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Math Help - Convergence of sequence of continuous random variables

  1. #1
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    Convergence of sequence of continuous random variables

    Let  X_1, X_2, .. be a sequence of absolutely continuous random variables such that  X_n has pdf

     f_n(x) = \begin{cases} 1 - \cos (2 \pi n x) & x \in [0,1], \\ 0 & otherwise \end{cases}.

    Show that the sequence converges in distribution. What happens when  f_n \rightarrow \infty ?
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  2. #2
    Moo
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    Hello,

    to check for the convergence of a sequence of rv, you mustn't see the convergence of the pdf, but the convergence of the mgf, cdf, or characteristic function.

    so for example, let's find the cdf :
    F_n(x)=\begin{cases} 0 & \text{if } x<0 \\ x-\frac{\sin(2\pi n x)}{2\pi n} & \text{if } x \in[0,1] \\ 1 &\text{if } x>1 \end{cases}

    by using the squeeze (or sandwich) theorem, we can show that \lim_{n\to\infty} \frac{\sin(2\pi n x)}{2\pi n}=0

    so the limiting cdf is F(x)=\begin{cases} 0 & \text{if } x<0 \\ x & \text{if } x \in[0,1] \\ 1 &\text{if } x>1 \end{cases}
    which is the cdf of a uniform distribution over [0,1]

    looks clear to you ?
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  3. #3
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    Quote Originally Posted by Moo View Post

    looks clear to you ?
    Yes I think so.

    Since we have the cdf, we can show that the sequence converges in distribution if the cdf converges weakly i.e.

     \lim_{n\to\infty} F_n = F(x)

    Is that essentially what you did here?
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  4. #4
    Moo
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    Yes, exactly

    converges weakly towards a uniform distribution.
    Last edited by Moo; August 30th 2009 at 05:59 AM.
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  5. #5
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    Im now only having trouble picturing what happens when f_n \rightarrow \infty. Can anyone please explain?
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  6. #6
    Moo
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    There is no point finding the convergence of f_n. I mean it's not useful for the convergence of random variables here.
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