Given $(X,Y)$ normal vector with: $$ E\left(X\right)=0,\,E\left(Y\right)=0,\,Var\left(X\right)=4,\,Var\left(Y\right)=1 $$ and $Cov\left(X,Y\right)=-1$, I was asked to to calculate $f_Z(2)$ where $Z=X+2Y$. In the solution they said that: $$ Z\sim N(E(X+2Y),Var(X+2Y)) $$ It made me think - Is it true to say that the sum of two random variables $X$ and $Y$ that have normal distribution is also normal distribution? I know that if $X$ and $Y$ are independent that is true and even $X+Y\sim N(\mu_X+\mu_Y,\sigma_X^2+\sigma_Y^2)$ but can we says the $X+Y$ is normal if we don't know that they are independent? If not, how did they got $Z\sim N(E(X+2Y),Var(X+2Y))$? If yes, is there a general formula of $X+Y\sim N(\mu_X+\mu_Y,\sigma_X^2+\sigma_Y^2)$ or something similar?
123415
I know how to prove the statement when two r.v. are independent. But how to prove it when the r.v. are correlated?  Edit:or counter example?
5168
I understand the proof that  $$Var(aX+bY) = a^2Var(X) +b^2Var(Y) + 2abCov(X,Y), $$  but I don't understand how to prove the generalization to arbitrary linear combinations.  Let $a_i$ be scalars for $i\in {1,\dots ,n}$ so we have a vector $\underline a$, and $\underline X = X_i,\dots ,X_n$ be a vector of correlated random variables. Then $$ Var(a_1X_1 + \dots a_nX_n) = \sum_{i=1}^n a_i^2 \sigma_i^2 + 2 \sum_{i=1}^n \sum_{j&gt;i}^n a_i a_j \text{ Cov}(X_i,X_j)$$  How do we prove this? I imagine there are proofs in the summation notation and in vector notation?
68380
Let $A$ and $B$ be two constant matrices and let $x$ and $ y$ be two random vectors, what is the general formula for $Var(Ax+By)$? I know the formula for when $x$ and $y$ are scalar random variables and $A$ and $B$ are constants, but what about the matrix case?
9232
Given   $$ f_{XY}(x,y) = \frac{1}{2\pi \sqrt{1-\rho^2}} \exp \left( -\frac{x^2 +y^2 - 2\rho xy}{2(1-\rho^2)} \right) $$  $Y = Z\sqrt{1-\rho^2} + \rho X$  And   $$ f_{XZ}(x,z) = \frac{1}{2\pi } \exp \left( -\frac{x^2 +z^2}{2} \right) $$  Show that $P(X&gt;0,Y&gt;0)= \frac{1}{4}+\frac{1}{2\pi}(\arcsin \rho) $  I'm supposed to use the fact that X and Z are independent standard normal random variables, but I don't quite understand how. Any help would be greatly appreciated.
53767
The question states:      Consider a set of random variables $X_i$, where $i=1,...n$. Each $X_i$ is   normally distributed with mean $0$ and variance $1$, i.e. $X_i$ are $\mathcal N(0,1)$.   What is the mean and the variance of the random variable $Y$, where   $Y=X_1+...+X_n$.   How do I do this?
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It's a little tricky to me, I am not sure how to proceed...my atempt was to stick strictly to the definitions given and the formula for the (continuous) expected value.  Prove that $E[e^{\mu+\sigma Z}\mathbb{1}_{Z&gt;-d}]=e^{\mu+\frac{\sigma ^2}{2}}\Phi(d+\sigma)$, where $\Phi$ is the standard normal distribution function, $Z  \sim N(0,1)$(i.e. standard normal distribution) and $\mathbb{1}_{Z&gt;-d}$ denotes the following indicator function of an event such that, $$\mathbb{1}_{Z&gt;-d}=1 \text{ when } Z &gt;-d \text{ and, } \mathbb{1}_{Z&gt;-d}=0 \text{  otherwise}$$ during your work, denote the standard normal density function as $\phi$.  I only got up to thinking $e^{\mu+\sigma Z}\mathbb{1}_{Z&gt;-d}$ is essentially just $e^{\mu+\sigma Z}$ if $Z&gt;-d$ for some $d$ and $0$ otherwise. And so, the expected value is defined (for continuous random variables) $E[X]= \int_{-\infty}^{\infty}xf(x)dx$ where $f(x)$ is the probability for each random variable $x$. I thought this would lead me to using $\phi$ and ultimately leading me to $\Phi$ in my final conclusion as in the RHS. But I am not sure about how to proceed, would someone please help me? It would be great if you could explain the steps to me...thank you very much in advance...
60260
Given a multivariate normal distribution  $f(x) = \frac{1}{\sqrt{(2 \pi)^n|\Sigma|}} \times \exp\left( -\frac{1}{2} (x-\mu)' \Sigma^{-1} (x-\mu)\right)$  how can I prove that $ (x-\mu)' \Sigma^{-1} (x-\mu) \sim \chi^2_{p} $?
