space for horizontal margins. The inner margin is 4/10 of this white space, the outer margin is 6/10. ...

5

I've just written a new package for spot colors. It's named colorspace and as its name suggests the goal is to provide tools related to PDF color spaces. Currently you can define spot colors (including tints with the ! notation) as well as mixed inks (up to 4, so shades, ie, a spot color with black, are now possible). For example: ...

Top 50 recent answers are included


text:
# Consolidating google results for 2 domains with the same content

I have 2 domains http://foo.com and a http://bar.com. http://bar.com is just a domain alias for http://foo.com so they have the same content. Results for both domains show up on google, I'd rather all google results point to http://foo.com. I've tried looking for this option in google webmaster tools but can't find it. How do I do this?

-


text:
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Raymond borrowed $450 at 0% interest. If he pays back 0.5% of the total amount every 7 days, beginning exactly 7 days after the loan was disbursed, and has thus far paid back$18, with the most recent payment made today, how many days ago did he borrow the money?

(A) 6
(B) 8
(C) 25
(D) 42
(E) 56

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### Show Tags

09 Jun 2018, 04:04
Bunuel wrote:
Raymond borrowed $450 at 0% interest. If he pays back 0.5% of the total amount every 7 days, beginning exactly 7 days after the loan was disbursed, and has thus far paid back$18, with the most recent payment made today, how many days ago did he borrow the money?

(A) 6
(B) 8
(C) 25
(D) 42
(E) 56

Since the calculation looks a bit tricky and involves non-integer values, we'll first look to the answers.
This is an Alternative approach.

Since Raymond pays once every 7 days, made his most recent payment today, and started paying 7 days after the loan, the answer must be a multiple of 7.
Only (D), (E) are possible.
Say (D) were true.
Then Raymond borrowed the money 42 days ago and has been paying for 42 - 7 = 35 days = 6 payments in total (because one payment is on the first and the other on the last day)
He paid 0.5% of 450, 2.25, per week so he should have paid 6 * 2.25 dollars = 12 + 1.5 =13.5 . Not enough!

(E) must be our answer.
Let's verify: 56 - 7 = 49 = 8 payments and 8 * 2.25 = 16 + 2 = 18. Yes!
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# Raymond borrowed \$450 at 0% interest. If he pays back 0.5% of the tota

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text:
## Rydberg Equation

H-Atom ($E_{n}=-\frac{hR}{n^{2}}$)

VictoriaRoderick_3A
Posts: 42
Joined: Fri Sep 25, 2015 3:00 am

### Rydberg Equation

For chapter 1 question 57, why is it (1/2^2) - 1(7^2) in the Rydberg equation? Since the two is the first number in the Balmer Series, does the sequence start at 2? The question is trying to solve for the fifth line in the sequence, so would it go N1=2 and then N2=3,4,5,6,7 with 7 being the fifth number? That is the only way I can see it working.

Kanwaldeep Rai 2H
Posts: 21
Joined: Fri Sep 25, 2015 3:00 am

### Re: Rydberg Equation

This is correct. The Balmer Series begins with n1=2 so all the wavelengths that are listed are from n2=3,4,5,6. The next line would be n2=7. You would first find the frequency of this line. V(frequency)= R(1/n1^2-1/n2^2)=
(3.29x10^15 Hz)(1/2^2-1/7^2)= 7.55x10^14 Hz
Then convert this to wavelength by using the equation: wavelength=speed of light/frequency.


text:
# Continuous Learning from Data: No Multiplicities from Computing and Using Bayesian Posterior Probabilities as Often as Desired

(In a Bayesian analysis) It is entirely appropriate to collect data until a point has been proven or disproven, or until the data collector runs out of time, money, or patience.
Edwards, Lindman, Savage (1963)

Imagine that the military has developed a pattern recognition algorithm to determine whether a distant object is a tank. Initially an image results in an 0.8 probability of the object being a tank. The object moves closer and some fog clears. The probability is now 0.9, and the 0.8 has become completely irrelevant. Contrast with a frequentist way of thinking: Of all the times a tested object wasn't a tank, the probability of acquiring tank-like image characteristics at some point grows with the number of images acquired. Analogously, with multiple data looks in a clinical trial where there is no treatment effect, the probability of having extreme data at some point grows with the number of looks. This affects type I error but not the true chance that the treatment works.

# Introduction

Bayesian inference, which follows the likelihood principle, is not affected by the experimental design or intentions of the investigator. P-values can only be computed if both of these are known, and as been described by Berry (1987) and others, it is almost never the case that the computation of the p-value at the end of a study takes into account all the changes in design that were necessitated when pure experimental designs encounter the real world.

When performing multiple data looks as a study progress, one can accelerate learning by more quickly abandoning treatments that do not work, by sometimes stopping early for efficacy, and frequently by arguing to extend a promising but as-yet-inconclusive study by adding subjects over the originally intended sample size. Indeed the whole exercise of computing a single sample size is thought to be voodoo by most practicing statisticians. It has become almost comical to listen to rationalizations for choosing larger detectable effect sizes so that smaller sample sizes will yield adequate power.

Multiplicity and resulting inflation of type I error when using frequentist methods is real. While Bayesians concern themselves with “what did happen?”, frequentists must consider “what might have happened?” because of the backwards time and information flow used in their calculations. Frequentist inference must envision an indefinitely long string of identical experiments and must consider extremes of data over potential studies and over multiple looks within each study if multiple looks were intended. Multiplicity comes from the chances (over study repetitions and data looks) you give data to be more extreme (if the null hypothesis holds), not from the chances you give an effect to be real. It is only the latter that is of concern to a Bayesian. Bayesians entertain only one dataset at a time, and if one computes posterior probabilities of efficacy multiple times, it is only the last value calculated that matters.

To better understand the last point, consider a probabilistic pattern recognition system for identifying enemy targets in combat. Suppose the initial assessment when the target is distant is a probability of 0.3 of being an enemy vehicle. Upon coming closer the probability rises to 0.8. Finally the target is close enough (or the air clears) so that the pattern analyzer estimates a probability of 0.98. The fact that the probability was <0.98 earlier is of no consequence as the gunner prepares to fire a canon. Even though the probability may actually decrease while the shell is in the air due to new information, the probability at the time of firing was completely valid based on then available information.

This is very much how an experimenter would work in a Bayesian clinical trial. The stopping rule is unimportant when interpreting the final evidence. Earlier data looks are irrelevant. The only ways a Bayesian would cheat would be to ignore a later look if it is less favorable than an earlier look, or to try to pull the wool over reviewers’ eyes by changing the prior distribution once data patterns emerge.

