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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 $\mathbb E\left[z^2\right]=1$.
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 LetY$be the random variable which obeys a log normal distribution and$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 ...