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.
Best Answer
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 $E \Big[ X_i \mid {\sum _{i=1}^n X_i}=m\Big]= \frac{m}{n}$.