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Math Help - Moment Generating Functions

  1. #1
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    Moment Generating Functions

    Let Y be a random variable that takes on positive integers.

    p(k) = \frac{1}{2}p(k-1), k = 1, 2, ... ,

    Where
    p(k) = P(Y = k) for\ k = 0, 1, 2, ... ,

    a) Find the probability that Y = 0
    b) Find the moment generating function of Y.
    c) Find the first and second moments

    Attempt:
    (A)
    I don't really know what to do with it. I understand that it is an infinite geometric series that will sum to 1. But I'm not sure how to find Y = 0.
    Using the formula for geometric series, I can find various p(k) in terms of p(0), but I don't know how to find p(0) itself.

    a_n=a_0r^n

    p(1)=\frac{1}{2}p(0)

    (B)
    m(t)=E(e^t^y)=\sum_{y=0}^\infty e^t^y\frac{1}{2}p(y-1)=\frac{1}{2}\sum_{y=1}^\infty e^t^yp(y-1)=...?

    I'm I on the right track? I'm not sure how to proceed after that.

    (C)
    That just involves finding the first and second derivative of the result I find in B, correct?
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  2. #2
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    Re: Moment Generating Functions

    You declare Y to be a variable that takes on positive integers. 0 is not a positive integer. So, the probability that Y=0 must be 0, right? But, that implies that the probability that Y=k must be 0 for all positive integers k. Did you mean Y is a variable that can be any nonnegative integer?

    Now, \sum_{k=0}^\infty p(k) = 1.

    p(1) = \dfrac{1}{2}p(0)
    p(2) = \dfrac{1}{4}p(0)
    And in general, p(k) = \dfrac{1}{2^k}p(0)

    Plugging that in, we have
    \sum_{k\ge 0} 2^{-k}p(0) = p(0)\sum_{k\ge 0}2^{-k} = 1.

    Now, you know \sum_{k\ge 0}\left(\dfrac{1}{2}\right)^k = 2, so 2p(0) = 1 implies p(0)=\dfrac{1}{2}.

    Does that help at all?
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  3. #3
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    Re: Moment Generating Functions

    Sorry, my mistake. It is indeed non-negative integers.

    That definitively makes a lot more sense now.
    Just the clarify my understanding, in the last step, the summation equals 2 because:
    \sum_{k\ge 0}\left(\dfrac{1}{2}\right)^k =\dfrac{1}{1-\frac{1}{2}}=\dfrac{1}{\frac{1}{2}}=2?

    Then for part B, would this be the solution?
    m(t)=E(e^t^k)=\sum_{k=0}^\infty e^t^k\left(\dfrac{1}{2}\right)^k=e^t \sum_{k=0}^\infty \left(\dfrac{e}{2}\right)^k=e^t\left(\frac{1}{1-\dfrac{e}{2}}\right)

    Thanks!
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  4. #4
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    Re: Moment Generating Functions

    Quote Originally Posted by RedXIII View Post
    Sorry, my mistake. It is indeed non-negative integers.

    That definitively makes a lot more sense now.
    Just the clarify my understanding, in the last step, the summation equals 2 because:
    \sum_{k\ge 0}\left(\dfrac{1}{2}\right)^k =\dfrac{1}{1-\frac{1}{2}}=\dfrac{1}{\frac{1}{2}}=2?

    Then for part B, would this be the solution?
    m(t)=E(e^t^k)=\sum_{k=0}^\infty e^t^k\left(\dfrac{1}{2}\right)^k=e^t \sum_{k=0}^\infty \left(\dfrac{e}{2}\right)^k=e^t\left(\frac{1}{1-\dfrac{e}{2}}\right)

    Thanks!
    Yes for the reason why the summation equals 2. No for part B.

    e^{tk} \neq e^te^k = e^{t+k}

    So,
    \begin{align*}m(t) & = E(e^{tk}) \\ & = \sum_{k=0}^\infty e^{tk}\left(\dfrac{1}{2}\right)^kp(0) \\ & = \sum_{k=0}^\infty \left(\dfrac{e^t}{2}\right)^kp(0) \\ & = \dfrac{\tfrac{1}{2}}{1-\tfrac{e^t}{2}} \\ & = \dfrac{1}{2-e^t}\end{align*}

    Then m(0) = \dfrac{1}{2-e^0} = 1

    m'(t) = \dfrac{e^t}{(2-e^t)^2}, m'(0) = \dfrac{e^0}{(2-e^0)^2} = 1

    m''(t) = \dfrac{e^t(2-e^t)^2 + 2e^{2t}(2-e^t)}{(2-e^t)^3} = \dfrac{e^t(e^t+2)}{(2-e^t)^3}, m''(0) = \dfrac{e^0(e^0+2)}{(2-e^0)^3} = 3
    Last edited by SlipEternal; October 18th 2013 at 06:05 PM.
    Thanks from RedXIII
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