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Thread: moment-generating function

  1. May 3rd 2008, 03:56 PM #1
    jorgeston
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    moment-generating function

    Y have pdf P(Y=k)=(k+1)p^2 q^k, with 0<p<1 and q=1-p and k=0,1,2,...
    Is easy prove that M_{Y}(t)=p^2 (1-qe^t)^2. Let X=Y_1+\cdots Y_{24}, with Y_{i} with same distribution of Y
    Calculate M_{X}(t)

    Note: M_Y (t)=E(e^{tk})
    Last edited by jorgeston; May 3rd 2008 at 06:41 PM.
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  2. May 3rd 2008, 04:08 PM #2
    mr fantastic
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    Quote Originally Posted by jorgeston View Post
    Y have pdf P(Y=k)=(k+1)p^2 p^k, with 0<p<1 and q=1-p and k=0,1,2,...
    Is easy prove that M_{Y}(t)=p^2 (1-qe^t)^2. Let X=Y_1+\cdots Y_{24}, with Y_{i} with same distribution of Y
    Calculate M_{X}(t)

    Note: M_Y (t)=E(e^{tk})
    I have no time now. I'll reply later if no-one else does.
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  3. May 3rd 2008, 04:46 PM #3
    tukeywilliams
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    X = 24Y . Note that M_{aY+b} = e^{bt} M_{Y}(at) . In this case, b = 0 and a = 24 . So M_{24Y}(t) = M_{X}(t) = p^2(1-qe^{24t})^{2} .
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  4. May 3rd 2008, 06:43 PM #4
    jorgeston
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    Quote Originally Posted by tukeywilliams View Post
    X = 24Y . Note that M_{aY+b} = e^{bt} M_{Y}(at) . In this case, b = 0 and a = 24 . So M_{24Y}(t) = M_{X}(t) = p^2(1-qe^{24t})^{2} .
    If you have reason, i don't understand why Y_1=Y_2=\cdots = Y_{24}

    If this is true, implies X=24Y, but i am confused
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  5. May 3rd 2008, 07:34 PM #5
    mr fantastic
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    Quote Originally Posted by jorgeston View Post
    If you have reason, i don't understand why Y_1=Y_2=\cdots = Y_{24}

    If this is true, implies X=24Y, but i am confused
    *Ahem* Didn't you say in your question:

    "with with same distribution of " ....?
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