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Let $X_1, \ldots , X_n$ be independent and identically distributed random variables. Calculate: $$E\left[ \frac{x_1 + \cdots + x_k}{x_1 + \cdots + x_n} \right]$$

I was wondering if I went about this the right way. $$E \left[ \frac{x_1 + \cdots + x_k}{x_1 + \cdots + x_n} \right]$$ $$= \frac{E[x_1 + \cdots + x_k]}{E[x_1 + \cdots + x_n]}, \text{ since each is i.i.d } \implies $$ $$\frac{\mu_1 + \cdots + \mu_k}{\mu_1 + \cdots + \mu_n}$$ $$= \frac{k\mu}{n\mu} = \frac{k}{n}$$ $$= 0 \text{ if $k = n$ and }\frac{k}{n} \text{ if n > k}$$

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    $\begingroup$ Shouldn't your first displayed equation involve $n$? $\endgroup$ Mar 8, 2020 at 21:20
  • $\begingroup$ That's right I typed that incorrectly $\endgroup$
    – Manny
    Mar 8, 2020 at 21:22
  • $\begingroup$ You wrote: "Let $X_1,\ldots,X_n$ be independent and identically distributed random variables." But then instead of $\displaystyle \operatorname E\left[ \frac{X_1+\cdots+X_k}{X_1+\cdots + X_n } \right]$ you wrote $\displaystyle \operatorname E\left[ \frac{x_1+\cdots+x_k}{x_1+\cdots + x_n }\right].$ That is not proper usage. If it is followed, how will one understand an expression like $\Pr(X\le x) \text{ ?} \qquad$ $\endgroup$ Mar 8, 2020 at 22:03
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    $\begingroup$ Your last line, implying $k/n=0$ if $k= n$, is a mistake. $\endgroup$ Mar 9, 2020 at 1:54
  • $\begingroup$ Question is incomplete; it should mention conditions for which $E\left[\frac{\sum_{i=1}^k X_i}{\sum_{i=1}^n X_i}\right]$ exists. See the discussions here and here for example. Did you mean to ask this same question? $\endgroup$ Mar 9, 2020 at 10:34

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It is unfortunately not true that $E[X/Y]=E[X]/E[Y]$ even if $X$ and $Y$ are independent.

However, we can show that $$\tag{*} \mathbb E\left[\frac{X_j}{X_1+\dots+X_n}\right]=\mathbb E\left[\frac{X_1}{X_1+\dots+X_n}\right]=a. $$ (this is due to the fact that $(X_1,\dots,X_n)$ and $(X_{\sigma(1)},\dots,X_{\sigma(n)})$, where $\sigma$ is a permutation, have the same law and $\sum_{i=1}^nX_{\sigma(i)}= \sum_{i=1}^nX_i$.

Also, note that $na=1$, by summing over $j$ in (*).

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    $\begingroup$ It is not instantly clear to me what distributions for $X_1$ are such that the expectations in (*) exist. $\endgroup$ Mar 8, 2020 at 21:51
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    $\begingroup$ @kimchilover The obvious ones are where $X_1 > 0$ a.s. or $X_1 < 0$ a.s. That's not a necessary condition. But on the other hand, if $X_1$ has a continuous density that is nonzero at some points $> 0$ and some points $< 0$, $X_1 + \ldots + X_n$ will have nonzero density at $0$ for some $n$. $\endgroup$ Mar 8, 2020 at 22:21

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