Expectation of function of multiple random variable

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).
• why do you think that we require this specific condition $\mathbb{E}[X^{-1}_1]$ is finite? Dec 21 '20 at 16:29
• @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. Dec 21 '20 at 16:40