Sum of Independent Random Variables - Moments

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November 19th 2009, 03:33 PM
azdang

Sum of Independent Random Variables - Moments

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

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

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

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

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

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!
November 19th 2009, 03:47 PM
Laurent

Quote:

Originally Posted by azdang

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

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

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

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

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

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

b) The "good" proof goes by expanding the cube in $E[(X_1+\cdots+X_n)^3]$ and noting that $E[X_i^2X_j]=0$ for any $i\neq j$ , and $E[X_iX_jX_k]=0$ for any distinct $i,j,k$ . Then the formula is straightforward. I let you try that.
November 19th 2009, 03:55 PM
azdang

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 :)
November 19th 2009, 04:26 PM
azdang

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

What I got down to was:

$\phi_{S_n}^{(3)}(0)=n((n-1)\phi_{X_i}''(0) + \phi_{X_i}^{(3)}(0))$
November 23rd 2009, 08:23 AM
Laurent

Quote:

Originally Posted by azdang

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

What I got down to was:

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

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

You should find (I simplify notations):
$\phi_S'=n\phi'\phi^{n-1}$ ,
$\phi_S''=n\phi''\phi^{n-1}+n(n-1)(\phi')^2\phi^{n-2}$ ,
$\phi_S'''=n\phi'''\phi^{n-1}+n\phi'\phi''\phi^{n-2}+2n(n-1)\phi'\phi''\phi^{n-2}+n(n-1)(n-2)(\phi')^3\phi^{n-3}$ .
and conclude from there using what I said.
November 23rd 2009, 09:40 AM
azdang

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