#StackBounty: #normal-distribution #convergence #metric #wasserstein Prove the existence of a fixed point of a certain mapping of distr…

Bounty: 50

Let $tilde{X}_0$ be some random variable on $mathbb{R}^n$, with a strictly positive p.d.f..

Define:
$$X_0:=(operatorname{var}{tilde{X}_0})^{-frac{1}{2}}(tilde{X}_0-mathbb{E}tilde{X}_0),$$
where we take the unique positive definite matrix square root.

Further, for all $kinmathbb{N}$ define:
$$tilde{X}{k+1}:=Phi^{-1}_n(F{X_k}(X_k)),$$
and:
$$X_{k+1}:=(operatorname{var}{tilde{X}k})^{-frac{1}{2}}tilde{X}_k,$$
where for all $zinmathbb{R}^n$:
$$Phi_n(z)=[Phi(z_1),dots,Phi(z_n)],$$ and:$$F
{X_k}(z)=[F_{X_{k,1}}(z_1),dots,F_{X_{k,n}}(z_n)],$$
where $Phi$ is the CDF of a standard Normal distribution, and for $iin{1,dots,n}$, $F_{X_{k,i}}$ is the CDF of the $i^mathrm{th}$ component of $F_{X_k}$.

It is easy to see that for all $kinmathbb{N}$, $X_k$ is mean zero and has covariance given by the identity matrix, and that $tilde{X}_{k+1}$ has standard Normal marginals.

I would like to prove that there is some $X$ such that $X_k$ converges in distribution to $X$ as $krightarrowinfty$, and (ideally) such that $tilde{X}_k$ also converges in distribution $X$ as $krightarrowinfty$.

The Banach fixed point theorem cannot be applicable as the mapping has more than one fixed point. E.g. with $n=2$, both the bivariate standard Normal distribution and the distribution with p.d.f. $(x,y)mapsto frac{1}{2}(1-Phi(max(|x|,|y|)))$ are fixed points.

Can you prove the convergence to a fixed point?


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#StackBounty: #normal-distribution #parameterization #flexible-model Generalized Normal Distribution

Bounty: 50

Is there a known distribution, $f(x|theta_1,theta_2,theta_3,theta_4)$, with the following properties:

  1. $E(X^n)=theta_n$ for $n in {1, 2, 3, 4}$.
  2. If $theta_3=0$ and $theta_4=3theta_2^2$, then $f$ is the normal density with mean $theta_1$ and variance $theta_2-theta_1^2$.
  3. $f$ is a continuous function in each parameter.
  4. (Any other regularity conditions that preclude contrived solutions.)

Essentially, I’m looking for something akin to Johnson’s SU distribution but with parameters that are more interpretable.

EDIT:
I consider the following to be an example of a contrived solution.

If $theta_3=0$ and $theta_4=3theta_2^2$, then
$$f(x|theta_1,theta_2,theta_3,theta_4)=dfrac 1 {sqrt{2pi}}exp(dfrac 1 2 dfrac {(x-theta_1)^2} {theta_2})$$
otherwise
$$f(x|theta_1,theta_2,theta_3,theta_4)=p(x|x_1,x_2,x_3,x_4)$$
where $p$ is the unique PMF with support over ${x_1,x_2,x_3,x_4}$ which satisfies Property 1 (provided a solution exists for the given $theta’s$). Obviously, this solution does not provide any modeling utility and it also doesn’t satisfy Property 3. It’s possible that other contrived solutions may exist which do satisfy Property 3. However, I’m only interested in ones which have utility for modeling – hence Property 4.


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#StackBounty: #hypothesis-testing #statistical-significance #normal-distribution #optimization #central-limit-theorem Picking a signifi…

Bounty: 50

Suppose you are running a casino and that you are responsible for ensuring that all the dice are fair to avoid lawsuits. In order to do this, you take a mean of 1000 throws of each die and perform a hypothesis test [using the central limit theorem, CLT] to see whether they are likely biased.

The average cost of a lawsuit is £240000, whilst the cost of a die is £3, so in order to minimise costs you would aim to have $240000P(textrm{Type II Error}) = 3alpha$ where $alpha$ is the significance level of the hypothesis test (and also the probability of a type I error). The cost of testing the die may be ignored.

Now, in order to find the optimal $alpha$ value, one must know the value of $P(textrm{Type II Error})$, something that can only be calculated if the actual mean of the die (which is what we are testing for in the first place) is known, so the optimal solution cannot be found. That being said, however, I’m sure scenarios like this arise rather often, so how are they usually dealt with?

tldr: How would you find a threshold value for the mean of a die above (or below) which it should be considered biased whilst also keeping $240000P(textrm{Type II Error}) approx 3P(textrm{Type I Error})$

Edit: It seems my choice of example is rather poor, as a die shouldn’t even be tested for fairness with a test like this. That being said, however, my question really concerns the tradeoff between Type I and Type II error, not the die in particular.


