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Math Help - probability generating function of a poisson random variable

  1. #1
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    probability generating function of a poisson random variable

    i should start by asking is anyone on the forum well versed in the coalescent theory?

    i have a problem in a derivation

    if the number of mutations is given by

     P_n(z) = E exp((z-1)\frac{\theta}{2}(nT_n + ..... + 2T_2))

    as one can see this is the probability generating function of a poisson random variable with random mean. where  \lambda = \frac{\theta}{2}(nT_n + ..... + 2T_2)

    now from here i can write this as a product

    so

     P_n(z) = \prod_{j=2}^n E\left(e^{(z-1)\frac{\theta}{2}_j T_j}\right)

    Now im confused what to do. If i have the probability generating function how do i find the density function (derivative?)

    the next step in the derivation should be


     P_n(z) = \prod_{j=2}^n \left(1 - (z-1)\cdot \frac{j \theta /2}{j(j-1)/2}\right)^{-1}

    I cant for the life of me figure out how this was found from the pgf. PLEASE ANY IDEAS ARE WELCOME, SO IF NO ONE KNOWS THE ANSWER SUGGEST ANYTHING.

    many thanks

    chogo
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    perhaps i should rather ask

    if i had

     P(z) = E\left(e^{(z-1)\lambda}\right)

    how would i remove the E
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  3. #3
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    Quote Originally Posted by chogo View Post
    perhaps i should rather ask

    if i had

     P(z) = E\left(e^{(z-1)\lambda}\right)

    how would i remove the E
    Are you sure that is the right formula? Because there is no random variable in e^{(z-1)\lambda}, it is a constant with respect to the expectation and  P(z) = E\left(e^{(z-1)\lambda}\right) = e^{(z-1)\lambda}. When specifying an expectation, the random variable involved should be made clear, for example, by writing P(z) = E_X z^X when the random variable is X.
    Last edited by JakeD; June 3rd 2007 at 07:56 AM.
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    thank you jakeD sorry yeah i should not have put the expectation there. My bad. Was just trying to write something simpler to see if anyone had any suggestions

    The expectation is valid in my original equation, where the random variable is  T_j

    the thing is how did the person who did that derivation remove the expectation? and arrive at the second formula.


    ->

    any suggestions?

    i thank you so much for your help
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    Quote Originally Posted by chogo View Post
    thank you jakeD sorry yeah i should not have put the expectation there. My bad. Was just trying to write something simpler to see if anyone had any suggestions

    The expectation is valid in my original equation, where the random variable is  T_j

    the thing is how did the person who did that derivation remove the expectation? and arrive at the second formula.


    ->

    any suggestions?

    i thank you so much for your help
    What is the distribution of the random variable T_j ? You haven't said. The expectation uses that distribution.

    Let M_{T_j}(t) be the moment generating function of T_j . Then comparing the first and second equations

    E_{T_j} (e^{(z-1)\theta T_j / 2j}) = M_{T_j}((z-1)\theta/2j ) = (1 - t 2j/(j-1))^{-1}

    where t = (z-1)\theta / 2j. Then

    M_{T_j}(t) = (1 - t 2j/(j-1))^{-1} = \frac{(j-1)/2j}{(j-1)/2j - t}

    which is the MGF of a Gamma(1,(j-1)/2j) distribution, that is, an exponential distribution with parameter \lambda = (j-1)/2j. So it appears that is the distribution of T_j. Is that correct?
    Last edited by JakeD; June 3rd 2007 at 03:08 PM.
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    My original comment won't work, but I have to wonder if "E" isn't supposed to be a " \Sigma?"

