Suppose $X_1, X_2 ...X_n$ are non-correlated and Gaussian random variables, all with mean value $\mu=0$ and variance=$\sigma$.  Is there an expression for the distribution of $Z=\max(X_1, X_2, ... X_n)$? If not, can we find its expected value $\textrm{mean(Z)}$ and $\textrm{variance(Z)}$?  Thanks a lot!
97446
Is there an exact or good approximate expression for the expectation, variance or other moments of the maximum of $n$ independent, identically distributed gaussian random variables where $n$ is large?  If $F$ is the cumulative distribution function for a standard gaussian and $f$ is the probability density function, then the CDF for the maximum is (from the study of order statistics) given by  $$F_{\rm max}(x) = F(x)^n$$  and the PDF is  $$f_{\rm max}(x) = n F(x)^{n-1} f(x)$$  so it's certainly possible to write down integrals which evaluate to the expectation and other moments, but it's not pretty. My intuition tells me that the expectation of the maximum would be proportional to $\log n$, although I don't see how to go about proving this.
102059
How is the notation $X\sim N(\mu,\sigma^2)$ read? Is it $X$ follows a normal distribution? Or $X$ is a normal distribution? Or perhaps $X$ is approximately normal..  What if there are several variables that follow (or whatever the words is) the same distribution? How is it written?
99395
How does the expected value of a continuous random variable relate to its arithmetic mean, median, etc. in a non-normal distribution (eg. skew-normal)?  I'm interested in any common/interesting distributions (eg. log-normal, simple bi/multimodal distributions, anything else weird and wonderful).  I'm looking mostly for qualitative answers, but any quantitative or formulaic answers are also welcome. I'd particularly like to see any visual representations that make it clearer.
75088
Say I have two independent random variables $X \sim N(u_1, \sigma_1)$ and $Y \sim N(u_2, \sigma_2)$. I want to get the conditional distribution of X given whether X is bigger than Y or not.  $P(X|X&lt;Y)$ = ... and  $P(X|X&gt;Y)$ = ...  I am thinking solving this in this way:  \begin{align} P(X|X&lt;Y) &amp;= \frac{P(Y&gt;X|X)P(X)}{P(Y&gt;X)} \\ &amp;= \frac{(1-\Phi(\frac{x-\mu_2}{\sigma_2}))N(\mu_1,\sigma_1)}{\Phi(\frac{\mu_2-\mu_1}{\sqrt{\sigma_2^2+\sigma_1^2}})}\\ P(X|X&gt;Y) &amp;= \frac{P(Y&lt;X|X)P(X)}{P(Y&lt;X)} \\ &amp;= \frac{\Phi(\frac{x-\mu_2}{\sigma_2})N(\mu_1,\sigma_1)}{1-\Phi(\frac{\mu_2-\mu_1}{\sqrt{\sigma_2^2+\sigma_1^2}})} \end{align}  My questions are:   (1) Whether above solution is correct  (2) How to get the mean and sd for $P(X|X&lt;Y)$ and $P(X|X&gt;Y)$ if they are still normal?
52067
I'm looking for someone to walk me through how to solve E[ZW] if Z = max[X,Y] &amp; W = min[X,Y]. X &amp; Y are independent, uniform variables over the interval [0,1].  I solved E[Z] = 2/3 and E[W] = 1/3.   E[ZW] should equal 1/4 but I keep getting the wrong answer. I have searched the forums and used other people's answers to try and solve it but have gotten the incorrect numbers.  Thank you in advance!
34219
In the vein of , I'm now at a roadblock on question 3 of this sheet:    (note: it's not my intention to ask every question I get stuck on here, merely the ones which have interesting general results; it just so happens that the two I've struggled with so far fit this criteria)  It goes as follows:  Let $N$ be a non-negative integer-valued random variable with mean $\mu_1$ and variance $\sigma_1^2$, and let $X_1, X_2, ...$ be random variables, each with mean $\mu_2$ and variance $\sigma_2^2$; furthermore, assume that $N, X_1, X_2,  . . .$ are independent. Without using generating functions, calculate the mean and variance of the random variable $S_N = X_1 + ... + X_N$ (when $N=0$ interpret $S_N$ as $0$).  I have the answer to calculating the mean, which I'll write up as an answer for future reference. Here's where I've got to with the variance bit:  $\mathbb{E}(S_N^2) = 0.P(N=0) + \sum_{r=1}^\infty\mathbb{E}((\sum_{i=1}^rX_i)^2).P(N=r)$  For $r=1$, we have the inner expectation as: $\mathbb{E}(X_1^2)$ which is equal to $\sigma_2^2+\mu_2^2$.  For $r &gt; 1$, we have the inner expectation equal to: $r(\sigma_2^2 + 2\mu_2^2)$  So $\mathbb{E}(S_N^2) = (\sigma_2^2+\mu_2^2)P(N=1) + (\sigma_2^2 + 2\mu_2^2)\sum_{r=2}^\infty rP(N=r)$  which is almost in the form where I can use the same trick as in calculating the mean, but not quite.  Any help is much appreciated.
