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Math Help - Moment generating function of the Binomial distribution

  1. #1
    Senior Member chella182's Avatar
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    [SOLVED] Moment generating function of the Binomial distribution

    Suppose that Y~Bin(n,p), with probability mass function given by

    p_Y(y)=P(Y=y)=\left( \begin{array}{c}n \\ y \end{array} \right) p^y(1-p)^{n-y}, y=0,1,...,n

    and 0 otherwise. Show that the moment generating function of Y is given by

    M_y(t)=\{pe^t+(1-p)\}^n


    I keep getting majorly lost with this. I've tried Googling to find a solution to this, since it's a fairly generic question, but to no avail. Can anyone show me how to show this?
    Last edited by chella182; August 23rd 2010 at 02:22 AM. Reason: Solved - just editing title
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by chella182 View Post
    Suppose that Y~Bin(n,p), with probability mass function given by

    p_Y(y)=P(Y=y)=\left( \begin{array}{c}n \\ y \end{array} \right) p^y(1-p)^{n-y}, y=0,1,...,n

    and 0 otherwise. Show that the moment generating function of Y is given by

    M_y(t)=\{pe^t+(1-p)\}^n


    I keep getting majorly lost with this. I've tried Googling to find a solution to this, since it's a fairly generic question, but to no avail. Can anyone show me how to show this?
    One way to do this is to consider X_i independent bernoulli trials, where the pdf is p(x)=p^x(1-p)^{1-x} for x=0,1. Also, recall that binomial is the sum of the bernoulli trials, in this case \displaystyle Y=\sum\limits_{i=1}^n X_i.

    The MGF of a bernoulli trial is \displaystyle M_{X_i}(t)=\sum\limits_{x=0}^{1}e^{tx}p^x(1-p)^{1-x}=pe^t+(1-p).

    Therefore, the MGF of the binomial is

    \begin{aligned}M_Y(t)&= M_{\sum\limits_{i=1}^n X_i}(t)\\ &=\prod_{i=1}^{n}M_{X_i}(t)\quad (by\,a\,theorem)\\ &=\prod_{i=1}^n\left[pe^t+(1-p)\right]\\ &= \left[pe^t+(1-p)\right]^n\end{aligned}

    which is the desired result.

    Does this make sense?
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  3. #3
    Senior Member chella182's Avatar
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    Yeah that's great thanks, should be sufficient if the question comes up again today
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