0
$\begingroup$

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.

$\endgroup$
1
1
$\begingroup$

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).

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.