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Math Help - Sum of Independent Random Variables - Moments

  1. #1
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    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!
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  2. #2
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    Quote Originally Posted by azdang View Post
    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.
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  3. #3
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    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
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  4. #4
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    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))
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  5. #5
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    Quote Originally Posted by azdang View Post
    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.
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  6. #6
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    Yes, I ended up actually redoing the problem and getting the answer just fine. I should have updated here. THank you though!
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