Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
is k greater or less than n? – ved Jul 1 '14 at 3:52
k can be less or equal to n – Ismael Jul 1 '14 at 3:53
up vote 5 down vote accepted

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 $0\lt \frac{X_i}{\sum}\lt 1$.

share|cite|improve this answer

Page 2 gives the solution to this:

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

share|cite|improve this answer
Cool, thanks for the pdf and the url. I'll take a look for the notes there! – Ismael Jul 1 '14 at 4:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.