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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.

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    I would be cautious regarding the *answers* posted at the linked question. I provided the link simply to connect the two questions since they are related. Also, you should reference the answer using the poster's name, since the ordering can change based on various factors (including *randomly*).2011-05-13

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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 $Z \sim N(0,1)$.

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    $E[X_1|Z_1] = \frac{Z_1}{1+\sigma^2}$ and *not simply $Z_1$* as written above! This is because conditioned on $Z_1$, $X_1$ and $Y_1$ are no longer independent. One can see $X_1,Y_1$ as a multivariate normal distribution with a diagonal covariance, and $X_1,Z_1$ as an affine transformation of $X_1,Y_1$, and then use the formula of conditional distributions of multivariate normal distribution to reach that result.2011-05-16