TITLE: Distribution of the maximum of noisy random variables
QUESTION [7 upvotes]: Let $X_1, \ldots, X_N$ be $N$ hidden random iid variables, all with the same standard distribution, let's say uniform $\mathcal{U}(0, 1)$ or Gaussians $\mathcal{N}(0, 1)$ (probably easiest).
I observe $N$ corresponding 'noisy' variables $Z_n = X_n + \mathcal{N}(0, \sigma^2)$.
I know how to derive the distribution of $X^* = \max({X_1, \ldots, X_N})$ (and correspondingly $Z^*$ if the $X$'s are Gaussians).
What I would like to know is how to compute the distribution (or at least the expectation) of $X_{\mathrm{argmax}_n(Z_n)}$.
Intuitively if $\sigma^2$ is small my observed variables will closely follow the hidden ones, and the distribution will be close to $X^*$, while if it is big, they will be dominated by the noise, and the distribution will be the original one of the $X$'s.

REPLY [2 votes]: Let's say $Z_n = X_n + Y_n$ with $X_n \sim N(0,1)$, $Y_n \sim N(0,\sigma^2)$,
and all $X_n$ and $Y_n$ independent.  Let $W = X_{{\rm argmax}_n Z_n} = \sum_n X_n \prod_{j \ne n} I_{Z_n > Z_j}$.  Thus $E[W] = N E[X_1 \prod_{j > 1} I_{Z_1 > Z_j}$.
Now $Z_n \sim N(0,1+\sigma^2)$. Moreover $X_1$ and $I_{Z_1 > Z_j}$ are conditionally independent given $Z_1$, so
$E[W] = N E[E[X_1 | Z_1] \prod_{j > 1} E[I_{Z_1 > Z_j}|Z_1]]
= N E[Z_1 \Phi(Z_1/\sqrt{1+\sigma^2})^{N-1}$.  This is $N \sqrt{1+\sigma^2} E[Z \Phi(Z)^{N-1}]$ where $Z \sim N(0,1)$.