12502
If $X_1, ..., X_n$ are independent identically-distributed random variables, what can be said about the distribution of $\min(X_1, ..., X_n)$ in general?
96591
Given that a random variable $X$ is a normally distributed variable with mean $\mu$ and variance $\sigma^2$. Why exactly is:  $$\mathbb{E}[|X|] = \mathbb{E}[X\mid X&gt;0]P(X&gt;0) - \mathbb{E}[X\mid X&lt;0]P(X&lt;0)$$  Because we have that:  $$\mathbb{E}[X\mid X&gt;0] = \frac{\int_0^\infty xf_X(x)\,dx}{P(X&gt;0)}$$  Therefore my definition immediately above must be incorrect?  Because if it is not we have that:  $$\mathbb{E}[|X|] = \int_0^\infty xf_X(x)\,dx - \int_{-\infty}^0 xf_X(x)\,dx$$  and where does the minus there come from anyway, and why?
48180
Given $x_1, \ldots, x_N$, independent and all distributed as a   Gaussian with mean $\mu$ and variance $\sigma^2$. Then, the average   $$\bar{x} = \frac{1}{N}\sum_{i=1}^Nx_i$$ is distributed as a Gaussian   with mean $\mu$ and variance $\frac{\sigma^2}{N}.$   This is a very well-known result. Anyway, I'm looking around to find a proof for this and I'm not having luck.
57761
Suppose that $\mathbf{X} \sim N_{2}(\mathbf{\mu}, \mathbf{\Sigma})$. Then the conditional distribution of $X_1$ given that $X_2 = x_2$ is multivariate normally distributed with mean:   $$ E[P(X_1 | X_2 = x_2)] = \mu_1+\frac{\sigma_{12}}{\sigma_{22}}(x_2-\mu_2)$$  and variance: $${\rm Var}[P(X_1 | X_2 = x_2)] = \sigma_{11}-\frac{\sigma_{12}^{2}}{\sigma_{22}}$$  It makes sense that the variance would decrease since we have more information. But what is the intuition behind the mean formula? How does the covariance between $X_1$ and $X_2$ factor into the conditional mean?
88502
Question  Suppose we have a multinomial distribution with $N$ possible outcomes, with probabilities $p_1,\ldots,p_N$. We sample this $n$ times, and denote the observed frequency of the $i$th outcome as $\xi_i$. In [1] the author claims that the distribution of the $\xi_i$ in the limit of large $n$ is:  $$p(\xi_1,\ldots,\xi_N)\propto\exp\left(-\frac{n}{2}\sum_{i=1}^N\frac{(\xi_i-p_i)^2}{p_i}\right).\;\;\;\;\;(1)$$  We can see immediately that this must be an approximation, as it assigns nonzero probabilities for $\xi_1+\cdots+\xi_N&gt;1$. However we can see that these have vanishing probability in the limit $n\rightarrow\infty$. My question is how do we derive (1) from the multinomial distribution, and show that they match in the $n\rightarrow\infty$ limit?  My thoughts  My first thought would be to appeal to the central limit theorem. The multinomial distribution has mean $\mu_i=p_i$ and covariance matrix $\Sigma_{ij}=\delta_{ij}p_i-p_ip_j$, so we would expect this in the large $n$ limit to be described by a multivariate Gaussian with mean $\mu$ and covariance $\frac{1}{n}\Sigma$. However, things are complicated by the fact that the multinomial covariance is singular (since $\xi_N$ is determined by the other $\xi_i$s), and so the multivariate Gaussian is not defined.  To address this, we may try and consider only the first $\xi_1,\ldots,\xi_{N-1}$, which have a non-singular covariance matrix and hence well-defined multivariate Gaussian distribution. Let's take the Binomial distribution $N=2$. The frequency $\xi_1$, this has mean $p_1$ and variance $p_1(1-p_1)$, so this would be described the the Gaussian: $$\propto\exp\left(-\frac{n}{2}\frac{(\xi_1-p_1)^2}{p_1(1-p_1)}\right).