In a paper I am reading it is written the following:
Let $X = (X_1, \dots, X_n) \sim E_n(\mu, \Sigma, \phi)$ be a elliptical-distributed random vector; let $S = \sum_{i = 1}^n X_i$. Then $$E[X_k \mid S=s] = \mu_k + \frac{\sigma_{k,S}}{\sigma_{S}^2}(s - \mu_S)$$ where $\mu_k$ is the mean of $X_k$, $\mu_S, \sigma^2_S$ are the mean and variance of $S$, and $\sigma_{k,S}$ is the covariance between $X_k$ and $S$
I have tried to prove this but without luck. Can anybody help?
Thoughts
I tried the following:
$$E[X_k \mid S=s] = \int xf_{X_k \mid S = s}(x) dx$$
Since $$f_{X_k \mid S = s}(x) = \frac{f_{(X_k, S)}(x,s)}{f_S(s)}$$
I get
$$E[X_k \mid S=s] = \frac 1{f_S(s)} \int x f_{(X_k, S)}(x,s) dx$$
I know the distribution of the vector $(X_k, S)$ (Since it is a linear transformation of the vector $X$, it is still elliptical distributed and I can compute it's mean and variance; the elliptical generator $\phi$ is still the same).
But how to simplify it further to get to the result stated in the paper?
The result is a special case of a more general result for conditional distribution of elliptical random variables; see theorem 2.18 of the book "Symmetric multivariate and related distributions" by Fang, Ng and Kotz for the statement and proof. Thanks to @gg in the comments for the resource!