There are two dependent normal variables with the same distribution and the correlation coefficient $\rho$: $X,Y \sim N(\mu, \sigma^2)$. I would like to get $P(X&gt;M|Y&gt;M)$. For that I need to know the conditional expectation and variance. But I can not find any information about the conditional normal variance in the case when $Y$ is smaller/bigger than some real number. There is not a comparison between two normal variables. And it's not the truncated normal distribution. I found the conditional expectation of $X$ given that $Y$ is bigger than $M$ is $E(X|Y&gt;M)= \mu + \rho \sigma \frac{\phi(\frac{M-\mu}{\sigma})}{1-\Phi(\frac{M-\mu}{\sigma})}$. But what is the conditional variance $var(X|Y&gt;M)$? Is it $(1-\rho^2)\sigma^2 $, as it would be in the case of $var(X|Y=M)$, where variance does not depend on $M$?
120850
I would like to find the conditional bivariate normal distribution. There are two dependent normal variables with the same distribution and the correlation coefficient $\rho$: $X,Y \sim N(\mu, \sigma^2)$. I would like to get $P(X|Y&gt;M)$. I found the conditional expectation of $X$ given that $Y$ is bigger than $M$: $E(X|Y&gt;M)= \mu + \rho \sigma \frac{\phi(\frac{M-\mu}{\sigma})}{1-\Phi(\frac{M-\mu}{\sigma})}$. But what is the conditional variance of $var(X|Y&gt;M)$? Is it $(1-\rho^2)\sigma^2 $, as it would be in the case of $var(X|Y=M)$, where variance does not depends on $M$? And is the conditional distribution $N(E(X|Y&gt;M),var(X|Y&gt;M))$?
24322
If X and Y are independent random variables both with the same mean (0) and variance, how about $X^2$ and $Y^2$? I tried calculating E($X^2Y^2$)-E($X^2$)E($Y^2$) but haven't been able to get anywhere.
58601
I have two independent variables: $X$ follows from standard Gaussian distribution $N(0,\sigma^2)$; $Y$ follows from Rademacher distribution, i.e., $Y$ can be either $-1$ or $1$ with the same probability.  My question is:  For a new random variable of $Z=XY$, What is the probability density function?   Or can we have some concentration inequalities for $Z$ (e.g., show how $Z$ approaches its expectation, etc.)?
57186
I have been reading this  about the maximum and minimum of two normal distributed variables.   Inside the paper there is the formula for the expectation of this the maximum of the two variables. It is given below:  $$ \DeclareMathOperator{\E}{\mathbb{E}}    \E(X) = \mu_1 \Phi\left( \frac{\mu_1-\mu_2}{\theta} \right) + \mu_2 \Phi\left( \frac{\mu_2-\mu_1}{\theta} \right) + \theta \phi\left( \frac{\mu_1-\mu_2}{\theta} \right) $$  Where:   $\mu_1$ is the mean of the first normally distributed random variable. $\mu_2$ is the mean of the second normally distributed random variable. $\Phi$ is the CDF of the standard normal. $\phi$ is the PDF of the standard normal. $\rho$ is the correlation co-efficient between the variables.  $\theta$ is $\sqrt{\sigma_{1}^2 + \sigma_{2}^2 + 2\rho\sigma_{1}\sigma_{2}}$     I am trying to use this to get the mean of the new distribution but I run in to what feels like an obvious problem.  If $\mu_{1} \neq \mu_2$ then the argument to the $\Phi$ function in either the first or second term of the sum will be negative. The standard CDF does not allow negative values‽   I feel like I have fundamentally misunderstood something here. Where have I gone wrong?
26564
In comments following  to a related question, Users ssdecontrol and Glen_b asked whether joint normality of $X$ and $Y$ is necessary for asserting the normality of the sum $X+Y$? That joint normality is sufficient is, of course, well-known. This supplemental question was not addressed there, and is perhaps worth considering in its own right.  Since joint normality implies marginal normality, I ask     Do there exist normal random variables $X$ and $Y$ such that   $X+Y$ is a normal random variable, but $X$ and $Y$ are not   jointly normal random variables?   If $X$ and $Y$ are not required to have normal distributions, then it is easy to find such normal random variables. One example can be found in my previous answer (link is given above). I believe that the answer to the highlighted question above is Yes, and have posted (what I think is) an example as an answer to this question.
