# Thread: Sum of Independent Random Variables - Moments

1. ## Sum of Independent Random Variables - Moments

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!

2. Originally Posted by 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!
a) You have $\displaystyle \phi_{S_n}(t)=\prod_{i=1}^n\phi_{X_i}(t)$, so you can derivate three time and let $\displaystyle t=0$, using $\displaystyle \phi_{X_i}'(0)=0$ (zero mean) and $\displaystyle \phi_{X_i}(0)=1$. But this is unnecessarily complicated...

b) The "good" proof goes by expanding the cube in $\displaystyle E[(X_1+\cdots+X_n)^3]$ and noting that $\displaystyle E[X_i^2X_j]=0$ for any $\displaystyle i\neq j$, and $\displaystyle E[X_iX_jX_k]=0$ for any distinct $\displaystyle i,j,k$. Then the formula is straightforward. I let you try that.

3. Thank you, Laurent. That second proof is definitely straightforward. However, the problem does specifically say to use characteristic functions, so I'm not sure I should be using this. I will continue trying to work on part a, so if you or anyone else has any suggestion, I would appreciate it

4. So, I tried to take the derivatives, using the fact that the $\displaystyle \phi_{S_n}(t)=(\phi_{X_i}(t))^n$ since the Xis are i.i.d. (I hope this is right?).

What I got down to was:

$\displaystyle \phi_{S_n}^{(3)}(0)=n((n-1)\phi_{X_i}''(0) + \phi_{X_i}^{(3)}(0))$

5. Originally Posted by azdang
So, I tried to take the derivatives, using the fact that the $\displaystyle \phi_{S_n}(t)=(\phi_{X_i}(t))^n$ since the Xis are i.i.d. (I hope this is right?).

What I got down to was:

$\displaystyle \phi_{S_n}^{(3)}(0)=n((n-1)\phi_{X_i}''(0) + \phi_{X_i}^{(3)}(0))$
If indeed the Xi's are i.i.d. (which was not specified in your first post), then you indeed have $\displaystyle \phi_{S_n}(t)=(\phi_{X_i}(t))^n$.
You did mistakes in your computation.

You should find (I simplify notations):
$\displaystyle \phi_S'=n\phi'\phi^{n-1}$,
$\displaystyle \phi_S''=n\phi''\phi^{n-1}+n(n-1)(\phi')^2\phi^{n-2}$,
$\displaystyle \phi_S'''=n\phi'''\phi^{n-1}+n\phi'\phi''\phi^{n-2}+2n(n-1)\phi'\phi''\phi^{n-2}+n(n-1)(n-2)(\phi')^3\phi^{n-3}$.
and conclude from there using what I said.

6. Yes, I ended up actually redoing the problem and getting the answer just fine. I should have updated here. THank you though!