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Math Help - sum of Bernoulli trials

  1. #1
    Senior Member Dinkydoe's Avatar
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    sum of Bernoulli trials

    Hoi, i dont see why the following is true

    Let \mathbb{P}(X_k =1)=\mathbb{P}(X_k=0)=1/2 ( i.i.d. Bernoulli trials)

    and consider Y = 3\sum_{k=1}^{\infty}4^{-k}X_k ( Then clearly 0\leq Y \leq 1 )

    Apparantly the distribution function F of Y is constant on the interval (1/4,3/4) and satisfies
    F(x) = 1-F(x) and

    for x< 1/4 it satisfies F(x)=2F(x/4)

    I'm a bit bewildered...:/
    Last edited by Dinkydoe; October 15th 2012 at 03:49 AM.
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  2. #2
    MHF Contributor
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    Re: sum of Bernoulli trials

    For the first one, can you show that the median occurs at that point? If your distribution is symmetric, can you show the mean occurs at that point? (Hint: your distribution has 1/2 for a 0 value and 1/2 for a 1 value for every X_k).
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  3. #3
    Senior Member Dinkydoe's Avatar
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    Re: sum of Bernoulli trials

    I think i got it now...

    P(X \leq 1/4) = P(X_1=0) = 1/2 and P(X\leq 3/4  ) = 1- \mathbb{P}(X\geq 3/4) = 1 - \mathbb{P}(X_1=1) = 1/2

    I think i see the other 2 as well...
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