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Math Help - Poisson Expectation Value

  1. #1
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    Poisson Expectation Value

    Let X be Poisson ( \lambda) with \lambda a positive interger. Show E{|X - \lambda|} = \frac{2 \lambda^\lambda e^{-\lambda}}{(\lambda - 1)!} and that \sigma^2=\lambda.

    We JUST started talking about Expectation Value, and I've never used it in any class before so I am completely confused. I've looked at the whole chapter, but I just cannot figure it out. I do know that for Poisson, EX = \lambda, but I don't understand what to do with the absolute value or the subtraction of lambda from X in the expectation.

    If it also helps, I know that P(X=k) for Poisson is equal to \frac{\lambda^k}{k!}e^{-\lambda}. Can anyone help me out? Thanks so much!
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  2. #2
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    Solution using Taylor Series Expansion

    Quote Originally Posted by azdang View Post
    Let X be Poisson ( \lambda) with \lambda a positive interger. Show E{|X - \lambda|} = \frac{2 \lambda^\lambda e^{-\lambda}}{(\lambda - 1)!} and that \sigma^2=\lambda.

    We JUST started talking about Expectation Value, and I've never used it in any class before so I am completely confused. I've looked at the whole chapter, but I just cannot figure it out. I do know that for Poisson, EX = \lambda, but I don't understand what to do with the absolute value or the subtraction of lambda from X in the expectation.

    If it also helps, I know that P(X=k) for Poisson is equal to \frac{\lambda^k}{k!}e^{-\lambda}. Can anyone help me out? Thanks so much!

    E(X) = Σ (from x=1 to ∞) (xλ^(x)e^(-λ))/x! = ∑ (x/x!)(λ^(x)e^(-λ)). Since x/x! = 1/(x-1)!, we can write
    = ∑ (1/(x-1)!)(λ^(x)e^-λ)
    = e^(-λ) ∑ (λ^x/(x-1)!), we can move e^(-λ) across because it is not affected by the index. Next is a little tricky but follow along...
    = e^-λ ∑ (from x=0 to ∞) (λ^x+1)/(x-1+1)!
    = e^-λ ∑ (λ^x)(λ)/x!
    = λe^-λ ∑ λ^x/x!, now we can use the Taylor Series Expansion of eλ, if you have forgotten what it looks like, you can find the solution in any calculus book, including the Stewart Calculus book if thatís what you have. Since eλ = ∑ λ^x/x!, we can state
    = λe^(-λ)e^λ
    = λe^(-λ+λ)
    = λe^0
    = λ1
    = λ. ■

    Sorry I don't know how to code. If you have any questions on anything just let me know. I hope that it makes sense. If you want to code what you understand the solution to be, I can look it over.
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  3. #3
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    Hi Yvonne. Thanks for getting back to me. If I'm not mistaked though, I think the work you just posted if just for E(X), which I noted I already know is \lambda. What I'm confused about is how to then find E{|X - |}, which should apparently be equal to and how to show that for E{|X - |}.
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  4. #4
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    Intro to Prob Theory Book

    I suggest using the following book to supplement your textbook:
    Introduction to Probability Theory by Hoel, Port, and Stone.
    ISBN: 0395244978

    Just look for it online at your school's library. I'm am certain they should have it. It has a very theoretical approach to probability so it contains a lot of proofs, instead of focusing on calculation like most modern textbooks. Anyway, I think this book should be somewhat useful. You just have to understand a lot of algebra to get through the calculations in the proofs.

    Best of luck to you this semester!
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