<s>[INST] Conditional Normal $X,Y,Z$ are standard normal R.V.
What is the value of $\operatorname{E}[X|X+Y+Z=1]$ and $\operatorname{Var}(X|X+Y+Z=1)$?
I think the first one should be $1/3$ by symmetry but don't know how to tackle the second part.
Any thought on this? Thanks a lot!
 [/INST] If $A$ and $B$ are jointly normal random variables, then the conditional density of $A$ given $B = b$ is a normal density with mean and variance as follows:
$$E[A\mid B = b] = \mu_A + \left.\left.\frac{\text{cov}(A,B)}{\sigma_B^2}
\right(b - \mu_B\right), ~~ 
\text{var}(A\mid B = b) = \frac{\sigma_A^2\sigma_B^2 
- (\text{cov}(A,B))^2}{\sigma_B^2}$$
where ${cov}(A,B)$ is the covariance
of $A$ and $B$.
If $X$, $Y$, $Z$ are jointly normal, then $X$ and $X+Y+Z$ are also
jointly normal with
$$\begin{align*}
E[X+Y+Z] &= \mu_X + \mu_Y + \mu_Z,\\
\text{var}(X+Y+Z) &= \sigma_X^2 + \sigma_Y^2+\sigma_Z^2
+2\text{cov}(X,Y)+2\text{cov}(X,Z)+2\text{cov}(Y,Z)\\
\text{cov}(X,X+Y+Z) &= \sigma_X^2 + 
\text{cov}(X,Y)+\text{cov}(X,Z)
\end{align*}$$
and plugging and chugging in the result of the previous
paragraph gives the desired answers for 
$E[X\mid X+Y+Z=1]$ and $\text{var}(X\mid X+Y+Z=1)$.
Of course, considerable simplifications occur in
the special case when $X$, $Y$, and $Z$ are independent standard normal
random variables.  We have $E[X+Y+Z]=0$, $\text{var}(X+Y+Z)=3$,
$\text{cov}(X,X+Y+Z)=1$, and so
$$E[X\mid X+Y+Z=1] = \left.\left.\frac{1}{3}\right(1-0\right) = \frac{1}{3},
~\text{var}(X\mid X+Y+Z=1) = \frac{1\times 3 - 1^2}{3} = \frac{2}{3}$$
exactly as found by others as well.
</s>