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Math Help - problem with calculating expected value

  1. #1
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    problem with calculating expected value

    Let X be Poisson with Parameter λ
    Compute the mean of (1+X)^-1.

    thanks
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  2. #2
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    Quote Originally Posted by adirh View Post
    Let X be Poisson with Parameter λ
    Compute the mean of (1+X)^-1.

    thanks
    Start by noting that E\left(\frac{1}{1+X} \right) = \sum_{x = 0}^{+\infty} \frac{1}{1 + x} \cdot \frac{\lambda^x e^{-\lambda}}{x!} = \sum_{x = 0}^{+\infty}\frac{\lambda^x e^{-\lambda}}{(x+1)!} = \frac{1}{\lambda} \sum_{x = 0}^{+\infty}\frac{\lambda^{x+1} e^{-\lambda}}{(x+1)!}.

    \sum_{k = 0}^{+\infty}\frac{\lambda^{k} e^{-\lambda}}{k!} = 1 will be useful in calculating the above sum ....
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  3. #3
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    thanks but...

    first of all thanks for the answer but...
    The answer (i have a only the final answer) is λ^-1(1- e^-λ) and
    you said that
    so isn't the sum above shoudnt be (1- e^-λ) instead of 1???d

    thanks again
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  4. #4
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    Quote Originally Posted by adirh View Post
    first of all thanks for the answer but...
    The answer (i have a only the final answer) is λ^-1(1- e^-λ) and
    you said that
    so isn't the sum above shoudnt be (1- e^-λ) instead of 1???d

    thanks again
    Try to look further than your nose ^^

    \frac{1}{\lambda} \sum_{x = 0}^{+\infty}\frac{\lambda^{x+1} e^{-\lambda}}{(x+1)!}
    Make the change k=x+1


    and 1=\sum_{k = 0}^{+\infty}\frac{\lambda^{k} e^{-\lambda}}{k!}=e^{-\lambda}+ \sum_{k=1}^\infty \frac{\lambda^k e^{-\lambda}}{k!}
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  5. #5
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    Quote Originally Posted by adirh View Post
    first of all thanks for the answer but...
    The answer (i have a only the final answer) is λ^-1(1- e^-λ) and
    you said that
    so isn't the sum above shoudnt be (1- e^-λ) instead of 1???d

    thanks again
    \frac{1}{\lambda} \sum_{x = 0}^{+\infty} \frac{\lambda^{x+1} e^{-\lambda}}{(x+1)!} = \frac{1}{\lambda} \left( \frac{\lambda e^{-\lambda}}{1!} + \frac{\lambda^{2} e^{-\lambda}}{2!} + \, .... \right) = \frac{1}{\lambda} \left( 1 - e^{-\lambda}\right) using \sum_{k = 0}^{+\infty}\frac{\lambda^{k} e^{-\lambda}}{k!} = 1.
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