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Let $X_1, \dots, X_n$ be iid standard normal random variables. Consider the vector $X = (X_1, \dots, X_n)$ and the vector

$Y = \frac{1}{\|X\|}(X_1, \dots, X_k)$

for $k < n$. What is $\mathbb{E}[\|Y\|^2] = \mathbb{E}\left[\frac{\sum_{i=1}^k X_i^2}{\sum_{i=1}^n X_i^2} \right]$?

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    $\begingroup$ By symmetry, the result is exactly $$\color{red}{\bf\frac{k}n}.$$ $\endgroup$ – Did Feb 10 '16 at 11:06
  • $\begingroup$ @Did Thanks, that's exactly the result. Could you please elaborate on your symmetry argument? My reasoning was: The numerator is a Chi-square random variable, with expectation $k$, and the numerator is also a Chi-square random variable with expectation $n$. However, it is not clear to me how to get $k/n$ from this (cc @ecstasyofgold) $\endgroup$ – rodms Feb 10 '16 at 14:27
  • $\begingroup$ @Cm7F7Bb Thanks for your suggestion, I will edit the question. $\endgroup$ – rodms Feb 10 '16 at 14:27
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    $\begingroup$ For every $i\ne j$, the distributions of $(X_i,\|X\|)$ and $(X_j,\|X\|)$ coincide hence $E(X_i^2\|X\|^{-2})=E(X_j^2\|X\|^{-2})$. Summing these from $i=1$ to $n$ yields the expectation of $(X_1^2+\cdots+X_n^2)\|X\|^{-2}=1$ hence $E(X_i^2\|X\|^{-2})=n^{-1}$ for every $i$ and you are done. $\endgroup$ – Did Feb 10 '16 at 14:30
  • $\begingroup$ @Did Thanks so much, amazing answer! :) $\endgroup$ – rodms Feb 10 '16 at 14:34
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I don't know the closed form of this but concentration of measure phenomenon ensures that the expectation will converge to $\frac{k}{n}$ exponentially fast in $k$. So using $k/n$ might be a reasonable approximation.

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