Calculate $\mathbb{E}[X+2Y|X-Y=a]$ where $X,Y$ are correlated standard normal distributions

Question

I was asked the following question during interview.

Given two normally distributed random variables $X$ and $Y$ with mean $0$ and variance $1,$ if there are correlated with $\rho$ where $-1\leq \rho\leq 1,$ then what is $$\mathbb{E}[X+2Y|X-Y=a]?$$

I evaluated it as follows: \begin{align*} \mathbb{E} [X+2Y |X-Y=a] & = \mathbb{E} [X+2(X-a) |X-Y=a] \\ & = \mathbb{E} [3X-a |X-Y=a] \\ & = 3\mathbb{E}(X) - a. \end{align*}

I think my evaluation above is not correct. However, I do not how to solve it.

Answer 1

Your evaluation is not correct (as you suspected).

The last equality fails because $X-Y$ and $3X-a$ are not independent.

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Observe that on base of bilinearity and symmetry of covariance: $$\mathsf{Cov}\left(X+Y,X-Y\right)=\mathsf{Cov}\left(X,X\right)-\mathsf{Cov}\left(X,Y\right)+\mathsf{Cov}\left(Y,X\right)-\mathsf{Cov}\left(Y,Y\right)=$$$$\mathsf{Var}X-\mathsf{Var}Y=1-1=0$$showing that $X+Y$ and $X-Y$ are uncorrelated, hence in this context of normal distributions independent.

Further we can write: $$X+2Y=\frac{3}{2}\left(X+Y\right)-\frac{1}{2}\left(X-Y\right)$$ so that: $$\mathbb{E}\left[X+2Y\mid X-Y=a\right]=\frac{3}{2}\mathbb{E}\left[X+Y\mid X-Y=a\right]-\frac{1}{2}\mathbb{E}\left[X-Y\mid X-Y=a\right]=$$$$\frac{2}{3}\mathbb{E}\left[X+Y\right]-\frac{1}{2}a=-\frac{1}{2}a$$

Answer 2

Let $U=X+Y$ and $V=X-Y$. Note that $EUV=E(X-Y)(X+Y)=E(X^{2}-Y^{2})=1-1=0$. Hence $U$ and $V$ are jointly normal independent random variables with mean $0$.

Now $E(X+2Y|V)=E(\frac {U+V} 2+2\frac {U-V} 2|V)=-\frac { V} 2$. Hence $E(X+2Y|X-Y=a)=-\frac {a} 2$.