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The variance of the resulting Gaussian is found by taking the second derivative of the exponent?

I've read the paper recently and wondered the same thing. The Cramér–Rao bound (see this and this) states that the variance of an unbiased estimator cannot be lower than the inverse Fisher information,...
Ivan Bilić's user avatar
1 vote
Accepted

How to find the power function given exponential distribution?

You are right. There is an error. The upper limit of integration must be $0.05$. The power is defined by the formula $\inf_{\theta \in \Theta_1} \mathbb{E}_{X \sim \theta}[T(X)]$ for any test $T$. In ...
温泽海's user avatar
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Questions on Box and whiskers diagram

Heropup wrote a pretty good answer, but I wanted to add an answer under the assumption that all the data points would be evenly distributed within their respective quartiles. Since each box and each ...
VV_721's user avatar
  • 184
1 vote
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Sub-gaussian norm

For any centered random variable $X$, let $\Sigma(X) = \{\sigma \geq 0 : X \text{ is } \sigma \text{ sub-Gaussian} \}$. Thus, $\|X\|_{vp} = \inf \Sigma(X)$. Let us begin with positive definiteness. ...
sudeep5221's user avatar
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2 votes

Evaluating $\lim_{n\to\infty}\sum_{i=1}^n \exp\left(\frac{-|x-X_i|^2}{2(\sigma/n)^2}\right)$

Thanks to some helpful suggestions, I was able to find a solution. For simplicity, I will prove it for the case of $\mathbb{R}^d$ and when $\rho$ is a K-Lipschitz continuous density. Define the ...
APP's user avatar
  • 136
2 votes

Questions on Box and whiskers diagram

Skewness is difficult to discern from a box-and-whisker plot, so I would not assume that in general one can assert extent of skew from such diagrams. In cases where we can make assumptions about the ...
heropup's user avatar
  • 139k
1 vote

concentration of maximum of gaussians

For the sake of completeness, I will prove here the inequality \begin{equation} P(\|X\|_\infty \geq \sqrt{2 \log (2n)}+t)\leq \frac{1}{2}\exp(-t^2 /2) \quad (t>0), \end{equation} which is slightly ...
Maurizio Barbato's user avatar
2 votes
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Method of Moment for Normal mixtures $p\cdot N(0, 1) + q\cdot N(\eta, 1)$

If $Y_0 \sim N(0,1)$ and $Y_\eta \sim N(\eta,1)$ and the $X_i$ are iid with the mixture distribution (a probability $p$ of following $Y_0$'s distribution and a probability $q$ of following $Y_\eta$'s ...
Henry's user avatar
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0 votes

On expectation of maximum of gaussians

For the sake of completeness, I give here the full proof of the existence of $M>0$ such that $\mathbb{E} \left[ \max_{i}|X_i|\right] \leq M \mathbb{E} \left[ \max_{i}X_i\right]$ for all $n \geq 2$ (...
Maurizio Barbato's user avatar
2 votes
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If $X\sim N(\mu, \sigma^2)$ and $\Phi$ is the CDF of a standard Normal random variable, what is the distribution of $\Phi(X)$?

Since $\ \Phi\ $ is strictly increasing, it has a strictly increasing inverse $\ \Phi^{-1}:(0,1)\rightarrow(-\infty,\infty)\ .$ Then, for $\ 0<x<1\ ,$ \begin{align} P(\Phi(X)\le x)&=P\big(X\...
lonza leggiera's user avatar
0 votes

Is $X Y/|Y|$, where $X, Y \text{ i.i.d. } \sim N(0, 1)$, normally distributed?

As an alternative method to the accepted answer, the proof can be completed by the following two steps: $U := Y/|Y|$ takes values $\pm 1$ almost surely, and $P(U = -1) = P(U = 1) = \frac{1}{2}$. If $...
Zhanxiong's user avatar
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0 votes

Optimizing entropy of conditional Gaussians

With Lagrange multipliers: $$J= \sum_i p_i \log (p_i \sigma_i) + \lambda \sum_i p_i + \beta \sum_i p_i \sigma_i$$ $$\frac{\partial J}{\partial p_i}= \log (p_i \sigma_i) + 1 +\lambda + \beta \sigma_i$$ ...
leonbloy's user avatar
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0 votes

Tsiatis Semiparametric Theory Chapter 2 Exercise 3(b) Use Change of Variable to Get a NEW Probability Density

I'm also reviewing this chapter. Since $E[XY^2]=E[(−X)(−Y)^2]=−E[XY^2]$, by symmetry $(X,Y)$ and $(-X,-Y)$ you can obtain your claim.
Michael Wang's user avatar
1 vote
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Linear combination of zero-mean sub-gaussians random variables

Notice that by independence and the sub-Gaussianity of the original variables (regarding independence, namely we are using that for independent random variables expectation of product is the product ...
LostStatistician18's user avatar
0 votes

What is a good approximation for the inverse of the cumulative distribution function?

Since you only need a precision of roughly 5 decimal places I'd suggest the approach used in an ancient HP pocket calculator: use an approximation for the cumulative distribution function and use it ...
m-stgt's user avatar
  • 374
4 votes
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What distribution does the height of both men and women follow?

That would be an example of a mixture distribution. And the mixture of two normals is not necessarily normal (in fact it is almost never normal). Also, the idea that men and women's heights follow ...
balddraz's user avatar
  • 7,710
3 votes

Evaluating the integral $\int_{-\infty}^\infty\Phi^2\left(\frac{x - a}{b}\right)\phi(x)dx$.

