# Proof by Characteristic Function

If $X_1, X_2, \ldots,X_n$ are independent random variables variables with expectation $0$ and finite third moments. Show that $$E((X_1+X_2+\cdots+X_n)^3) = EX_1^3+EX_2^3+\cdots+EX_n^3$$ using the aid of characteristic functions.

-
What does the notation $\mathbb{E}(X_1,X_2,\ldots,X_n)^3$ mean? A joint probability would be understandable, but what is joint expectation? – gt6989b Mar 19 '13 at 21:48
@user61147 : You must have meant $E((X_1+\cdots+X_n)^3)$. – Michael Hardy Mar 19 '13 at 21:50
Since the thing one is to prove in the first sentence is false (under the interpreation in my earlier comment) without the later qualification saying "note that", it doesn't make a lot of sense to put that "Note that" comment down below as an afterthought. – Michael Hardy Mar 19 '13 at 21:52
I edited the question with your corrections @MichaelHardy – user61147 Mar 19 '13 at 21:56

One way to do this is by a simple induction on the number of terms after proving it works when $n=2$. $$E((X_1+X_2)^3) = E(X_1^3)+3E(X_1^2X_2)+3E(X_1X_2^2)+E(X_2^3).$$ Because of independence this becomes $$E(X_1^3)+3E(X_1^2)E(X_2)+3E(X_1)E(X_2^2)+E(X_2^3).$$ Then the middle two terms are $0$ because each has a factor that is $0$.
Later edit: The characteristic function of random variable $X$ is $$\chi_X(t) = E\left( e^{itX} \right).$$ It follows that $$\chi_X'''(t) = -iE\left(X^3 e^{itX} \right),$$ so that $\chi_X'''(0) = -iE(X^3)$.
So we get $$\chi_{X_1+X_2}(t) = \chi_{X_1}(t)\chi_{X_2}(t),$$ and then $$\chi_{X_1+X_2}'''(t) = \chi_{X_1}'''(t)\chi_{X_2}(t) + 3\chi_{X_1}''(t)\chi_{X_2}'(t) + 3\chi_{X_1}'(t)\chi_{X_2}''(t) + \chi_{X_1}(t)\chi_{X_2}'''(t).$$ \begin{align} \chi_{X_1+X_2}'''(0) & = \chi_{X_1}'''(0)\chi_{X_2}(0) + 3\chi_{X_1}''(t)\chi_{X_2}'(0) + 3\chi_{X_1}'(0)\chi_{X_2}''(0) + \chi_{X_1}(0)\chi_{X_2}'''(0) \\[6pt] & = \chi_{X_1}'''(0)\cdot1 + 0 + 0 + 1\cdot\chi_{X_2}'''(0). \end{align} Then procede as above.