Given a multivariate Gaussian $\mathbf{X} \sim \mathcal{N}(\mathbf{\mu},\Sigma)$, I believe it is a difficult question to find a closed form for $$ \mathbb{E}[ \max\{X_1,\ldots,X_d\}].$$

However, the case I have at hand is perhaps combinatorially nicer: my Gaussian vector is $(X_1,\ldots,X_n)$ where $$X_j = Z_j - \frac{\sum_{i \neq j} Z_i}{n-1}$$

where $Z_1,\ldots,Z_n$ are i.i.d. standard normals.

In the $n = 2$ case, the vector $(X_1,X_2)$ is distributed like $\frac{1}{\sqrt{2}}(Z,-Z)$ where $Z$ is a standard normal, and thus the maximum is simply $\frac{1}{\sqrt{2}} \mathbb{E}[|Z|] = \frac{1}{\sqrt{ \pi}}.$

Can anything like this be done to compute $$c_n = \mathbb{E}\left[\max\{X_1,\ldots,X_n\} \right] $$

for general $n$? Note also that this is the same (up to scaling) of taking $(X_1,\ldots,X_n)$ to be i.i.d. normals conditioned on summing to $0$. It wouldn't be too surprising to me if this is possible to compute precisely (and someone has already done so!)

[this is cross-posted from math stack exchange]


For $S:=\sum_1^n Z_j$, we have $$X_j=Z_j-\frac{S-Z_j}{n-1}=-\frac{S}{n-1}+\frac{n}{n-1}\,Z_j, $$ whence $$\max_1^n X_j=-\frac{S}{n-1}+\frac{n}{n-1}\,\max_1^n Z_j $$ and $$E\max_1^n X_j=\frac{n}{n-1}\,EM_n,\quad M_n:=\max_1^n Z_j. $$ In turn, $$EM_n=\int_0^\infty [P(M_n>x)-P(M_n<-x)]\,dx =\int_0^\infty [1-\Phi(x)^n-\Phi(-x)^n]\,dx, $$ where $\Phi$ is the standard normal cdf. Alternatively, we can write $$EM_n=\int_{-\infty}^\infty x\, dP(M_n<x) =\int_{-\infty}^\infty x\, d(\Phi(x)^n) =n\int_{-\infty}^\infty x\, \Phi(x)^{n-1}\Phi'(x)\,dx; $$ cf. Robert Israel's answer.


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