Bivariate normal distribution and conditional expectation

Question

Let $X$ and $Y$ be two standard normal distributions with correlation $-0.72$. Compute $E(3X+Y\mid X-Y=1)$.

My solution: Conditioning on $X-Y=1$, we have $E(3X+Y\mid X-Y=1) = E(4Y+3\mid X-Y=1) = 3+4E(Y\mid X-Y=1) = 3$.

(1) Is my solution correct? My intuition is that the conditional density of $Y$ remains symmetric about 0 conditioning on $X-Y=1$.

(2) How to solve $E(Y\mid X-Y=1)$ more rigorously?

Thank you, guys!

Answer 1

A hint for your question (2): Define variables $U:=X+Y$ and $V:=X-Y$. Check that $\operatorname{Cov}(U,V)=0$. Since $U$ and $V$ are jointly Gaussian, a covariance of zero implies that $U$ and $V$ are independent. Now express $E(Y\mid X-Y)$ in terms of $U$ and $V$: $$E(Y\mid X-Y)=E\left(\textstyle\frac12(U-V)\mid V\right)$$ Simplifying the RHS will yield an expression in terms of $U$ and $V$, which you can re-express in terms of $X$ and $Y$.

Answer 2

Unfortunately, $$\operatorname{E}[Y \mid X-Y = 1] \ne 0.$$ You can see this if you look at this picture:

Here, the ellipses represent curves of constant bivariate density, and the blue line is the equation $X - Y = 1$. Consequently, along this line, the density is symmetric about $X + Y = 0$, which means that the expected value of $Y$ given that $X - Y = 1$ is $-1/2$, not $0$.