Covariance matrix of three random variables

Question

Let $X$, $Y$, $Z$ be random variables such that $X \sim N(0,1)$ and conditional on $X = x$, $Y$ and $Z$ are $i.i.d. N(x,1)$.

Find $E[X]$, $E[Y]$, $E[Z]$, $Var[X]$, $Var[Y]$ and $Var[Z]$. Find the covariance matrix of the random vector $(X,Y,Z)$.

My attempt:

Since we know $X\sim N(0,1)$, $E[X]$ and $Var[X]$ are easy to find $$E[X]=0$$ $$Var[X]=1$$

Given that conditional on $X=x$, $Y$ and $Z$ are i.i.d. $N(x,1)$. By Adam's Law, $$E[Y]=E[E[Y|X=x]]=E[X]=0=E[Z]$$

And using Eve's Law, $$Var[Y]=E[Var[Y|X]]+Var[E[Y|X]]=E[1]+Var[X]=2=Var[Z]$$

Denote the covariance matrix as $K$ $$K=\begin{bmatrix} Var[X] &Cov[X,Y] &Cov[X,Z] \\ Cov[X,Y] &Var[Y] &Cov[Y,Z] \\ Cov[X,Z] &Cov[Y,Z] &Var[Z] \end{bmatrix}$$ Since $Cov[X,Y]=E[(X-E[X])(Y-E[Y])]=E[XY]-E[X]E[Y]$ $$K=\begin{bmatrix} 1 &E[XY]-E[X]E[Y] &E[XZ]-E[X]E[Z] \\ E[XY]-E[X]E[Y] &2 &E[YZ]-E[Y]E[Z] \\ E[XZ]-E[X]E[Z] &E[YZ]-E[Y]E[Z] &2 \end{bmatrix}$$ $$K=\begin{bmatrix} 1 &E[XY] &E[XZ] \\ E[XY] &2 &E[YZ] \\ E[XZ] &E[YZ] &2 \end{bmatrix}$$ But how do I find $E[XZ]$, $E[XY]$, $E[YZ]$, ...?

Answer 1

But how do I find $E[XZ]$, $E[XY]$, $E[YZ]$, ...?

1. $E(XX) = E(X^2) = Var(X) + E(X)^2 = 1.$
2. $E(XY)= E(E(XY|X))= E(X E(Y|X)) = E(X X) =1.$ Similarly $E(XZ) = 1$.
3. $E(YZ) = E( E(YZ|X)) = E( E(Y|X) E(Z|X) ) = E(XX) = 1.$

Answer 2

In effect you have $X,V,W$ i.i.d. $N(0,1)$ with $Y=X+V, Z=X+W$, with $Y$ and $Z$ having identical but not independent distributions

so

• $E[X]=E[V]=E[W]=0$
• $E[X^2]=E[V^2]=E[W^2]=1$
• $E[XV]=E[XW]=E[VW]=0$

and you can conclude

• $E[Y]=E[Z]=E[X+W]=0+0=0$
• $E[Y^2]=E[Z^2]=E[X^2+2XW+W^2]=1+2\times 0+1=2$
• $E[XY]=E[XZ]=E[X^2+XW]=1+0=1$
• $E[YZ]=E[X^2+XV+XW+VW]=1+0+0+0=1$

making the covariance matrix $\begin{bmatrix} 1 &1 &1 \\ 1 &2 &1 \\ 1 &1 &2 \end{bmatrix}$