# Calculating the variance of the sum of two correlated variables

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?

• What are $X_1$ and $X_2$? I suspect the index means $X=\langle X_1,X_2\rangle$, but you should say so if so . Apr 19, 2020 at 0:44
• I suppose that is the case - sorry for not making it explicit. Apr 19, 2020 at 0:53
• You know from the means that $\mathsf E(X_1-X_2)=0$, but that does not say that $X_1-X_2=0$ for sure. You are not being asked to evaluate $\mathsf{Var}(X_1+X_2)$, rather you are being asked to evaluatethe conditional variance: $\mathsf{Var}(X_1+X_2\mid Z{=}0)$. Apr 19, 2020 at 1:46

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}