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Math Help - Help understanding how to use MGF to calculate first moment of Binomial Distribution

  1. #1
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    Help understanding how to use MGF to calculate first moment of Binomial Distribution

    I have it all in front of me, I'm just having a bit of trouble understanding the math.

    \phi(t)=E[e^{tX}]

    =\sum_{k=0}^{n}e^{tk}  {n}\choose{k}  p^{k}(1-p)^{n-k}
    =\sum_{k=0}^{n}  {n}\choose {k}  (pe^{t})^{k}(1-p)^{n-k}


    Then all of a sudden the summation and combination are gone and the book has

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

    It goes on to show the differentiation and whatnot, which I understand. I just don't understand what was done with the combination and the summation. Any help would be appreciated.
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  2. #2
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    Re: Help understanding how to use MGF to calculate first moment of Binomial Distribut

    Quote Originally Posted by downthesun01 View Post
    I have it all in front of me, I'm just having a bit of trouble understanding the math.

    \phi(t)=E[e^{tX}]

    =\sum_{k=0}^{n}e^{tk}  {n}\choose{k}  p^{k}(1-p)^{n-k}
    =\sum_{k=0}^{n}  {n}\choose {k}  (pe^{t})^{k}(1-p)^{n-k}


    Then all of a sudden the summation and combination are gone and the book has

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

    It goes on to show the differentiation and whatnot, which I understand. I just don't understand what was done with the combination and the summation. Any help would be appreciated.
    (pe^t+1-p)^2=( (pe^t) + (1-p) )^2

    Now consider the binomial expansion of (a+b)^n where a=pe^t and b=(1-p)

    CB
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  3. #3
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    Re: Help understanding how to use MGF to calculate first moment of Binomial Distribut

    Thank you for responding, but I don't know what you're talking about.
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  4. #4
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    Re: Help understanding how to use MGF to calculate first moment of Binomial Distribut

    Ok, I looked up binomial expansion. I don't understand the math, but apparently:

    (x+y)^{n}=\sum_{k=0}^{n} {n}\choose{k} x^{k}y^{n-k}

    I'll try to see if I can figure this all out now. Thanks
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