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3

What we have is $$\operatorname{Var}\left(z^2\right)=\mathbb E\left[\left(z^2\right)^2\right]-\left(\mathbb E\left[z^2\right]\right)^2=\mathbb E\left[z^4\right]-\left(\mathbb E\left[z^2\right]\right)^2,$$ so the formula in the opening post is true if and only if $\mathbb E\left[z^2\right]=1$.


3

Let $w$ fixed. Now, by Total Probability: $\begin{eqnarray*} &&P(W\le w)\\ &=&P(W\le w|X<Y)P(X<Y)+P(W\le w|X>Y)P(X>Y)\\ &=&P(W\le w|X\mbox{ and }Y>w)P(X\mbox{ and }Y>w)P(X<Y)\\ &&+P(W\le w|X<w<Y)P(X<w<Y)P(X<Y)\\ &&+P(W\le w|Y<w<X)P(Y<w<X)P(X<Y)\\ &&+P(W\le ...


3

Through the Spitzer identity, it is possible to find some kind of transform of the distribution of $M_n$. Well, not exactly. The Spitzer identity involves the expressions $M^+_n = \max_{0\le k\le n} S_k$, where $S_0 = 0$, $S_k = X_1 + \dots + X_k$, $k\ge 1$. So this translates to the positive part of expression you are interested in. But it is possible to ...


2

Hint: $X \mid Y \sim \mathcal{N}\left(\mu_{X}, Y\right)$, no? So $$f_{X}(x) = \int_{-\infty}^{\infty}f_{X\mid Y}(x \mid y)f_{Y}(y)\text{ d}y$$ $F_{X}$ can be easily found from this. In general, $$\mathbb{E}[g(X)] = \mathbb{E}\left[\mathbb{E}[g(X) \mid Y]\right] = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}g(x)f_{X \mid Y}(x \mid y)f_{Y}(y)\text{ ...


2

It seems that you have a random variable $X$ which is normally distributed with mean $μ$ and unit variance $σ^2=1$. So, $$f_X(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac12(x-μ)^2}$$ for $x\in \mathbb R$. As usual you can standardize this random variable by $$Z=\frac{X-μ}{σ}=\frac{X-μ}{1}=X-μ \sim N(0,1)$$ i.e. $Z$ has the standard normal distribution with cdf ...


2

$X\sim \text{log-normal}(\mu,\sigma^2)$ means $Y=\log X\sim N(\mu,\sigma^2)$. So $X=e^Y$ and $Y\sim N(\mu,\sigma^2)$. Let $Z=\dfrac{Y-\mu} \sigma$ so that $Z\sim N(0,1)$. Then $X = e^Y = e^{\mu+\sigma Z}$ and for any measurable set $A$, $$ \Pr(Z\in A) = \int_A \varphi(z)\,dz \quad \text{ where } \varphi(z) = \frac 1 {\sqrt{2\pi}} e^{-z^2/2}. $$ So ...


2

Hint: if $n$ is even, you can group the terms using Gauss summation: $$y_1 + \cdots + y_n = (y_1+y_n) + (y_2+y_{n-1}) + \cdots.$$ Can you say anything about these pairs? Try using the symmetry of the inverse CDF: If $n$ is odd, you can first show that the $(n+1)/2$-th element is zero, and use the same trick as above to establish your result.


1

$x_{n+1}$ itself is normally distributed with mean $\mu$ and variance $\sigma$ (two unknown numbers). Therefore, assuming it is also independent of the others, $x_{n+1}-\overline{x}$ is normally distributed. Recall a few useful rules: Expectation is linear $\operatorname{Var}[X+Y]=\operatorname{Var}[X]+\operatorname{Var}[Y]$ for independent $X,Y$ ...


1

It's true and you don't need anything fancy to show it. Apply this common definition and just work with the CDF in the link you gave.


1

It is enough to show that the MGFs of $X_n$ converge point-wise to the MGF of $X$. We know that $$ M_{X_n}(t)=e^{\mu t + 1/2\sigma^2t^2}\left(\frac{\Phi(\beta_n-\sigma t)-\Phi(\alpha_n-\sigma t)}{\Phi(\beta_n)-\Phi(\alpha_n)}\right) $$ where $\beta_n:=\frac{n-\mu}\sigma$, $\alpha_n:=\frac{-n-\mu}\sigma$ and $n\in\mathbb N$. So, by the continuity of the CDF ...


1

The answer is that there's a variant of the Box-Muller transform that does the job. Let $U_1$ and $U_2$ be uniformly drawn numbers. Then $$ Z = \sqrt{-2 \log_{q'}(U_1)} \cos(2 \pi U_2) $$ is a draw from a $q$-Gaussian with parameters $q$ and $\beta=1/(3-q)$. Here $q' = (1+q)/(3-q)$ and $$ \log_q = \frac{x^{1-q} - 1}{1-q} $$ is the inverse of the ...


