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Math Help - Show that...

  1. #1
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    Show that...

    I have no idea how to do this......
    Any help will be apreaciated


    Let \{ Xn\} _{n \ge 1}^{} be a sequence of iid random variabled such that E({X_1}) = \mu and V({X_1}) = {\sigma ^{^2}}.
    Consider the sequence of real numbers: {\{ {a_n}\} _{n \ge 1}}

    Proof that:

    P(\frac{{{S_n} - E({S_n})}}{{V({S_n})}} \le {a_n}) - \Phi ({a_n}) \to 0

    Where:

    {S_n} = \sum\limits_{i = 1}^n {{X_i}}


    Edit: I know I have to use Central limit theorem, but I'm doubtful cuz of that a_n thing inside PHI
    Last edited by FRMST; May 29th 2012 at 09:55 PM.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Show that...

    The Central limit theorem says:
    \frac{\mbox{S}_n - n\mu}{\sqrt{n}\sigma} \to Z with Z \sim N(0,1) for n \to +\infty
    in other words:
    \forall x \in \mathbb{R}: \lim_{n \to +\infty} P\left(\frac{\mbox{S}_n-n\mu}{\sqrt{n}\sigma}\leq x\right)=P(Z \leq x)=\Phi(x)
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  3. #3
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    Re: Show that...

    Thank you, but what happens when I have a sequence...
    The sequence is going to infinity... so, I don't know if I can use straight the central limit theorem

    EDIT:

    Oh... I got it...
    So, the sequence in this case would be:

    ( x - n*u )/( sqrt(n)*Sigma)


    Thank you buddy.
    Last edited by FRMST; May 30th 2012 at 06:05 AM.
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