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Let $X_1, X_2, \dots, X_n$ be $n$ positive iid random variables. Then show that $$E\left(\frac{\sum_{j=1}^k X_j}{\sum_{i=1}^{n} X_i}\right) = \frac{k}{n}.$$

Because of the linearlity of the expectation I known that $E\left(\frac{\sum_{j=1}^k X_j}{\sum_{i=1}^{n} X_i}\right) = \sum_{j=1}^k E\left(\frac{X_j}{\sum_{i=1}^{n} X_i}\right)$, so it's enougth to show $E\left(\frac{X_1}{\sum_{i=1}^{n} X_i}\right) = \frac{1}{n}$. But I'm unable to deal with the $X_i$ in the denominator.

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  • $\begingroup$ is k greater or less than n? $\endgroup$ – ved Jul 1 '14 at 3:52
  • $\begingroup$ k can be less or equal to n $\endgroup$ – Ismael Jul 1 '14 at 3:53
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Hint: In addition to the linearity property you mentioned, use the following facts:

$1$) By symmetry we have $E\left(\frac{X_i}{\sum}\right)=E\left(\frac{X_j}{\sum}\right)$.

$2$) $E\left(\frac{\sum}{\sum}\right)=E(1)=1$. This, $1$), and linearity forces $E\left(\frac{X_i}{\sum}\right)=\frac{1}{n}$.

Existence is not a problem since $0\lt \frac{X_i}{\sum}\lt 1$.

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Page 2 gives the solution to this:

http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-436j-fundamentals-of-probability-fall-2008/recitations/MIT6_436JF08_rec05.pdf

The crux of the problem is covered in the answer by André Nicolas.

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  • $\begingroup$ Cool, thanks for the pdf and the url. I'll take a look for the notes there! $\endgroup$ – Ismael Jul 1 '14 at 4:25

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