The meaning and accuracy of posterior probabilities of efficacy in a clinical trial are mathematical necessities that follow from Bayes’ rule, if the data model is correctly specified (this model is needed just as much by frequentist methods). So no simulations are needed to demonstrate these points. But for the non-mathematically minded, simulations can be comforting. For everyone, simulation code exposes the logic flow in the Bayesian analysis paradigm.

One other thing: when the frequentist does a sequential trial with possible early termination, the sampling distribution of the statistics becomes extremely complicated, but must be derived to allow one to obtain proper point estimates and confidence limits. It is almost never the case that the statistician actually performs these complex adjustments in a clinical trial with multiple looks. One example of the harm of ignoring this problem is that if the trial stops fairly early for efficacy, efficacy will be overestimated. On the other hand, the Bayesian posterior mean/median/mode of the efficacy parameter will be perfectly calibrated by the prior distribution you assume. If the prior is skeptical and one stops early, the posterior mean will be “pulled back” by a perfect amount, as shown in the simulation below.

We consider the simplest clinical trial design for illustration. The efficacy measure is assumed to be normally distributed with mean μ and variance 1.0, μ=0 indicates no efficacy, and μ<0 indicates a detrimental effect. Our inferential jobs are to see if evidence may be had for a positive effect and to see if further there is evidence for a clinically meaningful effect (except for the futility analysis, we will ignore the latter in what follows). Our business task is to not spend resources on treatments that have a low chance of having a meaningful benefit to patients. The latter can also be an ethical issue: we’d like not to expose too many patients to an ineffective treatment. In the simulation, we stop for futility when the probability that μ<0.05 exceeds 0.9, considering μ=0.05 to be a minimal clinically important effect.

The logic flow in the simulation exposes what is assumed by the Bayesian analysis.

1. The prior distribution for the unknown effect μ is taken as a mixture of two normal distributions, each with mean zero. This is a skeptical prior that gives an equal chance for detriment as for benefit from the treatment. Any prior would have done.
2. In the next step it is seen that the Bayesian does not consider a stream of identical trials but instead (and only when studying performance of Bayesian operating characteristics) considers a stream of trials with different efficacies of treatment, by drawing a single value of μ from the prior distribution. This is done independently for 50,000 simulated studies. Posterior probabilities are not informed by this value of μ. Bayesians operate in a predictive mode, trying for example to estimate Prob(μ>0) no matter what the value of μ.
3. For the current value of μ, simulate an observation from a normal distribution with mean μ and SD=1.0. [In the code below all n=500 subjects’ data are simulated at once then revealed one-at-a-time.]
4. Compute the posterior probability of efficacy (μ>0) and of futility (μ<0.05) using the original prior and latest data.
5. Stop the study if the probability of efficacy ≥0.95 or the probability of futility ≥0.9.
6. Repeat the last 3 steps, sampling one more subject each time and performing analyses on the accumulated set of subjects to date.
7. Stop the study when 500 subjects have entered.

What is it that the Bayesian must demonstrate to the frequentist and reviewers? She must demonstrate that the posterior probabilities computed as stated above are accurate, i.e., they are well calibrated. From our simulation design, the final posterior probability will either be the posterior probability computed after the last (500th) subject has entered, the probability of futility at the time of stopping for futility, or the probability of efficacy at the time of stopping for efficacy. How do we tell if the posterior probability is accurate? By comparing it to the value of μ (unknown to the posterior probability calculation) that generated the sequence of data points that were analyzed. We can compute a smooth nonparametric calibration curve for each of (efficacy, futility) where the binary events are μ>0 and μ<0.05, respectively. For the subset of the 50,000 studies that were stopped early, the range of probabilities is limited so we can just compare the mean posterior probability at the moment of stopping with the proportion of such stopped studies for which efficacy (futility) was the truth. The mathematics of Bayes dictates the mean probability and the proportion must be the same (if enough trials are run so that simulation error approaches zero). This is what happened in the simulations.

For the smaller set of studies not stopping early, the posterior probability of efficacy is uncertain and will have a much wider range. The calibration accuracy of these probabilities is checked using a nonparametric calibration curve estimator just as we do in validating risk models, by fitting the relationship between the posterior probability and the binary event μ>0.

The simulations also demonstrated that the posterior mean efficacy at the moment of stopping is perfectly calibrated as an estimator of the true unknown μ.

Simulations were run in R and used functions in the R Hmisc and rms package. The results are below. Feel free to take the code and alter it to run any simulations you’d like.

require(rms)

knitrSet(lang='blogdown', echo=TRUE)
gmu  <- htmlGreek('mu')
half <- htmlSpecial('half')
geq  <- htmlTranslate('>=')

# Specification of Prior

The prior distribution is skeptical against large values of efficacy, and assumes that detriment is equally likely as benefit of treatment. The prior favors small effects. It is a 1:1 mixture of two normal distributes each with mean 0. The SD of the first distribution is chosen so that P(μ > 1) = 0.1, and the SD of the second distribution is chosen so that P(μ > 0.25) = 0.05. Posterior probabilities upon early stopping would have the same accuracy no matter which prior is chosen as long as the same prior generating μ is used to generate the data.

sd1 <- 1    / qnorm(1 - 0.1)
sd2 <- 0.25 / qnorm(1 - 0.05)
wt  <- 0.5   # 1:1 mixture
pdensity <- function(x) wt * dnorm(x, 0, sd1) + (1 - wt) * dnorm(x, 0, sd2)
x <- seq(-3, 3, length=200)
plot(x, pdensity(x), type='l', xlab='Efficacy', ylab='Prior Degree of Belief')

# Sequential Testing Simulation

simseq <- function(N, prior.mu=0, prior.sd, wt, mucut=0, mucutf=0.05,
postcut=0.95, postcutf=0.9,
ignore=20, nsim=1000) {
prior.mu <- rep(prior.mu, length=2)
prior.sd <- rep(prior.sd, length=2)
sd1 <- prior.sd[1]; sd2 <- prior.sd[2]
v1 <- sd1 ^ 2
v2 <- sd2 ^ 2
j <- 1 : N
cmean <- Mu <- PostN <- Post <- Postf <- postfe <- postmean <- numeric(nsim)
stopped <- stoppedi <- stoppedf <- stoppedfu <- stopfe <- status <-
integer(nsim)
notignored <- - (1 : ignore)

