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Math Help - mgf adding two binomials

  1. #1
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    mgf adding two binomials

    We are to solve this problem using mgf. technique (transforms of rv)

    x1= b(4,1/2) and x2= b(6,1/3), They are independent.

    Let Y = X1+X2

    We are looking for the mean and variance.

    I assume that we need to proceed as follows:

    My(t) = E(e^tY) = ....= E(e^tx1)E(e^tx2).

    So I get (1/2 + 1/2 * e^t)^4 * (2/3 + 1/3 * e^t)^6

    Now I stuck. Dont i need to create an mgf I recognize (binomial I imagine) and pick out the mean and variance from the mgf parameters?

    Help?
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  2. #2
    MHF Contributor matheagle's Avatar
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    The sum of two indep binomials with the same parameter p is a binomial.
    (With N=n_1+n_2 and same p.)
    If the paramters are different I do not believe it's a binomial.
    MGFs are unique and since we cannot recognize this one as a binomial, it isn't.
    HOWEVER, you can get the mean and the second moment by differentiating with respect to t and setting t=0.
    And from that you can get the variance, but I don't know why you would use MGFs to find the mean and variance in this problem.
    This is a MOMENT generating function, so you can clearly get the MOMENTS by differentiating and setting t=0.
    That's how it got its name.
    Last edited by matheagle; February 15th 2009 at 09:03 PM.
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  3. #3
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    Thanks for the reply - I did forget about trying to differentiate the mgf.

    I ended up expanding the terms (1/2 + 1/2 * e^t)^4 * (2/3 + 1/3 * e^t)^6 and then looking at the "weights" on the e^tX terms to create a pmf which I then used to find the mean and variance.

    Thanks again for the other tip!
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