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Math Help - algebraic manipulation of probability generating functions

  1. #1
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    algebraic manipulation of probability generating functions

    I am revising for a statistics test and I have an urgent problem involving probability generating functions. I understand the concept its just the algebraic manipulation that gets me. My exam is only a few days away so if you could help me out that would be GREAT!

    I'm supposed to determine the probability distribution of X from the pgf below.

    Here is the working I have done so far:

    for (0<\alpha<1),
    \pi(z)=\frac{1-\alpha(1-z)}{1+\alpha(1-z)} = \frac{\frac{1-\alpha}{1+\alpha}+\frac{\alpha}{1+\alpha}z}{1-\frac{\alpha}{1+\alpha}z}

     = (\frac{1-\alpha}{1+\alpha}+\frac{\alpha}{1+\alpha}z)(1+[\frac{\alpha}{1+\alpha}z] +[\frac{\alpha}{1+\alpha}z]^2+...)

    therefore P(X=0)=\frac{1-\alpha}{1+\alpha}

    This matches exactly with what the lecturer had done exept I couldn't understand how he leapt to this next part:

    and for x=>1, P(X=x)=(\frac{1-\alpha}{1+\alpha})(\frac{\alpha}{1+\alpha})^x + (\frac{\alpha}{1+\alpha})(\frac{\alpha}{1+\alpha})  ^{x-1}

    =(\frac{\alpha}{1+\alpha})^x(\frac{1-\alpha}{1+\alpha}+1)=\frac{2\alpha^x}{(1+\alpha)^{  x+1}}

    Could someone please explain to me how he got this?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by zealot
    I am revising for a statistics test and I have an urgent problem involving probability generating functions. I understand the concept its just the algebraic manipulation that gets me. My exam is only a few days away so if you could help me out that would be GREAT!

    I'm supposed to determine the probability distribution of X from the pgf below.

    Here is the working I have done so far:

    for (0<\alpha<1),
    \pi(z)=\frac{1-\alpha(1-z)}{1+\alpha(1-z)} = \frac{\frac{1-\alpha}{1+\alpha}+\frac{\alpha}{1+\alpha}z}{1-\frac{\alpha}{1+\alpha}z}

     = (\frac{1-\alpha}{1+\alpha}+\frac{\alpha}{1+\alpha}z)(1+[\frac{\alpha}{1+\alpha}z] +[\frac{\alpha}{1+\alpha}z]^2+...)

    therefore P(X=0)=\frac{1-\alpha}{1+\alpha}

    This matches exactly with what the lecturer had done exept I couldn't understand how he leapt to this next part:

    and for x=>1, P(X=x)=(\frac{1-\alpha}{1+\alpha})(\frac{\alpha}{1+\alpha})^x + (\frac{\alpha}{1+\alpha})(\frac{\alpha}{1+\alpha})  ^{x-1}

    =(\frac{\alpha}{1+\alpha})^x(\frac{1-\alpha}{1+\alpha}+1)=\frac{2\alpha^x}{(1+\alpha)^{  x+1}}

    Could someone please explain to me how he got this?
    The pgf \pi (z) of a random variable X that takes values in \{0,\ 1,\ 2,\ ...\} is:

    \pi(z)=\sum_{n=0}^{\infty} P(X=n)z^n

    where P(X=n) is the probability that X takes a value n \in \{0,\ 1,\ 2,\ ...\}.

    So if you write the pgf as a power series in z the probability
    P(X=n) is equal to the coeficient of z^n in the expansion.

    What you lecturer has done is collect together the terms containing the same
    powers of z, and the resulting coefficients are the required probabilities.

    RonL
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