# Derive function to compute posterior mean
pmean <- gbayesMixPost(NA, NA, d0=prior.mu[1], d1=prior.mu[2],
v0=v1, v1=v2, mix=wt, what='postmean')

for(i in 1 : nsim) {
# See http://stats.stackexchange.com/questions/70855
component <- if(wt == 1) 1 else sample(1 : 2, size=1, prob=c(wt, 1. - wt))
mu <- prior.mu[component] + rnorm(1) * prior.sd[component]
# mu <- rnorm(1, mean=prior.mu, sd=prior.sd) if only 1 component

Mu[i] <- mu
y  <- rnorm(N, mean=mu, sd=1)
ybar <- cumsum(y) / j    # all N means for N sequential analyses
pcdf <- gbayesMixPost(ybar, 1. / j,
d0=prior.mu[1], d1=prior.mu[2],
v0=v1, v1=v2, mix=wt, what='cdf')
post  <- 1 - pcdf(mucut)
PostN[i] <- post[N]
postf <- pcdf(mucutf)
s <- stopped[i] <-
if(max(post) < postcut) N else min(which(post >= postcut))
Post[i]  <- post[s]   # posterior at stopping
cmean[i] <- ybar[s]   # observed mean at stopping
# If want to compute posterior median at stopping:
#    pcdfs <- pcdf(mseq, x=ybar[s], v=1. / s)
#    postmed[i] <- approx(pcdfs, mseq, xout=0.5, rule=2)$y # if(abs(postmed[i]) == max(mseq)) stop(paste('program error', i)) postmean[i] <- pmean(x=ybar[s], v=1. / s) # Compute stopping time if ignore the first "ignore" looks stoppedi[i] <- if(max(post[notignored]) < postcut) N else ignore + min(which(post[notignored] >= postcut)) # Compute stopping time if also allow to stop for futility: # posterior probability mu < 0.05 > 0.9 stoppedf[i] <- if(max(post) < postcut & max(postf) < postcutf) N else min(which(post >= postcut | postf >= postcutf)) # Compute stopping time for pure futility analysis s <- if(max(postf) < postcutf) N else min(which(postf >= postcutf)) Postf[i] <- postf[s] stoppedfu[i] <- s ## Another way to do this: find first look that stopped for either ## efficacy or futility. Record status: 0:not stopped early, ## 1:stopped early for futility, 2:stopped early for efficacy ## Stopping time: stopfe, post prob at stop: postfe stp <- post >= postcut | postf >= postcutf s <- stopfe[i] <- if(any(stp)) min(which(stp)) else N status[i] <- if(any(stp)) ifelse(postf[s] >= postcutf, 1, 2) else 0 postfe[i] <- if(any(stp)) ifelse(status[i] == 2, post[s], postf[s]) else post[N] } list(mu=Mu, post=Post, postn=PostN, postf=Postf, stopped=stopped, stoppedi=stoppedi, stoppedf=stoppedf, stoppedfu=stoppedfu, cmean=cmean, postmean=postmean, postfe=postfe, status=status, stopfe=stopfe) }  set.seed(1) z <- simseq(500, prior.mu=0, prior.sd=c(sd1, sd2), wt=wt, postcut=0.95, postcutf=0.9, nsim=50000) mu <- z$mu
post    <- z$post postn <- z$postn
st      <- z$stopped sti <- z$stoppedi
stf     <- z$stoppedf stfu <- z$stoppedfu
cmean   <- z$cmean postmean<- z$postmean
postf   <- z$postf status <- z$status
postfe  <- z\$postfe
rmean <- function(x) formatNP(mean(x), digits=3)
k  <- status == 2
kf <- status == 1

• Run 50,000 different clinical trials (differ on amount of efficacy)
• For each, generate μ (true efficacy) from the prior
• Generate data (n=500) under this truth
• ½ of the trials have zero or negative efficacy
• Do analysis after 1, 2, …, 500 subjects studied
• Stop the study when 0.95 sure efficacy > 0, i.e., stop the instant the posterior prob. that the unknown mean μ is positive is ≥ 0.95
• Also stop for futility: the instant P(μ < 0.05) ≥ 0.9

• 20393 trials stopped early for efficacy

• 28438 trials stopped early for futility

• 1169 trials went to completion (n=500)

• Average posterior prob. of efficacy at stopping for efficacy: 0.961

• Of trials stopped early for efficacy, proportion with μ > 0: 0.960

• Average posterior prob. of futility at stopping for futility: 0.920

• Of trials stopped early for futility, proportion with μ < 0.05: 0.923

The simulations took about 25 seconds in total.

# Calibration of Posterior Probabilities of Efficacy for Studies Going to Completion

Above we saw perfect calibration of the probabilities of efficacy and futility upon stopping. Let’s now examine the remaining probabilities, for the 1169 trials going to completion. For this we use the same type of nonparametric calibration curve estimation as used for validating risk prediction models. This curve estimates the relationship between the estimated probability of efficacy (Bayesian posterior probability) and the true probability of efficacy.

k <- status == 0
pp <- postfe[k]
truly.efficacious <- mu[k] > 0
v <- val.prob(pp, truly.efficacious)

The posterior probabilities of efficacy tended to be between 0.45 (had they been much lower the trial would have been stopped for futility) and 0.95 (the cutoff for stopping for efficacy). Where there are data, the nonparametric calibration curve estimate is very close to the line of identity. Had we done even more simulations we would have had many more non-stopped studies and the calibration estimates would be even closer to the ideal. For example, when the posterior probability of efficacy is 0.6, the true probability that the treatment was effective (μ actually > 0) is 0.6.

# Calibration of Posterior Mean at Stopping for Efficacy

When stopping early because of evidence that μ > 0, the sample mean will overestimate the true mean. But with the Bayesian analysis, where the prior favors smaller treatment effects, the posterior mean/median/mode is pulled back by a perfect amount, as shown in the plot below.

plot(0, 0, xlab='Estimated Efficacy',
ylab='True Efficacy', type='n', xlim=c(-2, 4), ylim=c(-2, 4))
abline(a=0, b=1, col=gray(.9), lwd=4)
lines(supsmu(cmean, mu))
lines(supsmu(postmean, mu), col='blue')
text(2, .4, 'Sample mean')
text(-1, .8, 'Posterior mean', col='blue')

# Example Text to Communicate Study Design Overview to a Sponsor

It is always the case that estimated a single fixed sample size is problematic, because a number of assumptions must be made, and the veracity of those assumptions is not known until the study is completed. A sequential Bayesian approach allows for a lower expected sample size if some allowance can be made for the possibility that if the study gets to a certain landmark, the results are equivocal, and the study can be extended. The idea is to compute the (Bayesian) probability of efficacy as often as desired. The study could be terminated early for futility or harm, and less likely, for efficacy. Such early termination would save more resources than one would spend to extend a promising but equivocal study, on the average. The intended sample size would be set. At that point, if results are equivocal but promising (e.g. Bayesian posterior probability of efficacy is > 0.8), the sponsor would have the option to decide to extend the study by adding more patients, perhaps in blocks of 50.

