8 votes

Average value of $\frac{x'A^2x}{x'A^3x}$ over surface of $n$-dimensional sphere

We can attempt to simplify this multidimensional integral as follows. The expression for the expectation value reads $$E[\frac{x^TA^2 x}{x^TA^3 x}]=\int d^nx \frac{e^{-|x|^2/2}}{(2\pi)^{n/2}}\frac{\...
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  • 7,988
7 votes
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Distribution and moments of $\frac{X_iX_j}{\sum_{i=1}^n X_i^2}$ when $X_i$'s are i.i.d $N(0,\sigma^2)$

Note that we may assume $X_i$'s are standard normal. Then \begin{align*} \mu_k &:= \mathbf{E}\left[ \biggl( \frac{ X_i X_j }{X_1^2 + \cdots + X_n^2} \biggr)^k \right] \\ &= \frac{1}{(2\pi)^{n/...
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7 votes

Distribution and moments of $\frac{X_iX_j}{\sum_{i=1}^n X_i^2}$ when $X_i$'s are i.i.d $N(0,\sigma^2)$

You can use the independence of $\color{blue}{U=\frac{X_iX_j}{\sum_{i=1}^n X_i^2}}$ and $V=\sum\limits_{i=1}^n X_i^2$ to find at least the mean and variance of $U$ for $i\ne j$. To argue somewhat ...
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  • 15.4k
7 votes
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Probability to get 170 1's

No. First, let's label each distribution so as to make clear which distribution I am referring to. Let $X$ be the original, exact Binomial distribution: $X\sim B\left(1000,\frac{1}{6}\right).$ Then ...
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5 votes
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Sum of normal random variables being not normal

Yes, sure. Take $X\sim\mathcal N(0,1)$ and then define $Y$ by $Y=X$ if $|X|>a$ and $Y = -X$ for $|X|<a$ (so basically you flip the center part around. Then we get: $$ \mathrm{cov}(X,Y) = \mathrm ...
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  • 3,884
5 votes
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Brownian Motion and Hitting Time expectation.

I do not know whether your expression is correct, but $$\begin{align} &\int_0^{\infty} t \cdot \frac{\alpha}{\sqrt{2\pi t^3}} e^{-\frac{1}{2t}\alpha^2} dt \\ \ge &\int_1^{\infty} t \cdot \frac{...
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  • 143k
5 votes
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Which algorithm to sample from the normal distribution $\mathcal{N}(0,1)$ is this?

This is the Marsaglia polar method, a variant of the Box–Muller transform. The very largest (and smallest) values of the result occur when the pair you're calling $(u,v)$ is as close to $0$ as ...
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5 votes
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Let $X,Y,Z$ be three random variables such that the correlation coefficients $\rho_{XY}=0.2, \rho_{YZ}=0.2$, what values can $\rho_{XZ}$ take?

Consider the covariance matrix of $(X,Y,Z)^T$, $C=\mathbb E[[X,Y,Z]^T[X,Y,Z]]$. (since the means are $0$) Compute the individual variance and standard deviation of $X,Y,Z$, call it $\sigma_X,\sigma_Y,\...
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4 votes
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Nice proof that an expectation vanishes?

In light of Robert Israel's answer, I would appreciate if someone can point out the flaw in my code. My simulations suggest the expectation is not zero. With $n=10^5$ samples and $\epsilon=1$, the ...
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  • 81.8k
4 votes
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Uniform convergence and exchanging limit and integral

If $f_n(x)\xrightarrow[n\to\infty]{}f(x)$ uniformly, then \begin{align*} \left\lvert\int_a^bf_n(x)\,\mathrm dx-\int_a^bf(x)\,\mathrm dx\right\rvert&\le\int_a^b\lvert f_n(x)-f(x)\rvert\,\mathrm dx\\...
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  • 2,816
4 votes

Why isn't a normal distribution defined on positive $x$ common

time until the next bus arrives for example It is not so obvious in the case you took as example. In particular, the most common distribution for random arrivals is a poisson and, as known, ...
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  • 31.6k
4 votes
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How to prove $X_n =e^{aS_n-bn} \to 0$ a.s.(where $S_n$ is the sum of standard normal random variables)?

You need to prove that your exponent converges a.s. to $-\infty$ to do that, observe that your exponent can be rewritten in $$aS_n-bn=n\left(a\underbrace{\overline{X}_n}_{\xrightarrow{a.s.}0}-b \...
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  • 31.6k
4 votes

$\mathbb{P}(X\le2Y)$ for independent normal distribution

Follow the hint provided by @Henry. Let $X,Y\sim \mathcal{N}(\mu,\sigma^2)$. By using characteristic functions and independence (recall if $X\perp Y$ then $E[f(X)g(Y)]=E[f(X)]E[g(Y)]$ for Borel $f,g$),...
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  • 8,846
4 votes

How do I calculate phi(z) to 4dp when my Normal Distribution tables are to 2dp?

You can get a good approximation with interpolation Your tables presumably give $\Phi(0.31)=0.6217$ though $0.6217195$ is closer $\Phi(0.32)=0.6255$ though $0.6255158$ is closer You want to go $23\%$...
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  • 143k
3 votes
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(Sheldon Ross) Proving the independence of sample mean and sample variance

It is obvious. Duh. Just kidding. So to add a bit on what the author said, there's a few things you have to put together to understand. First of all, $X_i - \overline X$ for $i=1,\cdots,n$ and $\...
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3 votes
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What is the variance of the product of a Bernoulli (0,1) and a normal random variable?

