Calculating the variance of the sum of two correlated variables

Question

I am currently struggeling with the following problem:

Let $X$ be a bivariate Normal random variable (taking on values in $R^2$) with mean $μ=(1,1)$ and covariance matrix $$\Sigma=\begin{bmatrix}3&1\\1&2\end{bmatrix}$$

What is the variance of the conditional distribution of $Y=X_1+X_2$ given $Z=X_1−X_2=0$?

I would have said that the anwser is $7$, as $Var(A+B) = Var(A) + Var(B) + 2Cov(A,B)$. However, the anwser is $6.666$. Why is that the case?

Also, does saying that $Z=0$ provide any new information? I thought we already knew this from the mean, regardless of what the covariance was. Or is that not the case?

Answer 1

You want $\mathsf {Var}(X_1+X_2\mid X_1=X_2)$ when $\left[\begin{smallmatrix}X_1\\X_2\end{smallmatrix}\right]\sim\mathcal{N}(\left[\begin{smallmatrix}1\\1\end{smallmatrix}\right],\left[\begin{smallmatrix}3&1\\1&2\end{smallmatrix}\right])$

$$\begin{align}\mathsf{Var}(X_1+X_2\mid X_1=X_2)&=\mathsf E((2X_1)^2\mid X_1=X_2)-\mathsf E(2X_1\mid X_1=X_2)^2\\[1ex]&=\dfrac{4\int_\Bbb R x^2\phi(\left[\begin{smallmatrix}x\\x\end{smallmatrix}\right])~\mathrm dx}{\int_\Bbb R\phi(\left[\begin{smallmatrix}x\\x\end{smallmatrix}\right])~\mathrm d x}-\dfrac{4\left(\int_\Bbb R x\phi(\left[\begin{smallmatrix}x\\x\end{smallmatrix}\right])~\mathrm d x\right)^2}{\left(\int_\Bbb R\phi(\left[\begin{smallmatrix}x\\x\end{smallmatrix}\right])~\mathrm d x\right)^2}\end{align}$$

Where $$\begin{align}\phi(\left[\begin{smallmatrix}x\\y\end{smallmatrix}\right]) &= \dfrac{\exp(-\tfrac 12\left[\begin{smallmatrix}x-1& y-1\end{smallmatrix}\right]\left[\begin{smallmatrix}3&1\\1&2\end{smallmatrix}\right]^{-1}\left[\begin{smallmatrix}x-1\\y-1\end{smallmatrix}\right])}{\sqrt{(2\pi)^2\left\lvert\begin{smallmatrix}3&1\\1&2\end{smallmatrix}\right\rvert}}\\[3ex]\phi(\left[\begin{smallmatrix}x\\x\end{smallmatrix}\right])&=\dfrac{\exp(-\tfrac 3{10}(x-1)^2)}{2\pi\sqrt{5}}\end{align}$$