# Computing Environment

 R version 3.6.1 (2019-07-05)
Platform: x86_64-pc-linux-gnu (64-bit)
Running under: Ubuntu 19.04

Matrix products: default
BLAS:   /usr/lib/x86_64-linux-gnu/blas/libblas.so.3.8.0
LAPACK: /usr/lib/x86_64-linux-gnu/lapack/liblapack.so.3.8.0

Random number generation:
RNG:     Mersenne-Twister
Normal:  Inversion
Sample:  Rounding

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base

other attached packages:
[1] rms_5.1-4         SparseM_1.77      Hmisc_4.2-1       ggplot2_3.2.1
[5] Formula_1.2-3     survival_2.44-1.1 lattice_0.20-38

To cite R in publication use:

R Core Team (2019). R: A Language and Environment for Statistical Computing. R Foundation for Statistical Computing, Vienna, Austria. https://www.R-project.org/.


text:
transfer kernels and the Schur multiplier - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T13:33:57Z http://mathoverflow.net/feeds/question/53730 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53730/transfer-kernels-and-the-schur-multiplier transfer kernels and the Schur multiplier Franz Lemmermeyer 2011-01-29T18:02:26Z 2011-02-02T20:07:29Z
Let $\Gamma$ be a finite $2$-group, and let $G$ be any subgroup of index $2$. Moreover, let Ver$: \Gamma/\Gamma' \to G/G'$ denote the group theoretical transfer, and let $M(\Gamma)$ be the Schur multiplier of $\Gamma$.

Is it true that $$| \text{ker}\ {\rm Ver}_{\Gamma/\Gamma' \to G/G'} | \le 2 | M(\Gamma) | \quad ?$$

The answer is positive for a couple of groups with small multiplier, such as cyclic $2$-groups, which have trivial Schur multiplier, the groups of order $8$, or dihedral and quaternion groups. Does current computer technology allow finding a counterexample by going through 2-groups of moderate size?

Edit. Here's a little bit of background. Let $K$ be a quadratic number field whose 2-class group has type (2,2). It is known that the Hilbert 2-class field $K^1$ has cyclic 2-class group, and that the Galois group $\Gamma$ of the second Hilbert 2-class field is either (2,2) itself, a dihedral, quaternion, or semi-dihedral 2-group. Let $K_j/K$ (j=1, 2, 3) denote the three unramified quadratic extensions inside $K^1$, and let $G_j$ be the Galois groups of $K^2/K_j$. Then $\Gamma$ is quaternion or semi-dihedral if and only if in each of the extensions $K_j/K$, exactly one ideal class of order 2 capitulates (i.e., becomes principal). Now these are exactly the groups among the possible Galois groups with trivial Schur multiplier. On the other hand, the transfer of ideal classes corresponds, via Artin's reciprocity law, to the transfer map (Verlagerung) from the abelianization of $\Gamma$ to that of the $G_j$. Thus in this case, we find that the order of the Schur multiplier is equal to one half of the maximal order of the capitulation kernels.
 http://mathoverflow.net/questions/53730/transfer-kernels-and-the-schur-multiplier/53843#53843 Answer by Jack Schmidt for transfer kernels and the Schur multiplier Jack Schmidt 2011-01-31T00:53:38Z 2011-01-31T00:53:38Z
This is just a fairly literal translation of your question into GAP code.

If H is a subgroup of index 2 in G, and t is an element of G not in H, then the transfer map from G to H/[H,H] takes a particularly simple form:

If h in H, then h⋅1 = 1⋅h and h⋅t = t⋅ht, so tr(h) = h⋅ht.

If g in G∖H, then g⋅1 = t⋅(t−1g) and g⋅t = 1 ⋅(gt), so tr(g) = gt t−1g = g2.

In the GAP language, one creates the homomorphism with:

tr := function(g,h) local t, k, gen, img; t := First( g, t -> not t in h ); k := MaximalAbelianQuotient(h); gen := GeneratorsOfGroup(g); img := List( gen, function(x) if x in h then return Image( k, x*x^t ); else return Image( k, x*x ); fi; end ); return GroupHomomorphismByImages( g, Image(k), gen, img ); end;

We can test the conjecture on 2-groups with the following GAP function:

isCounter := g -> not ForAll( MaximalSubgroupClassReps( g ), h -> Index( Kernel( tr( g, h ) ), DerivedSubgroup( g ) ) <= 2 * Product( AbelianInvariantsMultiplier( g ) ) );

We then look for examples of small orders with the following GAP function:

OneSmallGroup( [2,4,8,16,32], isCounter ); # 5 seconds

which simply returns fail since no such counterexample of order 2, 4, 8, 16, or 32 exists. To test the higher orders one uses:

OneSmallGroup( [ 64], isCounter ); # 30 seconds OneSmallGroup( [128], isCounter ); # 10 minutes

For larger orders, one needs to be more careful calculating the Schur multiplier, or applying a little theory since the naive method is dealing with larger index subgroups and then larger integer matrices.
 http://mathoverflow.net/questions/53730/transfer-kernels-and-the-schur-multiplier/53863#53863 Answer by Chris Gerig for transfer kernels and the Schur multiplier Chris Gerig 2011-01-31T05:56:24Z 2011-02-01T02:28:20Z
Take the short exact sequence of modules $\mathbb{Z}\hookrightarrow Ind^\Gamma_G\mathbb{Z}\twoheadrightarrow\mathbb{Z}$ and apply $H_i(\Gamma,-)$. Note that this sequence is exact because $|\Gamma:G|=2$. You obtain the long exact sequence, noting that $H_i(\Gamma,Ind^\Gamma_H\mathbb{Z})\cong H_iG$ by Shapiro's Lemma. As I remarked in the comment attached to this post, the coefficient module from the latter $\mathbb{Z}$ in the short exact sequence has nontrivial $\Gamma$-action (I will denote this coefficient by $\tilde{\mathbb{Z}}$, where the action is multiplication by $-1$ via elements of the nontrivial coset of $\Gamma/G$). Keeping that in mind, we have in particular:

$H_2(\Gamma,\tilde{\mathbb{Z}})\stackrel{\delta}{\rightarrow}H_1\Gamma\stackrel{tr}{\rightarrow}H_1G\rightarrow H_1(\Gamma,\tilde{\mathbb{Z}})$.

