Approach the covariance matrix

Question

I have this problem:

Let the random vector $\binom{X}{Y}\sim N_{2}(\binom{0}{0},\bigl(\begin{smallmatrix} 1 & \rho\\ \rho & 1 \end{smallmatrix}\bigr))$

1) Find the distribution of $Z=X+Y$.

2) Find the distribution of $W=X^2$, the mean and the variance.

3) Calculate $Cov(X,W)$ and $Cov(Z,W)$.

It's the first time that I have found myself analysing this type of function. What the covariance matrix involves? How do I manage it? Thanks in advance for any clarification!

Answer 1

I will not post full answers, so that You can practice a bit - as it is Your first encounter of this subject. Nevertheless I will point out some crucial facts, that are needed here if one need to handle this.

$X=(X_{1},\cdots, X_{k})^{T}$ is multivariate normal distributed if and only if every linear combination of the form $$Y=a_{1}X_{1}+\cdots+a_{k}X_{k}$$ is normally distibuted (well known, one dimensional distribution).

From that we see, that $Z=X+Y$ must be normally distributed. Recall, that if $Z$ is normally distributed, then its distribution is uniquely determined by its mean $\mu$ and variance $\sigma^{2}$. Variance is basically "Covariance matrix" - which in this case is one dimensional. And here comes the second fact.

If $C$ is covariance matrix of $X=(X_{1},\cdots, X_{k})^{T}$ and $a\in \mathbb{R}^{k}$, then $$\mathrm{Cov}(a^{T}X)=a^{T}Ca.$$

Knowing this You can proceed with point a). In b) You are asked about distribution of $W=X^{2}$, which depends only on distribution of $X\sim \mathrm{N}(0,1)$. This is a very famous distribution. This problem does not deal with any multivarious stuff.

Point c) $\mathrm{Cov}(X,W)=\mathrm{C}(X,X^{2})$. You can find it by basic calculations. From linearity of Covariance we have $$\mathrm{Cov}(Z,W)=\mathrm{Cov}(X+Y,X^{2})=\mathrm{Cov}(X,X^{2})+\mathrm{Cov}(Y,X^{2}).$$ Only the last part $\ \mathrm{Cov}(Y,X^{2})=\mathbb{E}(YX^{2}), \ $ is a little bit tricky.

The idea is that we can (without loss of generality) express $Y$ as $$Y=c_{1}X+c_{2}V$$ where $V\sim \mathrm{N}(0,1),$ $$c_{1}^{2}+c_{2}^{2}=1$$ and $X,V$ are independent.

Ask if You will need more explanations.

Answer 2

It is given that $X$ and $Y$ are jointly normal, $EX=EY=0$, $EX^{2}=EY^{2}=1$ and $cov(X,Y)=\rho$ which gives $EXY=\rho$.

$X+Y$ is normal with mean $0$ and its variance is $E(X+Y)^{2}=EX^{2}+EY^{2}+2EXY=1+1+2\rho$.

$P(W \leq w)=P-\sqrt w \leq X \leq \sqrt w)-\int_{-\sqrt w} ^{\sqrt w} f(x)dx$ where $f$ is the standard normal density.

$cov(X,W)=EX^{3}-EXEX^{2}=EX^{3}=0$ by symmetry of standard normal distribution. I will leave the calculation of $cov (Z,W)$ to you.