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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 $$ E(S_m / S_n ) = 1+ E(X_1)(m-n)E(\frac{1}{S_n}).$$

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

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