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Math Help - Poisson distribution question

  1. #1
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    Poisson distribution question

    Q has Poisson dist. Q will be observed tomorrow for once. If Q turns out to be a certain #, such as k, the coin will be tossed k times. Let X be the number of times tails come up in k tosses. If Q turns out to be zero, then X is also zero.

    1. Find the chance that X is equal to 1. This means that the chance tails comes up exactly once in k tosses. (The answer should be a formula involving parameters of Poisson dist.)

    2. Find the chance that X=n. (The answer should be a formula involving parameters of Poisson dist.)
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  2. #2
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    Walk through it. It's called exploration.

    Q is Poisson with parameter L, so E(Q=k) = (e^-L)(L^k)/k! for k in 0, 1, 2, ...

    X is Binomial with parameters n = k and p = 1/2. p(X=1) = k*(1/2)*(1/2)^(k-1) = k*(1/2)^k

    So, p(X=1,Q) = [(e^-L)(L^0)/0!]*0*(1/2)^0 + [(e^(-L))(L^1)/1!]*1*(1/2)^1 + [(e^(-L))(L^2)/2!]*2*(1/2)^2 + [(e^(-L))(L^3)/3!]*3*(1/2)^3 + ...

    = (e^(-L))*{[(L^0)/0!]*0*(1/2)^0 + [(L^1)/1!]*1*(1/2)^1 + [(L^2)/2!]*2*(1/2)^2 + [(L^3)/3!]*3*(1/2)^3 + ...}

    = (e^(-L))*{[(L^1)/1!]*1*(1/2)^1 + [(L^2)/2!]*2*(1/2)^2 + [(L^3)/3!]*3*(1/2)^3 + ...}

    = (e^(-L))*(L/2)*{1 + [L/2!]*2*(1/2) + [(L^2)/3!]*3*(1/2)^2 + ...}

    = [e^(-L)]*[e^(L/2) - 1]

    Do you recognize it as anything familair?

    Maybe examining the Moment Generating Function would be more instructive.
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  3. #3
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    I am not sure how you jumped from the second to the last part to the answer. Could you explain a little more? I did take a look at the Moment Generating Function, but I am not really sure how to relate it to this.
    Also, when X is 1, i see that it's simple. For part number 2, it gets really complicated, but is it the same concept? What should I look for in number 2?
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    Adding up infinite series is a nontrivial exercise. That's why I suggested the Moment Generating Functions. You might find a table with this one.

    \sum_{k=1}^{\infty}L^{k-1}\cdot \frac{1}{k!}\;=\;\frac{1}{L}\cdot \left(E^{L}-1\right)

    \sum_{k=1}^{\infty}L^{k-1}\cdot \frac{1}{k!}\;=\;\frac{1}{L}\cdot \left(E^{L}-1\right)

    I just used my CAS to tell me the sum. Perhaps some other soul will demonstrate it, but it escapes me at the moment.
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  5. #5
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    edit:didn't see previous post
    What should I look for in number 2?
    part 2 is the same concept.

    Whenever you are asked to do a sum involving a poisson distribution, you may find it helpful to remember these "tricks"
    1: can you express the sum in terms of another distribution function (possibly also posson), then use the fact that distribution functions always sum to 1?

    And/Or

    2: can you express the sum as a power series for e^{something}

    For this question (part 2):
    Hint #1 : You can write the binomial function as n!/(k!(n-k))
    Hint #2: if you wade through the algebra you may find the first "trick" useful at the end
    Last edited by SpringFan25; April 28th 2011 at 04:45 AM.
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  6. #6
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    Quote Originally Posted by TKHunny View Post
    Adding up infinite series is a nontrivial exercise. That's why I suggested the Moment Generating Functions. You might find a table with this one.

    \sum_{k=1}^{\infty}L^{k-1}\cdot \frac{1}{k!}\;=\;\frac{1}{L}\cdot \left(E^{L}-1\right)

    \sum_{k=1}^{\infty}L^{k-1}\cdot \frac{1}{k!}\;=\;\frac{1}{L}\cdot \left(E^{L}-1\right)

    I just used my CAS to tell me the sum. Perhaps some other soul will demonstrate it, but it escapes me at the moment.

    [where a = lambda to save typing]



    EDIT Solution removed....just slightly suspicious that this is assessed work...call me paranoid
    Last edited by SpringFan25; April 28th 2011 at 06:20 AM.
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