    -Dan
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    firstly JakeD and topsquark i cant thank you enough

    Top shark, yes its definitley an E

    yes T_j is assumed to be exponentially distributed

    also yes this term does cancel out



    and becomes

     P_n(z) = \prod^n_{j=2}\left(1-\frac{(z-1)\theta}{j-1}\right)^{-1}

    did he substitute a value for  T_j ?

    i dont know why the original equation which is of the form of a poisson generating function becomes what is is now, which you said is a gamma(1,1).

    im almost 100% certain this is not wrong, as its a very famous theory developed in mathematical biology and very well founded.

    thank you so much for you help again its really much appreciated. If you want i can write the entire derivation our for you guys, will this help?
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    what did you do here




    this seems to be exactly what i need
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    Quote Originally Posted by chogo View Post
    firstly JakeD and topsquark i cant thank you enough

    Top shark, yes its definitley an E

    yes T_j is assumed to be exponentially distributed

    also yes this term does cancel out



    and becomes

     P_n(z) = \prod^n_{j=2}\left(1-\frac{(z-1)\theta}{j-1}\right)^{-1}

    did he substitute a value for  T_j ?

    i dont know why the original equation which is of the form of a poisson generating function becomes what is is now, which you said is a gamma(1,1).

    im almost 100% certain this is not wrong, as its a very famous theory developed in mathematical biology and very well founded.

    thank you so much for you help again its really much appreciated. If you want i can write the entire derivation our for you guys, will this help?
    This post appears to be looking at my post before I finished editing it. So it responds to comments of mine I resolved and edited out.
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    Quote Originally Posted by JakeD View Post
    Let M_{T_j}(t) be the moment generating function of T_j . Then comparing the first and second equations

    E_{T_j} (e^{(z-1)\theta T_j / 2j}) = M_{T_j}((z-1)\theta/2j ) = (1 - t 2j/(j-1))^{-1}

    where t = (z-1)\theta / 2j.
    Quote Originally Posted by chogo View Post
    what did you do here




    this seems to be exactly what i need
    The explanations before and after this equation are needed. I used the definition of moment generating function and then equated corresponding parts of your first and second equations while substituting in the variable t.
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    thank your help is invaluable. but one last problem. Like you before i used the definition

     M(t) = E(e^{t,x})

    to go from  E_{t_j}(e^{(z-1)\theta T_j/2j}) = M_{t_j}((z-1)\theta T_j/2)

    but when you said you compared the forms of the first and second equations to arrive at the third equation im not fully satistifed.

    Is there a formal definition which says the moment generating function i have is equal to  (1-t2_j / (j-1))^{-1}

    where
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    Quote Originally Posted by JakeD View Post
    What is the distribution of the random variable T_j ? You haven't said. The expectation uses that distribution.

    Let M_{T_j}(t) be the moment generating function of T_j . Then comparing the first and second equations

    E_{T_j} (e^{(z-1)\theta T_j / 2j}) = M_{T_j}((z-1)\theta/2j ) = (1 - t 2j/(j-1))^{-1}

    where t = (z-1)\theta / 2j. Then

    M_{T_j}(t) = (1 - t 2j/(j-1))^{-1} = \frac{(j-1)/2j}{(j-1)/2j - t}

    which is the MGF of a Gamma(1,(j-1)/2j) distribution, that is, an exponential distribution with parameter \lambda = (j-1)/2j. So it appears that is the distribution of T_j. Is that correct?
    Quote Originally Posted by chogo View Post
    thank your help is invaluable. but one last problem. Like you before i used the definition

     M(t) = E(e^{t,x})

    to go from  E_{t_j}(e^{(z-1)\theta T_j/2j}) = M_{t_j}((z-1)\theta T_j/2)

    but when you said you compared the forms of the first and second equations to arrive at the third equation im not fully satistifed.

    Is there a formal definition which says the moment generating function i have is equal to  (1-t2_j / (j-1))^{-1}

    where
    I was trying to deduce what the distribution of T_j was because you didn't say what it was. So I deduced it was an exponential distribution with parameter \lambda = (j-1)/2j. Is this correct?

    Now you know what the distribution of T_j is. So you can say what the moment generating function for T_j is. I was working backwards to deduce the distribution; you can work forwards knowing the distribution.

    I got that  M_T (t) = \frac{\lambda}{\lambda - t} = (1 - t/\lambda)^{-1} when T \sim \text{Exp}(\lambda) from a probability text. See Exponential distribution - Wikipedia, the free encyclopedia .
    Last edited by JakeD; June 3rd 2007 at 05:19 PM.
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  13. #13
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    sheer brilliance! thanks alot for the help
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