Expected Value for Poisson

**joeyjoejoe** · August 31st 2011, 01:52 PM

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.

**chisigma** · September 2nd 2011, 05:14 AM

Originally Posted by joeyjoejoe

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

$\chi$ $\sigma$

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