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#StackBounty: #normal-distribution #clustering #gaussian-mixture Finding a Dominant Cluster

Bounty: 50

Super-basic question here:

I’m looking for a way to find the dominant cluster of a set of clusters (as in the first image):

enter image description here

This is not what I get when I run a Gaussian Mixture model with one component (it tries to cover everything):
enter image description here

I’m sure there’s a standard approach for doing this, I just don’t know what it’s called.


The approach I’m thinking of is to maximize the sum of likelihoods of all points under a normal distribution:

If $x in mathcal R^{Ntimes D}$ is my dataset

$mathcal L = sum_n det(Sigma)^{-1/2} expleft(-frac12 (x_n-mu)^T Sigma^{-1} (x_n-mu)right)$

and then find equations for $mu$ and $Sigma$ when $frac{partial mathcal L}{partial mu}=0$ and $frac{partial mathcal L}{partial Sigma}=0$, and solving with fixed-point iteration. What that’s led to so far, unless there’s an error in my implementation (possible), is that the cluster moves to the correct mean but then collapses over iterations towards zero variance. This I suppose makes sense, because under this formulation the maximum likelihood is obtained by having a zero-variance gaussian on one point.

Is there a name for this type of problem, and if so what is the common approach?


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#StackBounty: #normal-distribution #inference Estimating the distribution from percentile score, top 50 and curve

Bounty: 50

I’m trying to estimate how many people participated in a game competition and what my percentile score is based on some public data provided by the host.

The data I have is:

  • Top 50 leaderboard
  • 5th, 20th, 40th and 70th percentile score
  • A (not that clear) curveThe curve

(1) Assuming that the final score distribution is a normal distribution, is it possible to estimate the whole distribution?

  • Giving a score, is it possible to estimate the percentile score?
  • Is it possible to estimate the total entries?

(2) This curve doesn’t look like a trivial normal distribution. What’s the best guess of this distribution? Can it be two distributions added up? If this is the case, is it possible to separate them just using these data we have?


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#StackBounty: #machine-learning #neural-networks #normal-distribution #optimization Step-by-step construction of an RBF neural network

Bounty: 50

I would like to solve the following task as an exercise: Given is the data in the image below. Each output $Y_j$ of the network is defined as:

$()$ $Y_j = sum_{i}^N w_{ij} exp(-frac{||x-mu_i||^2}{2sigma_i^2})$

for the -ith neuron. The task is to draw an RBF network that perfectly classifies the data, with suitable means, covariances and weights. In a second step a point has to be taken and classified in the worked-out model.

My ideas We have two nodes in the input layer, one for each dimension. The hidden-layer has as many neurons as there are training samples. Each of these calculates the activation given by the exponential above. The output layer has three output nodes as there are three classes. I would determine $Sigma$ and $mu$ as follows: For each class and training-sample $x_i$ of this class, $mu_j$ is just the centroid of the training samples, $sigma = frac{1}{m}sum_i ||x_i-mu||$, and $Sigma = sigma*I_d$.

Questions: That does not seem to be correct however – if I have the same $Sigma$ and $mu$ for all the hidden nodes of the same class, each new test-input would result in the same activation for all these nodes..so what exactly are the particular $u_i$ and $sigma_i$? Also, for perfect classification, I would set all the weights which belong to the correct class to 1, and all the others to zero – would that make sense?

enter image description here


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#StackBounty: #distributions #normal-distribution #central-limit-theorem #sum Square roots of sums absolute values of i.i.d. random var…

Bounty: 50

In an earlier question, I asked about the limiting distribution of the square root of the absolute value of the sum of $n$ i.i.d. random variables each with finite non-zero mean $mu$ and variance $sigma^2$. The answer was (after a suitable location and scale adjustment) a standard normal distribution.

This time I am asking about the case where $mu=0$. I know and can demonstrate the answer, but I am interested in whether the resulting distribution has any other uses.

So suppose $X_1, X_2, ldots, X_n$ are i.i.d. random variables with zero mean and positive variance $sigma^2$. If $displaystyle Z=sqrt{left|sum_{i=1}^n X_iright|}$ then $displaystyle dfrac{Z}{sqrt[4]{n} , sqrt{sigma}} xrightarrow{d} W$ as $n$ increases, where $W$ has the distribution of the square root of a standard half-normal random variable.

  • The density of $W$ is $f(x)=sqrt{dfrac{8}{pi}},x,e^{-x^4/2}$ for $x ge 0$
  • The mean of $W$ is $dfrac{sqrt[4]{2};Gammaleft(frac34right)}{sqrt{pi}} approx 0.822179$
  • The variance of $W$ is $sqrt{dfrac2pi}{left(1- dfrac{Gammaleft(frac34right)^2}{sqrt{pi}}right)} approx 0.1219063$
  • The standard deviation of $W$ is then about $0.3491509$
  • So the mean and standard deviation of $Z$ could be approximated by those of $W$ scaled up by a factor of $sqrt[4]{n} , sqrt{sigma}$

The density of $W$ looks like the following chart. It has a very light tail, with $mathbb{P}(W>2) < 0.00007$

density square root of a standard half-normal random variable

Has anybody come across this distribution before?


