Let X be N($\mu$,Q) with $\mu$= (1,1) and Q= $\begin{bmatrix} 3 & 1 \\ 1 & 2 \\ \end{bmatrix}$. I would like to find the distribution of Y = $X_1 + X_2$ given Z = $X_1 - X_2 = 0$ .
Therefore I should obtain that Y = $2X_1$ or, equally, that Y = $2X_2$, but since they have a different variance I don't understand how it could be possible.
Continuing with the solution I should find $f_{Z=0}(y)$ being normal with $\mu=2$ and $\sigma^2=20/3$. But still I'm not able to obtain this result. Could someone help?
All variables in your problem are normal, therefore all the information you need is encoded in the covariance matrix (and means).
You can first compute: $$Var(Y) = Var(X_1+X_2) = 3 + 2 + 2 = 7$$ and $$Var(Z) = Var(X_1-X_2) = 3 + 2 - 2 = 3$$ and finally: $$Cov(Y,Z) = Cov(X_1+X_2,X_1-X_2) = 3 - 1 + 1 - 2 = 1$$ This gives joint distribution of $Y,Z$ being normal $ \mathcal{N}((2,0), \begin{bmatrix} 6 & 5 \\ 5 & 4 \\ \end{bmatrix} ) $
From this you can find pdf of $Y|Z=0$ explicitly and read off mean and variance.