Find expectation of $\frac{X_1 + \cdots + X_m}{X_1 + \cdots + X_n}$ when $X_1,\ldots,X_n$ are i.i.d
Solution 1:
Suppose $S_m=\sum\limits_{i=1}^{m} X_i$ and $S_n=\sum\limits_{i=1}^n X_i$.
Now, $$\frac{X_1+X_2+\cdots+X_n}{S_n}=1\,, \text{ a.e. }$$
Therefore,
$$ \mathbb E\left(\frac{X_1+X_2+\cdots+X_n}{S_n}\right)=1$$
Since $X_1,\ldots,X_n$ are i.i.d (see @SangchulLee's comments on main), we have for each $i$,
$$\mathbb E\left(\frac{X_i}{S_n}\right)=\frac{1}{n}$$
So for $m\le n$, $$\mathbb E\left(\frac{S_m}{S_n}\right)=\sum_{i=1}^m \mathbb E\left(\frac{X_i}{S_n}\right)=\frac{m}{n}$$