Let $X_1, ..., X_n$ be independent random variables with the same distribution such that $E(X_1^{-1})$ exists. Show that for $m \leq n$ the expected value
$E(\frac{S_m}{S_n}) = \frac{m}{n}$,
where $S_n = X_1 + ... + X_n$.
Writing this directly, I have:
$E(\frac{S_m}{S_n}) = E(\frac{X_1 + ... + X_m}{X_1 + ... + X_m + ... + X_n})$
I know that the expected value of a sum of independent random variables is the sum of their expected values and I understand that since they have the same distribution, their expected values are the same. However, I don't know how to deal with such fraction, any suggestions?
By linearity of expectation you know that
$$ \mathbb{E}[S_m/S_n] = \sum_{i=1}^m \mathbb{E}[X_i/S_n] = m \mathbb{E}[X_1/S_n]$$ because the $X_i$ are i.i.d. Finally note that $$ \mathbb{E}[S_n/S_n] = \mathbb{E}[1] = 1 = \sum_{i=1}^n \mathbb{E}[X_i/S_n]$$ and again since the $X_i$ are i.i.d., we notice that $$ \sum_{i=1}^n \mathbb{E}[X_i/S_n] = n \mathbb{E}[X_1/S_n] = 1 \Longrightarrow \mathbb{E}[X_1/S_n] = \frac1n$$ Combining these results, you get your claim.