Independent and identically distributed discrete random variables

Let $X_1 , X_2$, ... be independent and identically distributed random variables. I would like to show that if $S_m = X_1 + X_2 + ... + X_m$ , $S_n = X_1 + X_2 + ... + X_n$ and $m<n$, we have $E(S_m / S_n ) = m/n$. How should I do that?

Why don't we get $$E(S_m / S_n ) = \frac{1}{1+E(\frac{X_{m+1}+...+X_{n}}{X_{1}+...+X_{m}})},$$

what, since in general case $E(1 / X ) E(X )\neq 1$, doesn't seem to be equal $m/n$?

On the other hand, if $m>n$, we have $$E(S_m / S_n ) = 1+ E(X_1)(m-n)E(\frac{1}{S_n}).$$

• Symmetry, o sweet symmetry... $$E(S_mS_n^{-1})=\sum_{k=1}^mE(X_kS_n^{-1})=mE(X_1S_n^{-1})=mn^{-1}E(S_nS_n^{-1})=mn^{-1}$$ – Did Nov 20 '16 at 19:58