If $X_1,X_2 \sim N(\mu,\sigma^2)$ , then find $E[(X_1+X_2)^2|X_1=X_2]$

Question

If $X_1,X_2 \sim N(\mu,\sigma^2)$ , then find $E[(X_1+X_2)^2|X_1=X_2]$ It seems to me that the expectation will be $4 \mu^2$ but I am going wrong somewhere. How to tackle these types of problems where the condition given is a linear restriction on $X_1,X_2$?

Answer 1

The expression $E\,[(X+Y)^2\mid X=Y]$ is ambiguous, as user @Did justifies below.

I am treating it as $E\,[(X+Y)^2\mid X-Y=0]$ in my answer.

Assume that $(X,Y)$ is jointly normal with mean vector $(\mu,\mu)^T$ and dispersion matrix $\begin{bmatrix}\sigma^2 & \rho\sigma^2 \\ \rho\sigma^2 & \sigma^2 \end{bmatrix}$.

Define $U=X+Y$ and $V=X-Y$, so that $(U,V)$ is also jointly normal with mean vector $(2\mu,0)^T$ and dispersion matrix $\begin{bmatrix}2\sigma^2(1+\rho) & 0\\0 & 2\sigma^2(1-\rho) \end{bmatrix}$. As $\operatorname{cov}(U,V)$ vanishes, $U$ and $V$ are independently distributed.

Hence, $$E\,[(X+Y)^2\mid X-Y=0]=E\,[U^2\mid V=0]=E\,[U^2]=2\sigma^2(1+\rho)+4\mu^2$$