Let $X_1, X_2, \ldots, X_n$ be i.i.d. random variables from a distribution $D$ on the real numbers with mean $\mu$ and variance $\sigma^2$. Assume that the probability of $X_i = 0$ is $0$. What is the expectation $$ \mathbb{E}\left[\frac{(X_1 + X_2 + \cdots + X_n)^2}{X_1^2 + X_2^2 + \cdots + X_n^2}\right]? $$

Or, is it impossible to express in terms of $\mu$ and $\sigma^2$?

Comments: If we factor out $n^2$ from the numerator and $n$ from the denominator, it's equivalent to $$ \mathbb{E}\left[\frac{n^2 \cdot \text{AM}^2}{n \cdot \text{QM}^2}\right] = n \cdot \mathbb{E}\left[ \left(\frac{\text{AM}}{\text{QM}}\right)^2\right], $$ where AM is the arithmetic mean and QM is the quadratic mean. So the answer is positive and between $0$ and $n$, by QM-AM inequality.

This came up in the calculation of a different expectation. I tried calculating a few specific cases, but I haven't gotten an answer yet. It seems possible the answer doesn't just depend on $\mu$ and $\sigma^2$, but on the other hand, I might just be missing the right trick of how to apply i.i.d. Also the numerator relates nicely to the variance, so maybe a decomposition along those lines works.

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    $\begingroup$ As far as I can see the value depends on the actual distribution and not just the parameters $\mu$ and $\sigma$. $\endgroup$ – Kavi Rama Murthy Aug 20 '18 at 6:34
  • $\begingroup$ @KaviRamaMurthy Good! write an answer! :) $\endgroup$ – 6005 Aug 20 '18 at 18:16
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    $\begingroup$ If either $X_1$ or $X_2$ (not necessarily i.i.d.) are symmetric about the origin, then ${\mathbb E}\left[\frac{(X_1+X_2)^2}{X_1^2+X_2^2}\right]=1$, since you could equate this with ${\mathbb E}\left[\frac{(X_1+\varepsilon X_2)^2}{X_1^2+X_2^2}\right]$ where $\varepsilon$ is uniform on $\{1,-1\}$. (I mention this to save anyone else from trying to find a counterexample of this form - this extends to higher $n$ too, if I'm not mistaken if all but at most one $X_i$ is symmetric) $\endgroup$ – Milo Brandt Aug 20 '18 at 23:47

Let us consider all the distributions with mean $0$ and variance $1$ with only two elements in the output. It turns out that these are all of the form $$X=\begin{cases}a & \text{with probability }\frac{1}{1+a^2}\\ -1/a & \text{with probability}\frac{a^2}{1+a^2} \end{cases}.$$ where $a$ is non-zero. Then, you can compute the expectation $\mathbb E\left[\frac{(X_1+X_2)^2}{X_1^2+X_2^2}\right]$ as $$\frac{1}{1+2a^2+a^4}\cdot \frac{4a^2}{2a^2}+\frac{2a^2}{1+2a^2+a^4}\cdot \frac{(a-1/a)^2}{a^2+1/a^2}+\frac{a^4}{1+2a^2+a^4}\cdot \frac{4/a^2}{2/a^2}=2-\frac{2a^2}{1+a^4}.$$ If you evaluate this, for instance at $a=1$ and $a=2$, you get different numbers, so this depends more than just on mean and variance. More or less, mean and variance do not suffice to tell you how likely it is for $X_1$ to be small and $X_2$ to be large or vice versa - which is more or less what this quantity expresses.

  • $\begingroup$ Thanks! The example works but has a couple typos I think: last term should be $\frac{a^\color{red}{4}}{1 + 2a^2 + a^4} \cdot \frac{4/a^2}{\color{red}{2}/a^2}$. $\endgroup$ – 6005 Aug 21 '18 at 1:34
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    $\begingroup$ The expression then simplifies to $2 - \frac{2a^2}{a^4 + 1}$. On a graph you can see as expected it's always between $0$ and $2$ and that it approaches $2$ as $a \to 0$ or $a \to \pm \infty$. Minimum is $1$, when $a = \pm 1$. $\endgroup$ – 6005 Aug 21 '18 at 1:37
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    $\begingroup$ @6005 Ah, thanks for the simplification - I edited it in. I'm rather curious whether this expectation ever drops below $1$ - there seemed to be more structure than I expected in this expression when I was looking for a counterexample. $\endgroup$ – Milo Brandt Aug 21 '18 at 1:44
  • $\begingroup$ About whether the expectation drops below $1$, I asked a follow-up question for the $n = 2$ case (which hopefully generalizes). $\endgroup$ – 6005 Aug 24 '18 at 15:46

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