Monday, June 16, 2008

explaining the variance of the Poisson distribution

I'm reading a lot about probability and statistics these days. It's very interesting and practical. Many of the formulae and proofs involved are pretty ugly, but one of the cuter proofs is the proof of the variance of the Poisson distribution. I guess if you read this far, you know what 'variance' and 'Poisson distribution' are, but for a brief reminder, 'variance' is the expected value of the square of the difference between each point on the pdf and the mean, i.e., . The Poisson distribution is a special case of the binomial distribution which tends to be useful when events happen sporadically, but with more or less the same frequency over time. A common example of such a real-life distribution would be how often errors occur on a noisy connection.

Anyway, on to the explanation of the variance itself. Recall that, for the Poisson distribution, , where is the expected value multiplied by units of time. (I'm using instead of because the event in the Poisson distribution is said to happen times.) Also recall that the Taylor expansion of is . (And you can look up the proofs for those things on your own.) Letting in the Taylor expansion above equal , you'll notice that dividing the series by results in , the sum of the terms of the Poisson pdf. Appropriately, they add up to one. That'll be useful in a moment.

Now, consider trying to figure out the variance using the naïve method. (remember, using ) can't possibly work because ! So one would essentially expect zero. Since expectation of a constant is always that constant and different Poisson distributions obviously have different variances, that doesn't work. Putting the expectation in the form doesn't work either, because (skip because it adds nothing to the series) turns into an ugly sequence , which can't be manipulated into the Taylor expansion above. The solution is to find the expectation . The sequence for ( is the first where the factorial term is valid) converges neatly. is the same as . The bracketed term of course converges to one, as seen above, and so is (or ). Then is . So the variance of the Poisson distribution is the same as the mean!

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