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Math Help - Poisson distribution

  1. #1
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    Poisson distribution

    If X~Po(λ) and E(X^2) = 2, find λ.
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  2. #2
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    Re: Poisson distribution

    That's pretty straight forward, isn't it? The Poisson distribution is given by P(k)= \frac{\lambda^k}{k! e^k} where k is a positive integer. So the Expected value of X^2 is \sum_{k= 1}^\infty \frac{\lambda^k k^2}{k!e^k}= 2.
    What must \lambda be in order that that equation be true?
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  3. #3
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    Re: Poisson distribution

    Quote Originally Posted by HallsofIvy View Post
    X^2 is \sum_{k= 1}^\infty \frac{\lambda^k k^2}{k!e^k}= 2.
    That's exactly where I get stuck, I can't solve that, I don't know what to do after I expanded it in the form of (λ^1)(1^2)/1!(e^1) + (λ^2)(2^2)/2!(e^2) + (λ^3)(3^2)/3!(e^3) +..... = 2.
    Please give some hint or guide, thank you.
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    Re: Poisson distribution

    Can also try something like this.

    Since X~Po(λ), we know two things:

    1) E(X) = λ
    2) Var(X) = λ = E(X^2) - E(X)^2

    combining these two gives:
    λ = E(X^2) - λ^2
    E(X^2) = λ + λ^2

    If this is equal to 2 then its the same as solving the quadratic λ + λ^2 = 2
    Thanks from alexander9408
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