Conditional Expectation and Variance Question

Question

So I have a question I'm absolutely stumped with:

Given two random variables, $X$ and $Y$ , with common variance, $\sigma ^2$, where $\mathbb{E}(Y|X) = X + 1$, find $\rho (X,Y)$.

So I obviously need to find $Cov(X,Y)$ to work out correlation. And by extension of the law of total probability I know that,

$\mathbb{E}(Y)=\mathbb{E}(\mathbb{E}(Y|X))= \mathbb{E}(X)+1$.

Now, $cov(X,Y)=\mathbb{E}(XY)-\mathbb{E}(X)\mathbb{E}(Y)$,

which I don't see how I can use without $\mathbb{E}(EX)$, and the only other approach I can think of to use is

$Var(X+Y) = Var(X) + Var(Y) + 2Cov(X,Y) + Var(Y)= 2\sigma^2 +2Cov(X,Y)$; or,

$Var(Y) = \mathbb{E}(Var(Y|X)) + Var(\mathbb{E}(Y|X)) \Rightarrow \mathbb{E}(Var(Y|X))=0$

neither of which seem particularly usefule.

Answer 1

We can evaluate $\operatorname E(XY)$ using the law of total expectation. We have that $$ \operatorname E(XY)=\operatorname E(\operatorname E(XY\mid X))=\operatorname E(X\operatorname E(Y\mid X))=\operatorname E(X^2+X)=\operatorname EX^2+\operatorname EX $$ and $$ \operatorname{Cov}(X,Y)=\operatorname E(XY)-\operatorname E(X)\operatorname E(Y)=\operatorname EX^2+\operatorname EX-(\operatorname EX)^2-\operatorname EX=\sigma^2. $$ Hence, $\rho(X,Y)=1$.

Answer 2

You were on the right track. You have to use the additional formula about conditional expectation $$\mathbb{E}[f(X,Y)]=\mathbb{E}[\mathbb{E}[f(X,Y)|X]]$$ Thus $$\mathbb{E}[Y]=\mathbb{E}\mathbb{E}[Y|X]$$ If you use the formula of the covariance, you have this first equality: $$cov(X,Y)=\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]=\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[X+1]$$

Then you use again the conditionnal expectation for the first term: $$\mathbb{E}[XY]=\mathbb{E}[\mathbb{E}[XY|X]]=\mathbb{E}[X\mathbb{E}[Y|X]]=\mathbb{E}[X^2+X]$$

You plug in this second equality in the first to get: $$\mathbb{E}[XY]-\mathbb{E}[X]^2-\mathbb{E}[X]=\mathbb{E}[X^2]-\mathbb{E}[X]^2=\sigma^2$$