Showing the expectation of the third moment of a sum = the sum of the expectation of the third moment

Question

Let $X_{1},\cdots,X_{n}$ be independent, each with mean 0, and each with finite third moments. Show that $E\left\{\left( \sum_{i=1}^{n}X_{i}\right)^{3}\right\} = \sum_{i=1}^{n}E\left\{ X_{i}^{3} \right \}$.

Thanks in advance!

Answer 1

In the approach with characteristic functions, we have to use the fact that $\varphi_{S_n}(t)=\prod_{j=1}^n\varphi_{X_j}(t)$.

here is an other approach: we have, noting by $[n]$ the set $\{1,\dots,n\}$, $$\mathbb E\left(\sum_{i=1}^nX_i\right)^3=\sum_{(i,j,k)\in [n]^3}\mathbb E(X_iX_jX_k).$$ The set $[n]^3$ can be divided into the $(i,j,k)$ such that:

1. $i=j=k$;
2. two indexes coincide, but not the third one;
3. all the indexes are different.

The sum of the indexes in 2. is $0$ as the terms are of the form $\mathbb E(X_i^2X_j)$, which is equal by independence to $\mathbb E(X_i^2)\mathbb EX_j=0$, and the same remark applies to 3.

It's actually the phenomenon which will appear when you will compute the third derivative of a product (either there is a factor differentiated three times, or two times an an other factor one time, or three different factors).

Answer 2

Ok the idea is that $$E\{(\sum_{j=1}^n{X_i})^3\} = E\{\sum_{j=1}^n{X_i^3}+...\}$$

where ... contains the cross terms of the expansion of $(\sum_{j=1}^n{X_i})^3$, however all the cross term will be in the form $a_1X_iX_j^2$ or $a_2X_iX_jX_k$, where $a_1$ and $a_2$ are constants. Taking expectation, we get $$E\{a_1X_iX_j^2\}=a_1E\{X_i\}E\{X_j^2\}=0$$ by independence, and similarly $$E\{a_2X_iX_jX_k\}=a_2E\{X_i\}E\{X_j\}E\{X_k\}=0$$ Hence we have $$E\{(\sum_{j=1}^n{X_i})^3\} = E\{\sum_{j=1}^n{X_i^3}\} = \sum_{j=1}^nE\{{X_i^3}\}$$ after simplying the expectation of cross terms.

Answer 3

Hint( without characteristic functions):

Expand $(\sum_{k=1}^n a_i)^3$ and use facts, that $E X_1 X_2^2 = EX_1 \cdot E X_2^2$ and $E X_1 X_2 X_3 = E X_1 \cdot EX_2 \cdot E X_3$for independent random variables

Answer 4

I should choose for induction: $E\left[\left(\sum_{i=1}^{n}X_{i}\right)^{3}\right]=E\left[\left(\sum_{i=1}^{n-1}X_{i}\right)^{3}+3X_{n}\left(\sum_{i=1}^{n-1}X_{i}\right)^{2}+3X_{n}^{2}\sum_{i=1}^{n-1}X_{i}+X_{n}^{3}\right]$.

By induction:

$E\left[\left(\sum_{i=1}^{n-1}X_{i}\right)^{3}\right]=\sum_{i=1}^{n-1}E\left[X_{i}^{3}\right]$.

Further we have:

$E\left[X_{n}\left(\sum_{i=1}^{n-1}X_{i}\right)^{2}\right]=E\left[X_{n}\right]E\left[\left(\sum_{i=1}^{n-1}X_{i}\right)^{2}\right]=0$ since $E\left[X_{n}\right]=0$

and:

$E\left[X_{n}^{2}\sum_{i=1}^{n-1}X_{i}\right]=E\left[X_{n}^{2}\right]E\left[\sum_{i=1}^{n-1}X_{i}\right]=0$ since $E\left[\sum_{i=1}^{n-1}X_{i}\right]=\sum_{i=1}^{n-1}E\left[X_{i}\right]=0$.

So our final result is:

$E\left[\left(\sum_{i=1}^{n}X_{i}\right)^{3}\right]=\sum_{i=1}^{n-1}E\left[X_{i}^{3}\right]+E\left[X_{n}^{3}\right]=\sum_{i=1}^{n}E\left[X_{i}^{3}\right]$