44347
I am trying to understand the . When I try to apply it to the following two samples of data I am unable to understand how to interpret the results.  Let's say sample 1 is ${0, 5, 7, 12, 11, 17}$ and sample 2 is ${10 ,15 ,17 ,22 ,21 ,27}$. Here sample 2 $=$ sample 1 $+\ 10$ as you can see.  Both have the same standard deviation $\sigma_{2} = \sigma_{1}= 5.95539$ but $\mu_{2}=18.67$ and $\mu_{1}=8.66667$.  Now the coefficient of variation ${\sigma}/{\mu}$ will be different. For sample 2 it will be less than for sample 1. But how do I interpret that result? In terms of variance both are the same; only their means are different. So what's the use of the coefficient of variation here? It's just misleading me, or maybe I am unable to interpret the results.
1348
I am trying to prove the statement:  If $X\sim\mathcal{N}(0,\sigma_1^2)$ and $Y\sim\mathcal{N}(0,\sigma_2^2)$ are independent random variables, then $\frac{XY}{\sqrt{X^2+Y^2}}$ is also a Normal random variable.  For the special case $\sigma_1=\sigma_2=\sigma$ (say), we have the well-known result that $\frac{XY}{\sqrt{X^2+Y^2}}\sim\mathcal{N}\left(0,\frac{\sigma^2}{4}\right)$ whenever $X$ and $Y$ are independent $\mathcal{N}(0,\sigma^2)$ variables. In fact, it is more generally known that $\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}}$ are independent $\mathcal{N}\left(0,\frac{\sigma^2}{4}\right)$ variables. A proof of the last result follows by using the transformation $(X,Y)\to(R,\Theta)\to(U,V)$ where $x=r\cos\theta,y=r\sin\theta$ and $u=\frac{r}{2}\sin(2\theta),v=\frac{r}{2}\cos(2\theta)$. Indeed, here $U=\frac{XY}{\sqrt{X^2+Y^2}}$ and $V=\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}}$. I tried to imitate this proof for the problem at hand but it appears to get messy. If I haven't made any error, then for $(u,v)\in\mathbb{R}^2$ I end up with the joint density of $(U,V)$ as $$f_{U,V}(u,v)=\frac{2}{\sigma_1\sigma_2\pi}\exp\left[-\sqrt{u^2+v^2}\left(\frac{\sqrt{u^2+v^2}+v}{\sigma_1^2}+\frac{\sqrt{u^2+v^2}-v}{\sigma_2^2}\right)\right]$$ I have the multiplier $2$ above as the transformation is not one-to-one. So density of $U$ would be given by $\displaystyle \int_{\mathbb{R}}f_{U,V}(u,v)\,\mathrm{d}v$, which isn't readily evaluated. Now I am interested to know if there is a proof where I can only work with $U$ and don't have to consider some $V$ to show that $U$ is Normal. Finding the CDF of $U$ doesn't look so promising to me at the moment. I would also like to do the same for the case $\sigma_1=\sigma_2=\sigma$. That is, if $X$ and $Y$ are independent $\mathcal{N}(0,\sigma^2)$ variables then I wish to show that $Z=\frac{2XY}{\sqrt{X^2+Y^2}}\sim\mathcal{N}(0,\sigma^2)$ without using a change of variables. If somehow I can argue that $Z\stackrel{d}{=}X$, then I'm done. So two questions here, the general case and then the particular case. Related posts on Math.S.E: . . Edit. This problem is in fact due to L. Shepp as I found out in the exercises of An Introduction to Probability Theory and Its Applications (Vol. II) by Feller, alongwith a possible hint:  Surely, $U=\frac{XY}{\sqrt{X^2+Y^2}}=\frac{1}{\sqrt{\frac{1}{X^2}+\frac{1}{Y^2}}}$ and I have the density of $\frac{1}{X^2}$ at hand. Let's see what I could do now. Apart from this, a little help with the integral above is also welcome.
