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**azdang** Let $\displaystyle X_1,...,X_n$ be independent, each with mean 0, and each with finite third moments. Show that:

$\displaystyle E((\sum_{i=1}^{n}X_i)^3)=\sum_{i=1}^n E(X_i^3)$

It gives a hint to use characteristic functions, so here is what I tried doing. I used $\displaystyle S_n$ to represent the sum of the Xi's from 1 to n.

$\displaystyle E(X_i^3)$ = $\displaystyle i \phi_{X_i}^{(3)} (0)$ and $\displaystyle E((\sum_{i=1}^{n}X_i)^3)=i \phi_{S_n}^{(3)}(0)$

Then, we want to show that $\displaystyle \phi_{S_n}^{(3)}(0)=\sum_{j=1}^n\phi_{X_i}^{(3)} (0)$.

My first question is if what I have done so far is okay. My second question is, where would I go from here? I can't seem to see it. Thank you!