TITLE: Distribution of joint Gaussian conditional on their sum
QUESTION [8 upvotes]: Let $X = (X_1, X_2, \dots, X_n)$ be jointly Gaussian with mean vector $\mu$ and covariance matrix $\Sigma$. Let $S$ be their sum.
I know that the distribution of each $X_i \mid S = s$ is also Gaussian.
When $n=2$, I know that
$$
E\left( X_1\mid S = s \right) = s \frac{\sigma_1^2}{\sigma_1^2 + \sigma_2^2}
$$
and
$$
V\left(X_1\mid S = s \right) = \frac{\sigma_1^2\sigma_2^2}{\sigma_1^2 + \sigma_2^2}
$$
(see here and here). I could probably work out analogous expressions for an arbitrary $n$ if I sat down with a pencil and paper and worked at it for a bit.
What I want to know is, what is the distribution of $X$ given $S = s$?
I know that this can't be Gaussian, since the sum is bounded. It's clearly not Dirichlet or anything Dirichlet-esque, since the marginal distributions are Gaussian. But beyond that I don't have a clue.

REPLY [2 votes]: The distribution of $(\boldsymbol X | S = s)$ is still jointly normal but degenerate. Let $\boldsymbol T = (1, 1, \dots, 1)^t$ and let $\boldsymbol X$ and $\boldsymbol \mu$ also be column vectors. Then $(X_1, \dots, X_n, \boldsymbol T^t \boldsymbol X)$ is jointly normal as an affine transform of a jointly normal distribution, and we can use the general formula for a conditional distribution of components of a jointly normal distribution:
$$(\boldsymbol X | \boldsymbol T^t \boldsymbol X = s) \sim
\mathcal N \!\left(
 \boldsymbol \mu + \frac
   {s - \boldsymbol \mu^t \boldsymbol T}
   {\boldsymbol T^t \Sigma \boldsymbol T}
  \Sigma \boldsymbol T,
 \Sigma - \frac 1 {\boldsymbol T^t \Sigma \boldsymbol T}
  \Sigma \boldsymbol T (\Sigma \boldsymbol T)^t
\right).$$
$\boldsymbol T$ is an eigenvector of the covariance matrix with the eigenvalue $0$.