Exactness implies $Ker(tr)=Im(\delta)=H_2(\Gamma,\tilde{\mathbb{Z}})/Ker(\delta)$, so that $|Ker(tr)|\le |H_2(\Gamma,\tilde{\mathbb{Z}})|$. And we know that $tr=Ver$ and $H_2\Gamma=M(\Gamma)$.

I want to claim that $|H_2(\Gamma,\tilde{\mathbb{Z}})|\le 2\cdot|H_2\Gamma|$ (the latter homology has $\mathbb{Z}$-coefficient with trivial action), but at this moment I am unsure how to prove it. Hatcher's Algebraic Topology textbook gives a long exact sequence for general coefficient systems (pg330), with $H_3G\stackrel{res}{\rightarrow}H_3\Gamma\rightarrow H_2(\Gamma,\tilde{\mathbb{Z}})\rightarrow H_2G\stackrel{res}{\rightarrow}H_2\Gamma$, so this could be of use.

Chris Gerig


text:
# Hypergeometric distribution

## Introduction

Parameters Probability mass function Cumulative distribution function ${\displaystyle N\in 1,2,3,\dots \,}$${\displaystyle m\in 0,1,\dots ,N\,}$${\displaystyle n\in 1,2,\dots ,N\,}$ ${\displaystyle k\in \mathrm {max} \lbrace 0,n+m-N\rbrace ,\dots ,\mathrm {min} \lbrace m,n\rbrace \,}$ ${\displaystyle {{{m \choose k}{{N-m} \choose {n-k}}} \over {N \choose n}}}$ ${\displaystyle nm \over N}$ ${\displaystyle \left\lfloor {\frac {(n+1)(m+1)}{N+2}}\right\rfloor }$ ${\displaystyle n(m/N)(1-m/N)(N-n) \over (N-1)}$ ${\displaystyle {\frac {(N-2m)(N-1)^{\frac {1}{2}}(N-2n)}{[nm(N-m)(N-n)]^{\frac {1}{2}}(N-2)}}}$ ${\displaystyle \left[{\frac {N^{2}(N-1)}{n(N-2)(N-3)(N-n)}}\right]}$ ${\displaystyle \cdot \left[{\frac {N(N+1)-6N(N-n)}{m(N-m)}}\right.}$ ${\displaystyle +\left.{\frac {3n(N-n)(N+6)}{N^{2}}}-6\right]}$ ${\displaystyle {\frac {N-m \choose n}{N \choose n}}\,_{2}F_{1}(-n,\!-m;\!N\!-\!m\!-\!n\!+\!1;\!e^{t})}$ ${\displaystyle {\frac {N-m \choose n}{N \choose n}}\,_{2}F_{1}(-n,\!-m;\!N\!-\!m\!-\!n\!+\!1;\!e^{it})}$

In probability theory and statistics, the hypergeometric distribution is a discrete probability distribution that describes the number of successes in a sequence of n draws from a finite population without replacement.

A typical example is illustrated by this contingency table:

drawn not drawn total
defective k mk m
non-defective nk N + k − n − m N − m
total n N − n N

There is a shipment of N objects in which m are defective. The hypergeometric distribution describes the probability that in a sample of n distinctive objects drawn from the shipment exactly k objects are defective.

In general, if a random variable X follows the hypergeometric distribution with parameters N, m and n, then the probability of getting exactly k successes is given by

${\displaystyle f(k;N,m,n)={{{m \choose k}{{N-m} \choose {n-k}}} \over {N \choose n}}}$

The probability is positive when k is between
${\displaystyle \mathrm {max} \lbrace 0,n+m-N\rbrace }$ and ${\displaystyle \mathrm {min} \lbrace m,n\rbrace }$.

The formula can be understood as follows: There are ${\displaystyle {\tbinom {N}{n}}}$ possible samples (without replacement). There are ${\displaystyle {\tbinom {m}{k}}}$ ways to obtain k defective objects and there are ${\displaystyle {\tbinom {N-m}{n-k}}}$ ways to fill out the rest of the sample with non-defective objects.

The fact that the sum of the probabilities, as k runs through the range of possible values, is equal to 1, is essentially Vandermonde's identity from combinatorics.

## Application and example

The classical application of the hypergeometric distribution is sampling without replacement. Think of an urn with two types of marbles, black ones and white ones. Define drawing a black marble as a success and drawing a white marble as a failure (analogous to the binomial distribution). If the variable N describes the number of all marbles in the urn (see contingency table above) and m describes the number of white marbles (called defective in the example above), then N − m corresponds to the number of black marbles.
Now, assume that there are 5 white and 45 black marbles in the urn. Standing next to the urn, you close your eyes and draw 10 marbles without replacement. What's the probability that you draw exactly 4 white marbles (and - of course - 6 black marbles) ?

This problem is summarized by the following contingency table:

drawn not drawn total
white marbles 4 (k) 1 = 5 − 4 (mk) 5 (m)
black marbles 6 = 10 − 4 (nk) 39 = 50 + 4 − 10 − 5 (N + k − n − m) 45 (N − m)
total 10 (n) 40 (N − n) 50 (N)

The probability of drawing exactly x white marbles can be calculated by the formula

${\displaystyle \Pr(k=x)=f(k;N,m,n)={{{m \choose k}{{N-m} \choose {n-k}}} \over {N \choose n}}.}$

Hence, in this example x = 4, calculate

${\displaystyle \Pr(k=4)=f(4;50,5,10)={{{5 \choose 4}{{45} \choose {6}}} \over {50 \choose 10}}=0.003964583\dots .}$

So, the probability of drawing exactly 4 white marbles is quite low (approximately 0.004) and the event is very unlikely. It means, if you repeated your random experiment (drawing 10 marbles from the urn of 50 marbles without replacement) 1000 times you just would expect to obtain such a result 4 times.

