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Math Help - Expected Value for Poisson

  1. #1
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    Expected Value for Poisson

    A little context, here is the problem I am working on:

    For a random variable X, let g(X) be given by g(X) = abs(X - m(X)), where m(X) is the median of X. Find g(X) for a Poisson random variable with mean 1.5.

    So the easy part: the median for this X is 1, so g(X) = abs(X-1). If this were a continuous random variable, I'd integrate and be done with it, but Poisson is discrete.

    Is it the case that E(|X-1|) = sum(p(x)*|x-1|)?

    Thanks.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Expected Value for Poisson

    Quote Originally Posted by joeyjoejoe View Post
    A little context, here is the problem I am working on:

    For a random variable X, let g(X) be given by g(X) = abs(X - m(X)), where m(X) is the median of X. Find g(X) for a Poisson random variable with mean 1.5.

    So the easy part: the median for this X is 1, so g(X) = abs(X-1). If this were a continuous random variable, I'd integrate and be done with it, but Poisson is discrete.

    Is it the case that E(|X-1|) = sum(p(x)*|x-1|)?

    Thanks.
    The expected value of the discrete random variable g(X)= |X-1| is...

    E\{|X-1|\}= \sum_{n=0}^{\infty} |n-1|\ P\{X=n\} (1)

    If X is a Poisson random variable, then is...

    P\{X=n \}= e^{-\lambda}\ \frac{\lambda^{n}}{n!} (2)

    ... so that for \lambda=1.5 You have...

    E\{|X-1|\}= e^{-1.5}\ \sum_{n=0}^{\infty} |n-1|\ \frac{1.5^{n}}{n!}=e^{-1.5}\ \{1+\sum_{n=1}^{\infty} (n-1)\ \frac{1.5^{n}}{n!}\}=

    = e^{-1.5}\ \{1+ 1.5\ \sum_{n=1}^{\infty} \frac{1.5^{n-1}}{(n-1)!} - \sum_{n=1}^{\infty} \frac{1.5^{n}}{n!} \} = e^{-1.5}\ (.5\ e^{1.5} +2) = .94626032... (3)

    Kind regards

    \chi \sigma
    Last edited by chisigma; September 2nd 2011 at 06:15 AM.
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