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