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Thread: moment-generating function

  1. #1
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    moment-generating function

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

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

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

    If this is true, implies $\displaystyle X=24Y$, but i am confused
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    Quote Originally Posted by jorgeston View Post
    If you have reason, i don't understand why $\displaystyle Y_1=Y_2=\cdots = Y_{24}$

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

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