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  1. #1
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    mgf

    P(X = k) = \left(\begin{array}{cc}n\\k\end{array}\right)p^kq^  {n-k} k=0,...,n.
    Where n is a positive integer, 0 < p < 1 , q = 1 - p.

    How would I show the mgf of X is

    M_{x}(t) = (pe^t +q)^n\ \  -\infty < t < \infty


    Also, if Y = \Sigma_{1\le i \le m} X_{i}

    how would I derive the mgf of Y?
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  2. #2
    Moo
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    Hello,

    The mgf of a discrete distribution is :

    M_X(t)=\sum_{k=0}^\infty \mathbb{P}(X=k)e^{kt}

    So here, we have :

    \begin{aligned} M(t)<br />
&=\sum_{k=0}^n {n\choose k} e^{kt}p^k q^{n-k} \\<br />
&=\sum_{k=0}^n {n\choose k} (pe^t)^k q^{n-k}\end{aligned}

    Now just recall Newton's binomial formula :

    (a+b)^n=\sum_{k=0}^n a^k b^{n-k}



    As for the second one, if the Xi's are independent, then the mgf of Y is the product of the mgf of the Xi.
    If they're identically distributed, this will just be [M(t)]^m
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  3. #3
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    Quote Originally Posted by bigdoggy View Post
    P(X = k) = \left(\begin{array}{cc}n\\k\end{array}\right)p^kq^  {n-k} k=0,...,n.
    Where n is a positive integer, 0 < p < 1 , q = 1 - p.

    How would I show the mgf of X is

    M_{x}(t) = (pe^t +q)^n\ \ -\infty < t < \infty


    Also, if Y = \Sigma_{1\le i \le m} X_{i}

    how would I derive the mgf of Y?
    See here: http://www.mathhelpforum.com/math-he...tribution.html
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  4. #4
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    So if the mgf is

    M_{x}(t) = (pe^t +q)^n\ \  -\infty < t < \infty
    then the mgf of Y would be  (pe^t +q)^{mn}\ \mbox{where}\ Y = \Sigma_{1\le i \le m}X_{i} \mbox{ with }\ X_{i} independent??

    How would I find the distribution of Y,and does E[Y]= M'(0) and V[Y]= M''(0)??
    Last edited by bigdoggy; August 19th 2009 at 03:02 PM.
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  5. #5
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by bigdoggy View Post
    So if the mgf is

    M_{x}(t) = (pe^t +q)^n\ \ -\infty < t < \infty
    then the mgf of Y would be  (pe^t +q)^{mn}\ \mbox{where}\ Y = \Sigma_{1\le i \le m}X_{i} \mbox{ with }\ X_{i} independent??

    How would I find the distribution of Y,and does E[Y]= M'(0) and V[Y]= M''(0)??

    The point of using MGF's is that you hope to recognize the resulting one.
    Here you added m INDEPENDENT binomials.
    This techniques proves that the result is a binomial with same probability of success (p) and now the number of trials is nm.


    The MGF of X+Y where they are indep bin(n,p) and bin(m,p), SAME p, different sample sizes is

     (pe^t+q)^n(pe^t+q)^m=(pe^t+q)^{n+m}

    proving that X+Y is bin(n+m,p)


    So, for you the mean is (nm)p and the variance is (nm)pq.

    X_1+\cdots +X_m\sim Bin(nm,p)
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  6. #6
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    Why have you mentioned the mgf of X + Y?

    I came with the mgf of Y to be (pe^t +q)^{mn} by using the post from Moo, which used M_Y(t) = (pe^t +q)^{mn}

    Is this correct??

    Also is E[Y]= M'(0) and V[Y]= M''(0) correct??
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  7. #7
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by bigdoggy View Post
    Why have you mentioned the mgf of X + Y?

    I came with the mgf of Y to be (pe^t +q)^{mn} by using the post from Moo, which used M_Y(t) = (pe^t +q)^{mn}

    Is this correct??

    Also is E[Y]= M'(0) and V[Y]= M''(0) correct??
    NO, E[Y]= M'(0)=nmp BUT E[Y^2]= M''(0) SO V(Y)=nmpq
    AS I wrote 30 minutes ago.
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  8. #8
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    Quote Originally Posted by matheagle View Post
    NO, E[Y]= M'(0)=nmp BUT E[Y^2]= M''(0) SO V(Y)=nmpq
    AS I wrote 30 minutes ago.
    Do you mean no, the mgf of Y isn't (pe^t +q)^{mn}??
    What is it here??
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  9. #9
    Moo
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    It's (pe^t+q)^{mn}.. matheagle didn't deny it.

    The problem is your formula for the variance...

    We know that \mathbb{E}(X^k)=M^{(k)}(0) (k-th derivative)

    But V(X)=\mathbb{E}(X^2){\color{red}-[\mathbb{E}(X)]^2}=M''(0)-[M'(0)]^2, and not what you wrote.
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  10. #10
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    Thanks again Moo!!
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