Let $X_1,X_2,...,X_n$ be i.i.d random variables and $\mathbb{E}[X^{-1}_1]$ is finite. Define $S_n = \sum_{i=1}^{n}X_i$ . Show that for any $1 \leq a \leq b \leq n , $ $\mathbb{E}[\frac{S_a}{S_b}] = \frac{a}{b} $.

I don't think we can equate $S_n = nX_i$ that would make the problem too trivial. So my progress is turn the expectation into simpler parts like $\mathbb{E}[\frac{S_a}{S_b}] = \sum_{i=1}^{a}\mathbb{E}[\frac{X_i}{S_b}]$. My guess is that $\mathbb{E}[\frac{X_i}{S_b}]=\frac{1}{b}$ but was stuck proving it, the denominator thing is hard for me. Any Ideas or solutions? Thanks.


For $1 \le i \le b$, the random variables $X_i/S_b$ have the same distribution and thus have the same expectation. Since $\sum_{i=1}^b \frac{X_i}{S_b} = 1$, we must have $E[X_i/S_b]=\frac{1}{b}$.

Edit: The above argument overlooks some technical issues (assumes $S_b \ne 0$, assumes $E[X_i / S_b]$ is finite).

  • $\begingroup$ why do you think that we require this specific condition $\mathbb{E}[X^{-1}_1]$ is finite? $\endgroup$
    – Grentouce
    Dec 21 '20 at 16:29
  • $\begingroup$ @Grentouce I'm not really sure. I think it has something to do with division by zero and ensuring $E[S_a/S_b]$ is finite. But you can come up with examples where $E[X_1^{-1}]$ is finite and the denominator $S_b$ is still zero, so I'm not sure what is going on. $\endgroup$
    – angryavian
    Dec 21 '20 at 16:40

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