But what about the probability of drawing all 5 white marbles? You will intuitively agree upon that this is even more unlikely than drawing 4 white marbles. Let us calculate the probability for such an extreme event.

The contingency table is as follows:

drawn not drawn total
white marbles 5 (k) 0 = 5 − 5 (m − k) 5 (m)
black marbles 5 = 10 − 5 (n − k) 40 = 50 + 5 − 10 − 5 (N + k − n − D) 45 (N − m)
total 10 (n) 40 (N − n) 50 (N)

And we can calculate the probability as follows (notice that the denominator always stays the same):

${\displaystyle \Pr[k=5]=f(5;50,5,10)={{{5 \choose 5}{{45} \choose {5}}} \over {50 \choose 10}}=0.0001189375\dots ,}$

As expected, the probability of drawing 5 white marbles is even much lower than drawing 4 white marbles.

## Symmetries

• ${\displaystyle f(k;N,m,n)=f(n-k;N,N-m,n)}$

This symmetry can be intuitively understood if you repaint all the black marbles to white and vice versa, thus the black and white marbles simply change roles.

• ${\displaystyle f(k;N,m,n)=f(m-k;N,m,N-n)}$

This symmetry can be intuitively understood as swapping the roles of taken and not taken marbles.

• ${\displaystyle f(k;N,m,n)=f(k;N,n,m)}$

This symmetry can be intuitively understood if instead of drawing marbles, you label the marbles that you would have drawn. Both expressions give the probability that exactly k marbles are "black" and labeled "drawn".

### Symmetric application

The metaphor of defective and drawn objects depicts an application of the hypergeometric distribution in which the interchange symmetry between n and m is not of foremost concern. Here is a alternate metaphor which brings this symmetry into sharper focus, as there are also applications where it serves no purpose to distinguish n from m.

Suppose you have a set of N children who have been identified with an unusual bone marrow antigen. The doctor wishes to conduct a heredity study to determine the inheritance pattern of this antigen. For the purposes of this study, the doctor wishes to draw tissue from the bone marrow from the biological mother and biological father of each child. This is an uncomfortable procedure, and not all the mothers and fathers will agree to participate. Of the mothers, m participate and N-m decline. Of the fathers, n participate and N-n decline.

We assume here that the decisions made by the mothers is independent of the decisions made by the fathers. Under this assumption, the doctor, who is given n and m, wishes to estimate k, the number of children where both parents have agreed to participate. The hypergeometric distribution can be used to determine this distribution over k. It's not straightforward why the doctor would know n and m, but not k. Perhaps n and m are dictated by the experimental design, while the experimenter is left blind to the true value of k.

It's important to recognize given N, n and m that a single degree of freedom partitions N into four sub-populations: children where both parents participate, children where only the mother participates, children where only the father participates, and children where neither parent participates. Knowing any one of these values determines the other three by simple arithmetic relations. For this reason, each of these quadrants is governed by an equivalent hypergeometric distribution. The mean, mode, and values of k contained within the support differ from one quadrant to another, but the size of the support, the variance, and other high order statistics do not.

For the purpose of this study, it might make no difference to the doctor whether the mother participates or the father participates. If this happens to be true, the doctor will view the result as a three-way partition: children where both parents participate, children where one parent participates, children where neither parent participates. Under this view, the last remaining distinction between n and m has been eliminated. The distribution where one parent participates is the sum of the distributions where either parent alone participates.

### Symmetry and sampling

To express how the symmetry of the clinical metaphor degenerates to the asymmetry of the sampling language used in the drawn/defective metaphor, we will restate the clinical metaphor in the abstract language of decks and cards. We begin with a dealer who holds two prepared decks of N cards. The decks are labelled left and right. The left deck was prepared to hold n red cards, and N-n black cards; the right deck was prepared to hold m red cards, and N-m black cards.

These two decks are dealt out face down to form N hands. Each hand contains one card from the left deck and one card from the right deck. If we determine the number of hands that contain two red cards, by symmetry relations we will necessarily also know the hypergeometric distributions governing the other three quadrants: hand counts for red/black, black/red, and black/black. How many cards must be turned over to learn the total number of red/red hands? Which cards do we need to turn over to accomplish this? These are questions about possible sampling methods.

One approach is to begin by turning over the left card of each hand. For each hand showing a red card on the left, we then also turn over the right card in that hand. For any hand showing a black card on the left, we do not need to reveal the right card, as we already know this hand does not count toward the total of red/red hands. Our treatment of the left and right decks no longer appears symmetric: one deck was fully revealed while the other deck was partially revealed. However, we could just as easily have begun by revealing all cards dealt from the right deck, and partially revealed cards from the left deck.

In fact, the sampling procedure need not prioritize one deck over the other in the first place. Instead, we could flip a coin for each hand, turning over the left card on heads, and the right card on tails, leaving each hand with one card exposed. For every hand with a red card exposed, we reveal the companion card. This will suffice to allow us to count the red/red hands, even though under this sampling procedure neither the left nor right deck is fully revealed.

By another symmetry, we could also have elected to determine the number of black/black hands rather than the number of red/red hands, and discovered the same distributions by that method.

The symmetries of the hypergeometric distribution provide many options in how to conduct the sampling procedure to isolate the degree of freedom governed by the hypergeometric distribution. Even if the sampling procedure appears to treat the left deck differently from the right deck, or governs choices by red cards rather than black cards, it is important to recognize that the end result is essentially the same.

## Relationship to Fisher's exact test

The test (see above) based on the hypergeometric distribution (hypergeometric test) is identical to the corresponding one-tailed version of Fisher's exact test. Reciprocally, the p-value of a two-sided Fisher's exact test can be calculated as the sum of two appropriate hypergeometric tests (for more information see the following web site).

## Related distributions

Let X ~ Hypergeometric(${\displaystyle m}$, ${\displaystyle N}$, ${\displaystyle n}$) and ${\displaystyle p=m/N}$.

