Let $X_1, \ldots, X_n$ be i.i.d. random variables with expectation $a$ and variance $\sigma^2$, taking only positive values. Let $m < n$. Find the expectiation of $\displaystyle\frac{X_1 + \cdots + X_m}{X_1 + \cdots + X_n}$.
My attemps to solve this probles are rather straightforward. Denote $X = X_1 + \cdots + X_m$ and $Y = X_{m+1} + \dots + X_n$. So, $X$ has the expectation $ma$ and the variance $m\sigma^2$. And $Y$ has the expectation $(n-m)a$ and variance $(n-m)\sigma^2$. And also $X$ and $Y$ are independent. So we can compute the expectation by the definition $\mathbb{E}\displaystyle\frac{X}{X+Y} = \int\limits_{\Omega^2}\frac{X(\omega_1)}{X(\omega_1) + Y(\omega_2)}\mathbb{P}(d\omega_1)\mathbb{P}(d\omega_2)$. But we do not know the distribution, so we do not have chance to calculate it.
I would be glad to any help or ideas!
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 $m\le n$, $$\mathbb E\left(\frac{S_m}{S_n}\right)=\sum_{i=1}^m \mathbb E\left(\frac{X_i}{S_n}\right)=\frac{m}{n}$$