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#StackBounty: #probability #normal-distribution #z-score #odds Calculating Odds of Getting a Sample w/ a Specific Standard Deviation

Bounty: 50

Trying to calculate the odds on something and was getting myself confused. I’ll try to summarize into a simple problem with made up numbers.

Say a cannon fires projectiles with a population mean of 100 m/s and a standard deviation of 10 m/s, represented by a normal distribution.

I wanted to calculate the odds of firing off 15 rounds in a row that would have a standard deviation between 0 m/s and 2 m/s.

I basically calculated two z-scores:

Z1 = (101-100)/10 and Z2 = (99-100)/10.

Then assumed the probability of getting one round within that range was (using table for standardized z-scores):

P = P(X < Z1) – P(X < Z2)

To fire 15 rounds within that range, then I said P_15 = P^15.

Although, I feel more like I am calculating the odds of my sample to have more more like 3+ sigma (of 2 m/s), since with 1-sigma all the rounds from the sample don’t necessarily have to fall within the +/- 1 m/s range, just ~68% of them. But, I really would like the sample to have a 1-sigma between 0m/s and 2 m/s.

Question: what is the correct way to formulate this problem and what are the details of the calculation?

Thanks.


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#StackBounty: #normal-distribution #chi-squared #random-variable #bounds #quadratic-form Tails of products of random variables

Bounty: 100

Let $X$ be a non-negative random variable, and let $Y sim chi^2_n / n$ (so that $E(Y) = 1$). $X$ and $Y$ are independent. Note that $X$ and $Xcdot Y$ have the same mean, while $Xcdot Y$ has larger variance. Is it true that for every $t > E(X)$,
$$
Pr(X > t) < Pr(Xcdot Y > t),?
$$

My special case is $X = z^TKz/z^Tz$, when $zsim mathcal{N}(0,I)$ and $K$ is a symmetric positive definite matrix, although I suspect this hold generally.


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#StackBounty: #r #normal-distribution #lme4-nlme What to do when my data lack normality while using linear mixed models?

Bounty: 50

I have a question regarding the criteria I should follow. Briefly, when analyzing data using a linear mixed model, my data does not seem to comply with the Normality assumption.
Then, my question is, what do I do? Is there a non-parametric test I can use? (i.e. that doesn’t require Normal distribution of the errors)? In that case, which should I use? If not, what should I do?

Here I state briefly my script, data base and useful information.

  • Response variable: Vueltasmin, a numeric variable
  • Explanatory variables: Ojoviendisuestimulo and Velocidad, both factors with 2 levels each.
  • As I take several measures from each subject (Bicho) I added a random effect
  • I started with the model m1 which contains all my data points, then ended using m1bis model which excluded some data points whose Pearson residual is larger than 0.5.
m1 <- lmer(Vueltasmin~Ojoviendosuestimulo*Velocidad + (1|Bicho), data = Datos)
e1 <- resid(m1) 
pre1 <- predict(m1)  
plot(pre1, e1, xlab="Predichos", ylab="Residuos de pearson",
   main=" RE vs PRED",cex.main=.8 ) 
abline(0,0)
qqnorm(e1, cex.main=.9)   #QQ plot
qqline(e1)
shapiro.test(e1)   
(which(abs(e1)>0.5))

Datosm1bis <- Datos[-c(3, 4, 8, 11, 12, 15, 33, 36, 51, 62, 65, 
          66, 69, 70, 72, 90, 107, 108, 133, 144), 1:5]
m1bis <- lmer(Vueltasmin~Ojoviendosuestimulo*Velocidad + (1|Bicho), 
    data = Datosm1bis)
e1bis<-resid(m1bis) # residuos de pearson
pred1bis<-predict(m1bis) #predichos
plot(pred1bis, e1bis, xlab="Predichos", 
 ylab="Residuos de pearson",
main="Gráfico de dispersión de RE vs PRED",cex.main=.8 ) 
#Grafico RP vs predichos para ver ausencia de patrones
shapiro.test(e1bis)

A glimpse of my data base is as follows: (Please ignore the Ganancia column)

Bicho   Ojoviendosuestimulo Velocidad   Vueltasmin  Ganancia
92  Preferido   1   1   1
93  Preferido   1   0.916666667 0.916666667
94  Preferido   1   1.333333333 1.333333333
95  Preferido   1   0   0
96  Preferido   1   0.833333333 0.833333333
97  Preferido   1   0.833333333 0.833333333
98  Preferido   1   0.5 0.5
100 Preferido   1   0   0
101 Preferido   1   0.5 0.5
102 Preferido   1   0.333333333 0.333333333
103 Preferido   1   1.083333333 1.083333333
104 Preferido   1   0   0
105 Preferido   1   0.333333333 0.333333333
106 Preferido   1   0.333333333 0.333333333
108 Preferido   1   1.333333333 1.333333333
109 Preferido   1   0.833333333 0.833333333
110 Preferido   1   0.083333333 0.083333333
111 Preferido   1   1   1
92  Antipreferido   1   -0.083333333    -0.083333333
93  Antipreferido   1   -0.25   -0.25
94  Antipreferido   1   -0.083333333    -0.083333333
....... continues....


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