88043
We are given two independent standard normal random variables $X$ and $Y$. We need to find out the M.G.F of $XY$.  I tried as follows : \begin{align} M_{XY}(t)&amp;=E\left(e^{(XY)t}\right)\\&amp;=\int_{- \infty}^{\infty}\int_{- \infty}^{\infty}e^{(XY)t}f_X(x)f_y(y)dxdy \\ &amp;=\dfrac{1}{2\pi}\int_{- \infty}^{\infty}\int_{- \infty}^{\infty}e^{(xy)t}e^{\frac{-x^2}{2}}e^{\frac{-y^2}{2}}dxdy \\ &amp;= \dfrac{1}{2\pi}\int_{- \infty}^{\infty}\int_{- \infty}^{\infty}e^{-\dfrac{(x-ty)^2}{2}}e^{\dfrac{-t^2y^2}{2}}e^{\frac{-y^2}{2}}dxdy \\ &amp;= \dfrac{1}{\sqrt{2\pi}}\int_{- \infty}^{\infty}e^{\dfrac{-t^2y^2}{2}}e^{\frac{-y^2}{2}}dy \\ &amp;= \dfrac{1}{\sqrt{2\pi}}\int_{- \infty}^{\infty}e^{-y^{2}(\frac{1}{2}+t^2)}dy \\ &amp;= \dfrac{1}{\sqrt{1+2t^2}} \end{align} Is this correct ?
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?
19356
I'm trying to understand better the difference between  covariance and correlation, besides the fact that the correlation coefficient is a dimensional and has values between $-1$ and $1$.   One unclear point is the following: the correlation coefficient can highlight only a linear relation between two variables, but does  covariance only work for linear relations too? Supposing that the relation is not linear, is the covariance not zero? Is it bigger than in the case in which the relation is linear?   Cauchy inequality says $$|\sigma_{xy}|&lt;\sigma_{x} \sigma_{y}$$  Which seems to say that the covariance is maximum when there is a perfect linear relation.    So I do not get what does happen if the relation is not linear, and, if covariance is not a suitable parameter in that cas I don't get the reason for that, since simply looking at the definition:  $$\sigma_{xy}=\frac{\sum(x-\bar{x})(y-\bar{y})}{N}$$  it looks like that this should be non zero for any kind of relation (like parabolic).  So can this be a difference between the use of covariance and correlation coefficient? (The first is useful for any kind of relation, the second one only for linear ones).
98658
I am working on a paper in which I'd need to use the two following conditional expectations:   $E(X_{1}|a \leq X_{2} \leq b)$ $E(X_{1}|a \leq X_{2} \leq b, a \leq X_{3} \leq b)$   where $X_{1}, X_{2}, X_{3}$ come from a trivariate normal distribution and $a$ and $b$ are some real numbers ($X_{2}$ and $X_{3}$ have the same bounds).   I have checked several textbooks and references on the web, including those mentioned in previous related posts (e.g., Tallis 1961), but did not find anything relevant. Could anyone help me derive these conditional expectations, or else point me to any good reference(s) I may have missed?
118458
Suppose we have  $$X_1 \sim \textrm{unif}(n,0,1),$$ $$X_2 \sim \textrm{unif}(n,0,1),$$  where $\textrm{unif}(n,0,1)$ is uniform random sample of size n, and   $$Y=X_1,$$  $$Z = 0.4 X_1 + \sqrt{1 - 0.4}X_2.$$  Then the correlation between $Y$ and $Z$ is $0.4$.  How can I extend this to three variables: $X_1$, $X_2$, $X_3$?
16546
Suppose, $X\sim N(\mu,\sigma^2)$.   Can anyone help in finding the following : $\text{Var } \bigg(\Phi\big(\frac{X + c}{d}\big) \bigg)$ ? Here, c and d are positive.  Here, $\Phi(x)$ is the "Cumulative Distribution Function" of the above-mentioned normal distribution.  Thanking you in advance.