\;\;\;\;\;(2)$$ The expression (1) gives: $$\propto\exp\left(-\frac{n}{2}\left(\frac{(\xi_1-p_1)^2}{p_1}+\frac{(\xi_2-p_2)^2}{p_2}\right)\right).\;\;\;\;\;(3)$$ If we substitute $\xi_2\rightarrow 1-\xi_1$, $p_2\rightarrow 1-p_1$ into (3), we can verify that this gives the same answer as (2). I have verified that this also works for $N=4$.   I'm sure that if I just bashed out the algebra for general $N$ we would get agreement between the central limit theorem and (1) when we restrict the latter to $\xi_1+\cdots+\xi_N=1,p_1+\cdots+p_N=1$. However, how can we start with the multinomial distribution and derive (1) as a limit which is valid everywhere? One idea would be to say that (1) goes to zero as $n\rightarrow\infty$ when you are not on that plane, however I am a bit uncomfortable with this as it goes to zero everywhere except the mean as $n\rightarrow\infty$, so I don't know if that argument is good enough.  [1] Wootters, William K. "Statistical distance and Hilbert space." Physical Review D 23.2 (1981): 357.
50765
I have a random variable $X(a) = \log(a)$ where a is normal distributed $\mathcal N(\mu,\sigma^2)$. What can I say about $E(X)$ and $Var(X)$? An approximation would be helpful too.
3712
If $Y \sim N(\mu,\sigma^2)$ is normally distributed, then $X=\mathrm{e}^Y$ is lognormally distributed. To get the log-$\mu$ and log-$\sigma$ of this lognormal distribution you calculate $$\sigma^2 = \ln\left( \frac{\mathit{Var}}{E^2} + 1 \right)$$ and $$\mu = \ln(E)-\frac{\sigma^2}{2}$$.  But how is this in the case of a bivariate Normaldistribution, $Y \sim N((\mu_1, \mu_2),\Sigma)$, $\Sigma=((\sigma_1,\rho),(\rho,\sigma_2))$ How do I get $\mu, \sigma, \rho$? And more general in the case of a multivariate distribution?
123566
Given that $a$~$\mathcal{N}(\mu, \Sigma)$ and $\mu$~$\mathcal{N}(0, \sigma^2 I)$, Is $p(a|\Sigma, \sigma)=\int p(a|\mu, \Sigma)p(\mu|\sigma)d\mu$ again a Normal distribution? And how to estimate the mean and variance respectively? Thanks a lot!
121789
I am having a small doubt regarding maximum of random variables. I have $$Z= \max\{ X_1, X_2,\dots X_p, \dots X_N\}$$ where all $X_i$ are independent, identically distributed. Now, If for sure, I know that $$\{X_1,\dots,X_p\}&gt;X_{p+1}&gt;\dots&gt;X_N$$ the can i say that the CDF of $Z$,  $$\begin{align*}P(Z&lt;z)&amp;= P\left(\max\{ X_1, X_2,\dots X_p, \dots X_N\}&lt;z\right)\\&amp;=P\left(\max\{ X_1, X_2,\dots X_p\}&lt;z\right)\\&amp;=\left[P(X_1&lt;z)\right]^{N-p-1}\end{align*}$$  How to find CDF of $Z$ in this case ? Is there any theorem that reduce the size of total variables for certain condition  ?
62581
Let $X \sim N (0, 1)$ and $Y ∼ N (0, 1)$ be two independent random variables, and define $Z = \min(X, Y )$. Prove that $Z^2\sim\chi^2(1),$ i.e. Chi-Squared with degree of freedom $1.$  I found the density functions of $X$ and $Y,$ as they are normally distributed. How would one use the fact that $Z = \min(X,Y)$ to answer the question? Thanks!
83429
If $$Y=\sum_{i=1}^{N}X_i^2$$ where $X_i \sim \mathcal{N}(0,\sigma^2)$, i.e. all $X_i$ are i.i.d. normal random variables of zero mean with same variances, then $$Y \sim \Gamma\left(\frac{N}{2},2\sigma^2\right).$$  I know the chi-squared distribution is a special case of the gamma distribution, but could not derive the chi-squared distribution for random variable $Y$. Any help, please?