If $X \sim B(1, p)$ and $Y \sim N(\mu, \sigma^2)$ are independent, you don't have to determine the CDF of $Z = XY$ for merely determining its variance. All you need is to use the expectation property $...
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  • 12.1k
3 votes

(Sheldon Ross) Proving the independence of sample mean and sample variance

It says in the previous sentence or two that $Y$ is independent of the $X_i$, so it is independent of $X_i-\bar X$ (bc $\bar X$ is a function of $X_i$). Since $Y, X_i -\bar X$ has the same ...
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  • 10.2k
3 votes
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Why is the poisson distribution more symmetrical with a higher mean?

The Poisson distribution takes values on $0,1,2,3,\ldots$. No matter what the mean is, it isn't symmetric because to the right of the mean, it has a tail tending to infinity, while to the left it gets ...
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  • 81.8k
3 votes
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Is the sum of randomly chosen half of $n$ values approximately normal?

Consider $y_i = 2^{-i}$ and $x_i = a_n(y_i - b_n)$ for all $i\geq 1$, where $a_n$ and $b_n$ are chosen to satisfy $$ \sum_{i=1}^{n} x_i = 0 \qquad\text{and}\qquad \frac{1}{n-1}\sum_{i=1}^{n} x_i^2 = 1....
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3 votes
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Does $\frac{\overline X - \mu}{S/\sqrt n}$ converge to $N(0,1)$?

The sample variance $S^2$ is consistent for $\sigma^2$. By the Mann-Wald (continuous mapping) theorem, since the function $g: \mathbb{R}_{\geq 0} \to \mathbb{R}$ defined by $g(x) = \sqrt{x}$ is ...
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  • 18.2k
3 votes

Does $\frac{\overline X - \mu}{S/\sqrt n}$ converge to $N(0,1)$?

Yes, this is the Slutsky's theorem https://en.wikipedia.org/wiki/Slutsky%27s_theorem. $$\frac{\bar{X}-\mu}{S/\sqrt{n}}=\frac{\bar{X}-\mu}{\sigma/ \sqrt{n}}\frac{\sigma}{S}\rightarrow\mathcal{N}(0,1)\...
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3 votes
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Inequality related to standard normal distribution function

For all $x<0$, $$\Phi(x):=\int_{-\infty}^x\frac{\mathrm e^{-\frac{t^2}2}}{\sqrt{2\pi}}\,\mathrm dt\le\int_{-\infty}^x\frac{t\,\mathrm e^{-\frac{t^2}2}}{x\sqrt{2\pi}}\,\mathrm dt=-\frac{\Phi'(x)}{x}....
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  • 2,816
3 votes

Why is there no UMVUE for $\mu$ with two samples from $N(\mu, \sigma^2)$, $N(\mu, \tau^2)$?

Joint density of $\boldsymbol X=(X_i)_{1\le i\le m}$ and $\boldsymbol Y=(Y_j)_{1\le j\le n}$ is $$f_{\theta}(\boldsymbol x,\boldsymbol y)\propto \frac1{\sigma^m \tau^n}\exp\left[-\frac1{2\sigma^2}\...
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  • 15.4k
3 votes

Distribution and moments of $\frac{X_iX_j}{\sum_{i=1}^n X_i^2}$ when $X_i$'s are i.i.d $N(0,\sigma^2)$

Due to the symmetry $\sigma_1=...=\sigma_n=\sigma$, it is enough to consider $i=1$ and $j=2$. While my first thought was n-dimensional spherical coordinates also, you can immediately do the integrals ...
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  • 5,393
3 votes
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Normal distribution of the mean of a uniform distribution

Hint: $P(-a<X<a)=2\Phi(a)-1$. These are the inner (orange) areas of the graph below. For more detailed explanation see here. Additional: In your case the standardized value is $a=\frac{0.53-0.5}...
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3 votes
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$\frac{N_1}{\sqrt{N_{1}^{2} + N_2^2}} \perp \frac{N_2}{\sqrt{N_{1}^{2} + N_2^2}}$ where $N_1, N_2 \sim \mathcal{N}(0,1)$ are independent?

Hint: To prove that $X$ and $Y$ are uncorrelated, you need to prove that the covariance is null, in other words $$Cov(X,Y) = \mathbb{E}(XY)-\mathbb{E}(X)\mathbb{E}(Y) = 0 \tag{1}$$ If we can prove ...
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  • 6,712
3 votes
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$X\sim N(\mu,\sigma^{2})$. It is also known that $P(X ≤ 160) = \frac{1}{2}$ and $P(X ≤ 140) = \frac{1}{4}.$ What are the values of $µ$ and $σ$?

From $P(X\leq 160)=1/2$, we deduce that the normally distributed random variable $X$ has mean $\mu=160$. Then, from $P(X\leq 140)=1/4$, we get $$P\Big(\frac{X-160}{\sigma}\leq \frac{-20}{\sigma}\Big)=\...
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3 votes

Minimize the expected value of the product of 2 normally distributed variables

The formula is correct, but your application of the formula is not. The correct expectation using this formula is $$\begin{align} \operatorname{E}[X^5 Y^3] &= \sigma_1^5 \sigma_2^3 \frac{5! 3!}{...
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  • 113k
3 votes

Understanding confidence intervals and percentiles

Since this is clearly homework, I won't do all the work for you. Instead I will try to guide you in the right direction. Part b): When two events $A$ and $B$ are assumed independent (such as choosing ...
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  • 803

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