I am trying to solve the following question but I am totally lost on how to approach it. I know the classical division of joint pdf to marginal pdf is indeed the correct answer but I believe there has to be an easier way to solve the question.
Say we have a joint Gaussian pdf:
$$ X\sim\mathcal{N}(\begin{bmatrix} 0 \\ 0 \\ 2 \end{bmatrix} , \begin{bmatrix} 1 & 1 & 2 \\ 1 & 4 & 4 \\ 2 & 4 & 16 \end{bmatrix} ) $$
Question is, $$E[X_1 | X_2 + X_3 = 0]$$
What I did so far is to find the pdf of $X_2 + X_3 = Z \sim \mathcal{N}(2, 28)$, now I know that formally,
$$f_{X_1 | X_2 + X_3}(X_1 | X_2 + X_3) = \frac{f_{X_1, X_2,X_3}(X_1, X_2,X_3)}{f_Z(z)}$$
But is there an easier way to calculate this other than doing the algebra? Another question is, do I have to put $x_1 + x_2 = 0$ everywhere I see after I calculate the marginal pdf division above due to the condition?
I guess an easy solution is as follows;
Lets say we have a new bivariate Gaussian such that; $$Y = \begin{bmatrix} X_1 \\ Z \end{bmatrix} \sim \mathcal{N}(\begin{bmatrix} 0 \\ 2 \end{bmatrix}, \begin{bmatrix} 1 & 3 \\ 3 & 28 \end{bmatrix}), $$
Covariance is $3$ because, $E[X_1Z] = E[X_1(X_2 + X_3)] = 3$. Then we can just use the following formula;
$$E[X_1 | Z = 0] = \mu_1 + \Sigma_{12}\Sigma_{22}^{-1}(0-\mu_2) = \frac{-3}{14}$$
without using the formal conditional pdf method.