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Math Help - Weak law of large numbers and CF

  1. #1
    Member mabruka's Avatar
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    Weak law of large numbers and CF

    I am asked to prove the weak law of large numbers using characteristic functions. For simplicity i can assume that the r.v. are simmetryc (so their CF is real).

    Let X_1,X_2,... r.v. i.i.d. , with a common mean and variance \mu and  \sigma^2.

    The first think i would like to prove is that

    Var \left ( \frac{1}{n} \sum_{1}^{n} X_i \right ) \underset{n\rightarrow \infty}{\longrightarrow } 0


    So i define Z_n= \sum_{0}^{n}\frac{X_n}{n} as the partial sum. Its characteristic functions is (they are all independent):

    \varphi_{Z_n}(t) =\varphi_{\sum_{0}^{n}\frac{X_n}{n}}(t)=
    \prod_{0}^{n} \varphi_{\frac{X_i}{n}} = \left [ \varphi_{X/n} \right ]^n

    Im interested in the variance (second moment) so i am to diferentiate two times the CF and evaluate at 0. After computing the derivatives carefully using chain rule i get:

     \varphi^{(2)}_{Z_n}(0)=\frac{n-1}{n} \mu^2 + \frac{\sigma^2}{n^2} which obviously does not goes to 0 (unless \mu=0 which it has not to be true)

    So i dont know if i am doing something wrong.

    This is only one of the 3 statements that i understand to be "the weak law of large numbers", the other two are as follow:

    2) Z_n \overset{\mathcal{L}_2}{\longrightarrow}\mu (in L2 norm)

    3) Z_n \overset{P}{\longrightarrow}\mu (in probability)


    Any comments will be much apreciated. thank you
    Last edited by mabruka; February 10th 2010 at 06:21 PM.
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  2. #2
    Member mabruka's Avatar
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    i found in wikipedia good info about number 3)

    Proof of the law of large numbers - Wikipedia, the free encyclopedia



    but none about the other two.
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  3. #3
    Moo
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    Hello,

    But \varphi^{(2)}_{Z_n}(0)=-\mathbb{E}(Z_n^2), while Var(Z_n)=\mathbb{E}(Z_n^2)-[\mathbb{E}(Z_n)]^2

    And that missing part will make your thing go to 0 !
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  4. #4
    Member mabruka's Avatar
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    true!! i sometimes think second moment and variance are the same!


    thank you!
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