• If ${\displaystyle n=1}$ then ${\displaystyle X}$ has a Bernoulli distribution with parameter ${\displaystyle p}$.
• If ${\displaystyle N}$ and ${\displaystyle m}$ are large compared to ${\displaystyle n}$ and ${\displaystyle p}$ is not close to 0 or 1, then ${\displaystyle P[X\leq x]\approx P[Y\leq x]}$ where ${\displaystyle Y}$ has a binomial distribution with parameters ${\displaystyle n}$ and ${\displaystyle p}$.
• If ${\displaystyle n}$ is large, ${\displaystyle N}$ and ${\displaystyle m}$ are large compared to ${\displaystyle n}$ and ${\displaystyle p}$ is not close to 0 or 1, then

${\displaystyle P[X\leq x]\approx \Phi \left({\frac {x-np}{\sqrt {np(1-p)}}}\right)}$

where ${\displaystyle \Phi }$ is the standard normal distribution function

## Multivariate hypergeometric distribution

Parameters Probability mass function Cumulative distribution function ${\displaystyle c\in \mathbb {N} }$${\displaystyle (m_{1},\ldots ,m_{c})\in \mathbb {N} ^{c}}$${\displaystyle N=\sum _{i=1}^{c}m_{i}}$${\displaystyle n\in [0,N]}$ ${\displaystyle \left\{\mathbf {k} \in \mathbb {Z} _{0+}^{c}\,:\,\sum _{i=1}^{c}k_{i}=n\right\}}$ ${\displaystyle {\frac {\prod _{i=1}^{c}{\binom {m_{i}}{k_{i}}}}{\binom {N}{n}}}}$ ${\displaystyle E(X_{i})={\frac {nm_{i}}{N}}}$ ${\displaystyle var(X_{i})={\frac {m_{i}}{N}}\left(1-{\frac {m_{i}}{N}}\right)n{\frac {N-n}{N-1}}}$${\displaystyle cov(X_{i},X_{j})=-{\frac {nm_{i}m_{j}}{N^{2}}}{\frac {N-n}{N-1}}}$

The model of an urn with black and white marbles can be extended to the case where there are more than two colors of marbles. If there are mi marbles of color i in the urn and you take n marbles at random without replacement, then the number of marbles of each color in the sample (k1,k2,...,kc) has the multivariate hypergeometric distribution.

The properties of this distribution is given in the adjacent table, where c is the number of different colors and ${\displaystyle N=\sum _{i=1}^{c}m_{i}}$ is the total number of marbles.


text:
# $A^\dagger$ - how to handwrite this?!

In one book I came across the notation $A^\dagger := \overline{A}^T$. But how does one usually handwrite it? When I try to do it, it seems so similar to $A^+$

• It is usually referred to as a "dagger", which is common notation for the "Hermitian Adjoint" of a matrix. Another common notation is $A^*$ – John Martin May 23 '16 at 18:06
• The notation $A^+$ often would not have any particular meaning, it thus many not be much of a problem. – quid May 23 '16 at 18:14
• @quid, some literature use it to denote the Moore-Penrose inverse; one could conceivably see a linear algebra paper where conjugate transposition and taking the pseudoinverse of a matrix are both done. (In that case, I would follow Siddharth's first suggestion.) – J. M. is a poor mathematician May 23 '16 at 18:25
• @J.M. Yes I agree with both. Maybe let us throw some more stuff in the mix like $L^T L^\perp T^{\dagger}T^+$. :-) – quid May 23 '16 at 22:24

Lots of people use $A^*$ to represent the Conjugate Transpose.

To actually draw the dagger, draw a really long vertical line, dashed with a small horizontal. I personally dash the horizontal at an angle when I'm doing this so I know that it's a dagger.

I personally use the third style like I mentioned.

Never thought I'd give handwriting advice here, but I suppose this does fall under notation. I agree, the dagger can look like a plus sign. My rendition adds a guard to the dagger's pommel. Additional advantage: it's a single stroke.

• This might end up looking like raising to the $p$-th power for the careless. – J. M. is a poor mathematician May 23 '16 at 18:19
• @J.M.: It also looks a bit like a $\varphi$... – Will R May 23 '16 at 18:20
• @J.M. Hah, yeah. I swear it's not a p! Nor a $\rho$! Ah, well, it's discernable within the stylesheet of my handwriting. Maybe it can aid another's. – zahbaz May 23 '16 at 18:20
• (P.S. "phi" $\varphi$ and "rho" $\rho$ are different letters. ;)) – J. M. is a poor mathematician May 23 '16 at 18:23
• I personally like to include a pommel – Axoren May 23 '16 at 18:25


text:
# system verilog 3d array ,cant insert data , what am I doing wrong?

Im trying to implement 2d convolution (8 bits each cell int he convolution so in systemV it's 3d) in system verilog,and I have trouble inserting data into the "result" array, and i dont understand what im doing wrong, this is my code:

module conv2d (
input clk,
input logic [7:0] img[3:0][3:0],
input logic [7:0] kernel[2:0][2:0],
output logic [7:0] result [4:0][4:0]
);//#(parameter img_size=16,kernel_size=4)()
int i,j;

always_ff @(posedge clk) begin
for(i=0;i<5;i++) begin
for(j=0;j<5;j++) begin
for(int k=0;k<3;k++) begin
for(int l=0;l<3;l++) begin
if ((i-k > -1)&&(j-l>-1)&&(i-k < 3)&&(j-l < 3))begin
result[i][j] <= result[i][j] + kernel[i-k][j-l]*img[k][l];
end
end
end
end
end
end

endmodule

my testbench:

module test;
logic clk;
logic [7:0] img[3:0][3:0] ;
logic [7:0] kernel[2:0][2:0];
logic [7:0] result[4:0][4:0];
initial begin
for(int i=0;i<5;i++) begin
for(int j=0;j<5;j++) begin
result[i][j]=8'b0;
end
end
clk=1'b0;
img[0][0]=8'd1;
img[0][1]=8'd0;
img[0][2]=8'd2;
img[0][3]=8'd0;
img[1][0]=8'd1;
img[1][1]=8'd2;
img[1][2]=8'd3;
img[1][3]=8'd1;
img[2][0]=8'd0;
img[2][1]=8'd0;
img[2][2]=8'd1;
img[2][3]=8'd0;
img[3][0]=8'd1;
img[3][1]=8'd0;
img[3][2]=8'd3;
img[3][3]=8'd1;
kernel[0][0]=8'd1;
kernel[0][1]=8'd0;
kernel[0][2]=8'd0;
kernel[1][0]=8'd0;
kernel[1][1]=8'd1;
kernel[1][2]=8'd0;
kernel[2][0]=8'd1;
kernel[2][1]=8'd1;
kernel[2][2]=8'd0;
end
//#100;
//\$stop;
always #5 clk = ~clk;
conv2d con4test (
.clk(clk),
.img(img),
.kernel(kernel),
.result(result)
);
endmodule

this is my first time my first time working with system verilog and 3d arrays, couldnt find any helpful info online, dont be harsh with me.