70859
We have a multivariate normal vector $\mathbf{X} \sim \mathcal{N}(\boldsymbol{\mu}, \boldsymbol{\Sigma})$, where $\mathbf{X} =  \left[ \begin{matrix} X_1 \\ X_2 \\ \dot\\\\ \dot\\\\ X_n \end{matrix} \right]$,    $\boldsymbol{\mu} =  \left[ \begin{matrix} \mu_1 \\ \mu_2 \\ \dot\\\\ \dot\\\\ \mu_n \end{matrix} \right]$, and  $\boldsymbol{\Sigma} =  \begin{bmatrix} \sigma_1^2 &amp; \sigma_1\sigma_2 &amp; \dots &amp; \sigma_1\sigma_n   \\ \sigma_2\sigma_1 &amp; \sigma_2^2 &amp; \dots &amp; \sigma_2\sigma_n  \\ \dot\\\\ \dot\\\\ \sigma_n\sigma_1 &amp; \sigma_n\sigma_2  &amp; \dots &amp; \sigma_n^2 \end{bmatrix} $ is the covariance matrix.  Y is a random scalar of the form $Y = \beta_0 + \boldsymbol{\beta^TX} + \epsilon$, where $\epsilon \sim \mathcal{N}(0, \sigma_\epsilon^2)$  How can we go about deriving the joint probability density function(pdf) of the random vector $(Y, \boldsymbol{X^T})$?  The marginal multivariate pdf of $\boldsymbol{X}$ is given by  $ pdf(\boldsymbol{x}) = \frac{1}{\sqrt{(2\pi})^n. \begin{vmatrix} \boldsymbol{\Sigma} \end{vmatrix}}.exp\left[ \frac{-1}{2}\boldsymbol{(x-\mu)}^T \boldsymbol{\Sigma^{-1}}\boldsymbol{(x-\mu)} \right]$  The conditional probability density function of $Y \,\mid \, \boldsymbol{X=x}$ is given by  $pdf(y\mid \boldsymbol{X=x}) = \frac{1}{2\pi\sigma_\epsilon}.exp\left[  \frac{-1}{2\sigma_\epsilon^2}(y - \beta_0 - \boldsymbol{\beta^Tx})^2  \right]   $  To compute the joint probability density function, we can use the fact that   $pdf(y, \boldsymbol{x})=  pdf(y\mid \boldsymbol{X=x}).pdf(\boldsymbol{x})$  I am unable to do the witty manipulation required in the algebra to arrive at the result the joint distribution will have a probability density function of a normal distribution. All questions previously asked seem to address the problem of deriving the conditional density function given the joint probability density function, for example,   Edit:  Multiplying the pdfs, we get $pdf(y, \boldsymbol{x}) = \frac{1}{\sqrt{(2\pi})^n. \begin{vmatrix} \boldsymbol{\Sigma} \end{vmatrix}}. \frac{1}{2\pi\sigma_\epsilon}.exp\left[ \frac{-1}{2}\boldsymbol{(x-\mu)}^T \boldsymbol{\Sigma^{-1}}\boldsymbol{(x-\mu)} + \frac{-1}{2\sigma_\epsilon^2}(y - \beta_0 - \boldsymbol{\beta^Tx})^2 \right]$   Like the comment below mentions, most manipulations dealing with normal pdfs involve completing squares; but since this specific algebraic expression contains $\boldsymbol{x}$ in both terms within the exp term, I'm unsure on how to go about completing the squares.
76920
Suppose for two random variables $X_1$ and $X_2$ $\sigma(X_1) = \sigma(X_2)$. Why $E[Y | X_1] = E[Y | X_2]$ a.e. ?  the set where $X_1= X_1(\omega)$ can be different from the set $X_2= X_2(\omega) $. How can then conditional expectation conditioned on the event can be equal ?
8153
I have 2 random variables $X,Y$ with mean 0 and variance 1, their correlation is $\rho$. I need to prove this inequality  $$\mathrm E\left({\max}\left\{X^2 , Y^2\right\} \right) \leq 1 + \sqrt{1 - \rho^2}$$ I need some pointers as to how to solve this problem. Thanks!
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.