28039
Let $\mathbf{X}\sim \mathcal{N}(0,I)$ be an $n-$dimensional standard Gaussian vector. We know that  $$\begin{align} ||\mathbf{X}||^2&amp;=\sum_{i=1}^nX_i^2\sim\chi^2_n\\ ||\mathbf{X}||&amp;=\sqrt{\sum_{i=1}^nX_i^2}\sim\chi_n\\ \end{align}$$  Suppose the components are correlated, i.e., we consider $\mathbf{Y}\sim \mathcal{N}(0,\Sigma)$ where $\Sigma$ is positive definite. We know that $||\mathbf{Y}||^2$ has the . Is there also a corresponding generalized chi distribution for $||\mathbf{Y}||$? If so, is there some asymptotic result on the mean of such a distribution for $n\to\infty$? This is related to       But here the components of the random vector are neither independent or identically distributed.
44122
Can you help me?  Find the conditional expectation $\mathbb{E}[X|Y]$ if $(X,Y)$ possesses a bivariate normal distribution.  Is $\mathbb{E}[X|Y=y]=\mu_X+\sigma_X\rho(\frac{\displaystyle y-\mu_Y}{\displaystyle \sigma_Y})$ the solution?  My question: Is the same $\mathbb{E}[X|Y=y]$ and $\mathbb{E}[X|Y]$?
51403
Let $X\sim N(0,1)$ and let $Z$ be a random variable independent of $X$ such that:  \begin{equation} \Pr(Z=z) = \begin{cases} \frac{1}{2} &amp; \mbox{if $z = -1$ or $z=1$}, \\ \\ 0 &amp; \mbox{otherwise}. \end{cases} \end{equation}   Define $Y = ZX$. Show that $Y \sim N(0,1)$. Determine $\operatorname{Cor}(X,Y)$. Are $X$ and $Y$ independent?   Progress:  I'm basically stuck at the beginning. I got as far as $P(ZX &lt; y) = P(Z &lt; y/X)$, which I'm not even sure is correct. I'm also not clear on what it means by defining $Y = ZX$ - does it mean I need to state the probability density function or something? And how do I show that $Y \sim N(0,1)$?
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.
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 $N(E(X|Y&gt;M),var(X|Y&gt;M))$?
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.
80741
I'm trying to understand my teacher's solution to this problem:     Let $X_1,X_2,...$ sequence of iid random variables with expected value   $\mu$. Define $Y_n = \frac{2}{n(n+1)} \sum_{j=1}^{n} jX_j$. Prove that   $Y_n \to_\mathbb{P} \mu$.   His solution:  Let $\sigma^2 = \rm{Var}(X_i)$.  $E(Y_n) = \frac{2}{n(n+1)} \sum_{j=1}^{n} jE(X_n) = \mu$.  $\displaystyle\rm{Var}(Y_n) = \frac{4}{n^2(n+1)^2} \sum_{j=1}^{n} j^2 \rm{Var}(X_n) = \frac{4\sigma^2}{n^2(n+1)^2}\bigg(\frac{n(n+1)(2n+1)}{6}\bigg) = \frac{2\sigma^2}{3}\bigg(\frac{2n+1}{n(n+1)}\bigg) \leq \frac{2\sigma^2}{3}\bigg(\frac{2}{3}\bigg) = \sigma^2 ,\forall n$  Then it follows by Tchebychev's Weak Law of Large Numbers.  My doubt is in the last step. Isn't  Tchebychev's WLLN only true for uncorrelated random variables? In this case, $\mathbb{E}(Y_n Y_m) \neq \mu^2$, since they aren't independent, so $\rm{Cov}(Y_n,Y_m) \neq 0$, right?  Thanks.  Edit: As u/clarinetist noted, there's a problem in this exercise, since the WLLN states the convergence of the arithmetic mean of random variables, whose expected value isn't $\mu$.
117051
Let $(X, Y)$ have a normal distribution with mean $(\mu_X, \mu_Y)$, variance $(\sigma_X^2, \sigma_Y^2)$ and correlation $\rho$. I want to know the corresponding marginal densities.  All I found so far was the well-known density expressions for $X\sim N(\mu_X, \sigma_X^2)$ and $Y\sim N(\mu_Y, \sigma_Y^2)$, but isn't that just for $X \perp Y$? Shouldn't $\rho$ appear in the expressions? How do I calculate the marginals of any $X$ and $Y$? I think using the definition will end up in an integral that cannot be solved analytically…
19186
Let $X_1$ and $X_2$ be two normal random variables. Write $X_1\sim N(\mu_1, \sigma^2_1)$ and $X_2\sim N(\mu_2, \sigma^2_2)$, to fix ideas.   Consider the corresponding log-normal random variables: $Z_1 = \exp(X_1)$, $Z_2 = \exp(X_2)$.  Question: what is the distribution of the product of the two random variables, i.e., the distribution of $Z_1Z_2$?  If the normal random variables $X_1, X_2$ are independent, or they have a bivariate normal distribution, the answer is simple: we have $Z_1Z_2 = \exp(X_1+X_2)$ with the sum $X_1+X_2$ normal, hence the product $Z_1Z_2$ is still lognormal.  But suppose that $X_1, X_2$ are generally $not$ independent, say with correlation $\rho$. What can we say about the distribution of $Z_1Z_2$?
30482
I read from my textbook that $\text{cov}(X,Y)=0$ does not guarantee X and Y are independent. But if they are independent, their covariance must be 0. I could not think of any proper example yet; could someone provide one?