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Let $X_i$ be IID and $\bar{X} = \sum_{i=1}^{n} X_i$. $$ E\left[\frac{X_i}{\bar{X}}\right] = \ ? $$ It seems obvious, but I am having trouble formally deriving it.

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Let $X_1,\dots,X_n$ be independent and identically distributed random variables and define $$\bar{X}=\frac{X_1+X_2\dots+X_n}{n}.$$

Suppose that $\Pr\{\bar{X}\ne 0\}=1$. Since the $X_i$'s are identically distributed, symmetry tells us that, for $i=1,\dots n$, the (dependent) random variables $X_i/\bar{X}$ have the same distribution: $$ \frac{X_1}{\bar{X}} \sim \frac{X_2}{\bar{X}} \sim \dots \sim \frac{X_n}{\bar{X}}. $$ If the expectations $\mathrm{E}[X_i/\bar{X}]$ exist (this is a crucial point), then $$ \mathrm{E}\left[ \frac{X_1}{\bar{X}} \right] = \mathrm{E}\left[ \frac{X_2}{\bar{X}} \right] = \dots = \mathrm{E}\left[ \frac{X_n}{\bar{X}} \right], $$ and, for $i=1,\dots,n$, we have $$ \begin{align} \mathrm{E}\left[ \frac{X_i}{\bar{X}} \right] &= \frac{1}{n} \left( \mathrm{E}\left[ \frac{X_1}{\bar{X}} \right] + \mathrm{E}\left[ \frac{X_2}{\bar{X}} \right] + \dots + \mathrm{E}\left[ \frac{X_n}{\bar{X}} \right] \right) \\ &= \frac{1}{n}\,\mathrm{E}\left[ \frac{X_1}{\bar{X}} + \frac{X_2}{\bar{X}} + \dots + \frac{X_n}{\bar{X}} \right] \\ &= \frac{1}{n}\,\mathrm{E}\left[ \frac{X_1+X_2+\dots+X_n}{\bar{X}} \right] \\ &= \frac{1}{n}\,\mathrm{E}\left[ \frac{n\bar{X}}{\bar{X}} \right] \\ &= \frac{n}{n}\,\mathrm{E}\left[ \frac{\bar{X}}{\bar{X}} \right] = 1. \end{align} $$

Let's see if we can check this by simple Monte Carlo.

x <- matrix(rgamma(10^6, 1, 1), nrow = 10^5)
mean(x[, 3] / rowMeans(x))

[1] 1.00511

Fine, and the results don't change much under repetition.

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  • 3
    $\begingroup$ (+1) The conclusion that $E[X_i/\bar X]$ does not exist is true, but requires a subtler argument than any of those you have yet linked to, because $X_i$ and $\bar X$ are not independent. $\endgroup$ – whuber Oct 14 '17 at 22:26
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    $\begingroup$ @whuber: Can you expand this a little bit, Bill? I mentioned the dependence of $X_i$ and $\bar{X}$ in one of the comments to the linked question. Also, Xi'an's answer addresses the $n=2$ case with a simple transformation. He also gave the distribution of $X_i/\bar{X}$ in one of his comments. Thank you for your thoughts on this. $\endgroup$ – Zen Oct 14 '17 at 23:35
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    $\begingroup$ @whuber: I think my explanation works out since$$X_i/\bar{X}=n/\{1+X_2/X_1+\cdots+X_n/X_1\}$$which is $n/\{1+(n-1)Z\}$, $Z$ being a standard Cauchy. No dependence involved. $\endgroup$ – Xi'an Oct 15 '17 at 12:24
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    $\begingroup$ @Xi'an: did you use here that (consider the $n=3$ case), since $U=X_2/X_1$ and $V=X_3/X_1$ are standard Cauchy, then $(U+V)/2$ is also standard Cauchy? But that's not true because $U$ and $V$ are not independent, right? $\endgroup$ – Zen Oct 15 '17 at 13:44
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    $\begingroup$ @Zen: However, $(X_2+\cdots+X_n)$ and $X_1$ are independent Normal variates, hence $(X_2+\cdots+X_n)/X_1$ is a Cauchy, if with a $\sqrt{n-1}$ scale rather than $n-1$. $\endgroup$ – Xi'an Oct 15 '17 at 14:37

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