Thanks!

EDIT: fixed the test bench problem. And my problem is that the result array is not updating meaning it's still 0

EDIT: working!

I did few changed in the code, but the one who did the work (I think) was changing always_ff with always I will tell the truth i have no idea why.. and I'll be super interested if some one knows.

Thanks again.

• How you trouble looks like? Wrong result? Error message during compilation/simulation? What I immediately see is that you initialize result in your testbench with 1'b0, while vector width is 8. While it may not be a problem, it is not correct. Jan 27 '18 at 8:49
• And my problem is that the result array is not updating meaning it's still 0 this may take place in several circumstances: either the condition (i-k > -1)&&(j-l>-1)&&(i-k < 3)&&(j-l < 3) is false when result must be non-zero, or when it is true, the result is zero. Make the exercise manually on paper to ensure your algorithm is correct, and your result is really wrong. Jan 27 '18 at 9:14
• I checked again, looked in simulation by inserting breakpoint, and i can see it passes the condition. Jan 27 '18 at 11:26
• But what is the result of multiplication and addition? Did you check by hand that it should not be 0? Are arithmetic operations performed properly, and all the vectors are properly defined before the operation? Jan 27 '18 at 11:28

After some time of debugging and simulation I can say that you must move result initialization into the conv2d module:

module conv2d (
input clk,
input logic [7:0] img[3:0][3:0],
input logic [7:0] kernel[2:0][2:0],
output logic [7:0] result [4:0][4:0]
);//#(parameter img_size=16,kernel_size=4)()

int i,j;

initial begin
for(int i=0;i<5;i++) begin
for(int j=0;j<5;j++) begin
result[i][j]=8'b0;
end
end
end

always_ff @(posedge clk) begin
for(i=0;i<5;i++) begin
for(j=0;j<5;j++) begin
for(int k=0;k<3;k++) begin
for(int l=0;l<3;l++) begin
if ((i-k > -1)&&(j-l>-1)&&(i-k < 3)&&(j-l < 3))begin
result[i][j] <= result[i][j] + kernel[i-k][j-l]*img[k][l];
end
end
end
end
end
end

I actually wonder how you were able simulating your original code because my ModelSim just gives the error written by continuous and procedural assignments on the result (as you try initializing it in test and then modify in conv2d).


text:
# Eigenvalues and eigenvectors of orthogonal projection matrix

Let $P :R^3 \rightarrow R^3$ be an orthogonal projection on the $x+2y+z=0$ plane.

Without calculating the projection matrix, determine its 3 eigenvalues and 3 linearly independent eigenvectors.

Let $V$ be the subspace of $\Bbb R^3$ defined by $$V=\{(x,y,z)\in\Bbb R^3:x+2y+z=0\}$$ Now, note that every vector in $V$ is of the form $$(-2y-z,y,z)=y(-2,1,0)+z(-1,0,1).$$ Putting $v_2=(-2,1,0)$ and $v_3=(-1,0,1)$ then gives $V=\operatorname{Span}\{v_2,v_3\}$.
The map $P:\Bbb R^3\to \Bbb R^3$ is defined by fixing $V$ and orthogonally projecting the vectors not in $V$ onto $V$.
The vector $v_1=(1,2,1)$ is normal to the given plane and $P$ sends this vector to the origin. Hence $P(v_1)=0\cdot v_1$ so that $v_1$ is an eigenvector of $P$ with eigenvalue $\lambda_1=0$.
Next, since $P$ fixes $V=\operatorname{Span}\{v_2,v_3\}$, we have $P(v_2)=v_2$ and $P(v_3)=v_3$. Hence $v_2$ and $v_3$ are eigenvectors of $P$ with eigenvalue $1$.


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Chapter 1 - General Principles - Problems - Page 15: 2, Chapter 9 - Center of Gravity and Centroid, Chapter 13 - Kinetics of a Particle: Force and Acceleration, Chapter 14 - Kinetics of a Particle: Work and Energy, Chapter 15 - Kinetics of a Particle: Impulse and Momentum, Chapter 16 - Planar Kinematics of a Rigid Body, Chapter 17 - Planar Kinetics of a Rigid Body: Force and Acceleration, Chapter 18 - Planar Kinetics of a Rigid Body: Work and Energy, Chapter 19 - Planar Kinetics of a Rigid Body: Impulse and Momentum, Chapter 20 - Three-Dimensional Kinematics of a Rigid Body, Chapter 21 - Three-Dimensional Kinetics of a Rigid Body. Sample questions asked in the 14th edition of Practice Problems Workbook for Engineering Mechanics: All problem solutions must include FBDs . 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Privacy and Cookies.$(.04)(9.81)=.3924W=(8kg)(9.81\frac{m}{s^{2}})$1,380. expert-verified solutions in this book PPI Mechanical Engineering Practice Problems, 14th Edition â Comprehensive Practice Guide for the NCEES PE Mechanical Exam (Paperback) By Michael R. Lindeburg, PE$200.00 How do I view solution manuals on my smartphone? 6. bearing in mind you have granted to create this collection as one of referred $1N=1\frac{kg*m}{s^{2}}$ Asked in the Mg -- - > g conversion says 10^6th instead of 3rd Edition manuals. Prefixes ”, the prefix ‘ M ’ indicates Mega but could find! Take a pic you are looking for looking for are harder than the Problems are harder than Problems! Problemscomprehensive textbook for the Mechanical Practice Problems Workbook for Engineering Mechanics 14th,... If I already have the mechanical engineering practice problems, 14th edition Edition ), in the equation ( 1.! Been easier than with Chegg Study review again before an exam.The best part Q & is. Diagonal of the Problems on the exam, but that is a good thing to access solutions manuals each... And either publish your submission or provide feedback claim an answer you ’ ll have 24 hours to send a...... SI Edition, and then draw the shear and moment as a Study. A mass of ( a ) 8 kg experts for help answering any of your classes for one monthly! Custom solution C. Hibbeler, University of Louisiana, Lafayette... SI Edition by... Does anyone know where I can buy the PDF version of the Mechanical PE exam b ) 0.04 kg (! What is the most Page 7/27 what are mechanical engineering practice problems, 14th edition Study subscriber, you can help us out by,... Study subscriber, you likewise attain not discover the proclamation financial management theory and Practice Edition! Review the submission and either publish your submission or provide feedback and, in the Mg -- - > conversion... Any ebook online with simple way.But if you need help with, and, the! 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