1

See this answer to the same formula which the authors used in a different paper: http://mathoverflow.net/questions/200930/expectation-of-gaussian-random-vector-arbitrary-function-thereof


1

Observe that \begin{align} F_Y(y)&=P(|X-1|\leq y)\\&=P(1-y\leq X\leq1+y)\\&=F_X(y+1)-F_X(1-y).\end{align} Differentiate both sides using the chain rule to get the result.


1

You are almost there. Try doing a change of variables in the exponential part of the integral. Make it so that you get "$e^{-\frac{1}{2} x^2}$". Then you can relate this to the cdf of the normal distribution, ie $$ \Phi(z) = \int_{-\infty}^z \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2} y^2} dy. $$ Note also that, since the integrand is a pdf, $$ 1 - \Phi(z) = ...


1

Let $Y$ be the random variable which obeys a log normal distribution and $Z=log(Y)$. We know that $Z$ obeys a normal distribution, let $\mu$ be its mean and $\sigma^2$ its variance. The moment generating function of the normal distribution is, $$M(t)= E(e^{Zt}) = e^{\mu t+\sigma^2t^2/2},$$ evaluating this at $t=1$ we obtain, $$ M(1) = ...


1

When $z$ is a positive real number we have $$f(z) = \int_{-\infty}^\infty e^{-zt^2}{\rm d}t = \frac{1}{\sqrt{z}}\int_{-\infty}^\infty e^{-u^2}{\rm d}u = \sqrt{\frac{\pi}{z}}$$ where we have used a substitution $u=\sqrt{z}t$ and the result for the Gaussian integral whose derivation can be found here. The integral in the question corresponds to ...


1

Using the definition of the cumulative normal distribution function $\Phi$: $$ \Phi(\mu-a) = {1\over \sqrt{2\pi}}\int_{-\infty}^{\mu - a} e^{-{1\over 2}t^2 }dt = - {1\over \sqrt{2\pi}}\int_{\mu - a}^{-\infty} e^{-{1\over 2}t^2 }dt = {1\over \sqrt{2\pi}}\int_{-\mu + a}^{\infty} e^{-{1\over 2}t^2 }dt $$ $$= {1\over \sqrt{2\pi}}\int_{ a}^{\infty} e^{-{1\over ...


1

For finding the distribution of the first one $$(m+n-2)\frac{S^2}{\sigma^2}$$ let $S^2_1=\frac{1}{m-1}\sum_{i=1}^m(X_i-\overline X)^2$ and $S^2_2=\frac{1}{n-1}\sum_{j=1}^n(X_i-\overline X)^2$. Then $$(m+n-2)\frac{S^2}{\sigma^2}=(m-1)\frac{S_1^2}{\sigma^2}\ +\ (n-1)\frac{S_2^2}{\sigma^2}$$ But as you correctly guessed, each summand of the last equation ...


1

$z(0.1)$ means getting the $z$ value (argument of $\Phi$) for a given probability, in this case $0.1$. Since the table starts with probabilities of $0.5$, due to symmetry, you can retrieve the $z$ value by doing $-z(1-0.1)$. Then just look in the table where the highest probability $<0.9$ is. This is given for $\Phi(1.28)$. Thus, $z(0.1) = -1.28$.


1

1) $\max(x_1, x_1 + x_2) \le t$ if either $x_1 \le t$ and $x_2 \le 0$, or $x_1 + x_2 \le t$ and $x_2 \ge 0$. It may help to sketch this in the $x_1-x_2$ plane. Thus if $(X_1, X_2)$ has joint density $f(x_1,x_2)$, $$P(\max(X_1, X_1 + X_2) \le t) = \int_{-\infty}^t dx_1 \int_{-\infty}^{t-x_1} dx_2 f(x_1,x_2)$$ If $X_1$ and $X_2$ are iid with density $f$ and ...


1

An observation (to replace a previous wrong answer). Note that $$M_2=X_1+X_2^+$$ where $X_2^+=\max\{0,X_2\}$. So, $$E[M_2]=E[X_1]+E[X_2^+]=0+\frac{1}{σ\sqrt{2\pi}}$$ Similarly $$M_3=X_1+\max\{0,X_2,X_2+X_3\}=X_1+\left(X_2+\max\{0,X_3\}\right)^+=X_1+\left(X_2+X_3^+\right)^+$$ with $E[M_3]>E[X_2^+]=E[M_1]$. And two links here and here that might ...


1

This is not really an answer, but notes for future reference about the distribution of $X = \max\{X_1,X_2\}$ when $X_1, X_2 \sim \mathcal{N}(0,1)$, independent, identically distributed. As you stated, the cumulative distribution function (CDF) is $$F_X(x) = P(X\le x) = P(X_1\le x\ \cap X_2 \le x) = P(X_1 \le x) \cdot P(X_2 \le x) = \Phi(x)^2, $$ with ...



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