Expectation of quotient of random variables Let $X_1,...X_n$ be independent, identically distributed and nonnegative random variables, and let $k\le n$. Compute: $$E\left[{\sum_{i=1}^k X_i\over \sum_{i=1}^n X_i}\right].$$ This question has already been asked: Expectation of random variables ratio.
The thing is that my teacher told me that the solution in the link wasn't really a "solution" the correct thing to do is to compute the conditional expectation of $${\sum_{i=1}^k X_i\over \sum_{i=1}^n X_i}$$ given that $\sum_{i=1}^n X_i=m$ where $m$ is a positive integer, but I have no idea how to do this, I would really appreciate if you can help me with this problem.
 A: $\begin{align}
\text{Let }\qquad M &=\sum_{i=1}^n X_i & \text{Define a new random variable}
\\[2ex]
\mathsf E(\sum_{i=1}^k X_i / \sum_{i=1}^n X_i)
 & = \mathsf E(\mathsf E(\sum_{i=1}^k X_i / M\mid M)) & \text{Law of Iterated Expectation}
\\[1ex]
 & = \mathsf E(\sum_{i=1}^k \mathsf E(X_i/M\mid M)) & \text{Linearity of Expectation}
\\[1ex]
 & = \mathsf E(\sum_{i=1}^k 1/n) & \{X_i\}\sim\text{ i.i.d.}
\\[1ex]
 & = k/n
\end{align}$
A: You can prove as follows:
$$\begin{align}
\mathsf E \Big[ \frac{\sum _{i=1}^k X_i}{\sum _{i=1}^n X_i} \Big]
&  =
\sum _{m=0}^{\infty} \mathsf E \Big[ \frac{\sum _{i=1}^k X_i}{\sum _{i=1}^n X_i} \mid {\sum _{i=1}^n X_i}=m\Big]\mathsf P({\sum _{i=1}^n X_i}=m)
\\[1ex] & = 
\sum _{m=0}^{\infty} \mathsf E \Big[ \frac{\sum _{i=1}^k X_i}{m} \mid {\sum _{i=1}^n X_i}=m\Big]\mathsf P({\sum _{i=1}^n X_i}=m)
\\[1ex] & = 
\sum _{m=0}^{\infty} \frac{1}{m} \sum _{i=1}^{k}\mathsf E \Big[ X_i \mid {\sum _{i=1}^n X_i}=m\Big]\mathsf P({\sum _{i=1}^n X_i}=m)
\\[1ex] & = 
\sum _{m=0}^{\infty} \frac{1}{m} \frac{km}{n} \mathsf P({\sum _{i=1}^n X_i}=m)
\\[1ex] & = \frac{k}{n} \sum _{m=0}^{\infty}\mathsf P({\sum _{i=1}^n X_i}=m)
\\[1ex] & = \frac{k}{n}
\end{align}$$
Note: I have used identical distribution of $X_i$s and this fact that $E \Big[ {\sum _{i=1}^n X_i} \mid {\sum _{i=1}^n X_i}=m\Big] = m$ to conclude that $E \Big[ X_i \mid {\sum _{i=1}^n X_i}=m\Big]= \frac{m}{n}$.