Let $\Sigma$ be a $2\times 2$ covariance matrix: $$\tag1 \Sigma=\pmatrix{\Sigma_{11}&\Sigma_{12}\\\Sigma_{21}&\Sigma_{22}}\,,\quad\Sigma_{12}=\Sigma_{21}\, $$ of which the corresponding ...
Kurt G.'s user avatar
  • 14.9k
4 votes

Evaluating the integral $\int_{-\infty}^\infty\Phi^2\left(\frac{x - a}{b}\right)\phi(x)dx$.

For slight type saving, let me reparametrize $I_2(a, b)$ as \begin{align*} J_2(a, b) = \int_{-\infty}^\infty\Phi(a + bx)^2\phi(x)dx \end{align*} Differentiating $J_2(a, b)$ with respect to $a$ gives (&...
Zhanxiong's user avatar
  • 14.2k
0 votes

Normally distributed R.V. depends only on numeric parameters.

Claim: The distribution of $$Z := (X-a)^2+(Y-b)^2$$ only depends on $a,b>0$ via $r:=\sqrt{a^2+b^2}>0.$ Proof. Let us interpret the claim in terms of planar geometry. We write $(x,y)$ for a point ...
Mero Wingz's user avatar
1 vote

Geometric Brownian Motion problem - Compute $\mathbb{P}(X_3 < 3)$

Whereas $X_{t} \sim GBM(\mu, \sigma^2)$, another way to solve it is to remember that $$ X_{t+s} = X_{t} \ e^{\left( \mu-\frac{\sigma^2}{2} \right)s \ + \ \sigma(W_{t+s} - W_{t})}, $$ then $$X_{1+2} = ...
Victor Nunes's user avatar
0 votes

Gaussian with prior $p(\mu, \Sigma) \propto |\Sigma|^{-(d+1)/2}$. Why is the posterior $\Sigma \mid y \sim \text{Inv-Wishart}(S^{-1})$?

Your calculations are correct, it's just the definition I suppose. Look at the top of the page 73 of the Gelman book where $p(\Sigma)∝ \exp(−\frac{1}{2}Tr(\Lambda_0\Sigma^{−1}))$ and in the previous ...
Zain's user avatar
  • 111
1 vote
Accepted

concentration of norm of non-central Gaussian around its mean

Let $f(X) := \sum_{i = 1}^n \|X_i - a\|$. We first prove that $f$ is $\sqrt{n}$-Lipschitz: $$ \begin{aligned} f(X) - f(Y) &= \sum_{i = 1}^n \|X_i - a\| - \|Y_i - a\| \\ &\leq \sum_{i = 1}^n \|...
VHarisop's user avatar
  • 3,900
0 votes

Unitary invariance of the Gaussian Unitary Ensemble

Because the trace is invariant wrt rotation of factors in product arguments $$\text{Tr}( a \ b \ c) = \sum_{ikl}a_{ik}\ b_{kl}\ c_{li} = \sum_{ikl} c_{li}\ a_{ik}\ b_{kl} $$ so with $U^\star = U^{-1}...
Roland F's user avatar
  • 2,683
1 vote
Accepted

Range centered at mean of a normal distribution

You've got some errors. The statement $$\Pr(20 - c \leq X \leq 20 + c) = 0.899 \implies 2 \Pr(X \leq 20 + c) = 0.899$$ is wrong. $$2 \Pr(X \leq 20 + c) > \Pr(X \leq 20 + c) > \Pr(20 - c \leq X \...
msantama's user avatar
  • 1,096
0 votes
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Asymptotic normality of the square root of the Unbiased Estimate of Variance from I.I.D. samples of $N(a\sigma, \sigma^2)$

You're almost there. Seeing that you know that $$\sqrt{n}(S^2 - \sigma^2) \to N(0, 2\sigma^4)$$ in distribution, we may now apply the delta method. If $g(x) = \sqrt{x}$, then $g^\prime (\sigma^2) = \...
Jose Avilez's user avatar
  • 13.2k
0 votes

How should independent random variables be distributed to have their product be distributed as a gaussian?

Since $Z^2$ is Gamma distributed when $Z\sim N(0,1)$, with shape parameter $1/2$ and since the log of a Gamma variable is infinitely divisible, the answer is easy. More specifically, if $Y$ is Gamma ...
Letac Gérard's user avatar
1 vote

How should independent random variables be distributed to have their product be distributed as a gaussian?

Here is a partial answer inspired by Robert Israel's idea. Actually, computations will turn out to be carried more easily easier with a positive random variable, that is why we will consider the sign ...
Abezhiko's user avatar
  • 8,551
3 votes

How should independent random variables be distributed to have their product be distributed as a gaussian?

If $X$ has a normal distribution with mean $0$, we can write $X = S \exp(L)$ where $S = \pm 1$, each with probability $1/2$, and $S$ and $L$ are independent. The product of $n$ iid random variables $...
Robert Israel's user avatar
1 vote
Accepted

Finding a derivative for option pricing

If we let $g(s)=\frac{\ln\left(\frac S k\right)}{\sigma\sqrt{t}}$ and $R$ be the normal distrubution $$ \frac{\partial}{\partial s}\left(e^{-dt}N(g(s))\right)=e^{-dt}R(g(s))g'(s)=e^{-dt}R(g(s))\frac{1}...
Masd's user avatar
  • 884
1 vote
Accepted

The gap distribution of the random variables.

Let $F(x)$ be the CDF of your continuous random variable $X_j$ and $f(x)$ its density; you have $n$ of them and $i\in\{2,3,\dots,n\}$. For given two real numbers $a<b$, the probability that $k:=i-1$...
van der Wolf's user avatar
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