Re: Variance question

Robin R wrote:
thanks. I was a bit confused as I found an article where given
Poisson sample they calculte the mean and states that as variance.
I used to do the var/cov approach of matlab that is (x-mu_sample)^2
approach.
Now I am worried whether i am right. the Matlab var/cov does not
talk about the probability densiity functions

Hi Robin -

The Poisson distribution has only one parameter. Forget about what it represents for a second; the maximum likelihood estimator of that parameter is the sample mean. Now it just so happens that the mean of a Poisson distribution with parameter lambda is lambda, and it also happens that the variance of that distribution is lambda. Therefore the MLEs for the mean and the variance of the distribution are both the sample mean, lambda_mle.

The sample variance is also a perfectly reasonable estimator of the variance of a Poisson distribution, it just isn't the MLE. Consider the following similar situation:

Maximum likelihood says that to estimate the mean of a lognormal distribution (not the "mean parameter" mu, but the actual mean itself, exp(mu+.5*sigma^2)), you should log your data and compute the sample mean and variance, i.e. fit a normal to log(x), and then compute exp(mu_hat + .5*sigma_hat^2). But you can just as easily compute the sample mean of your original data. Both are valid, but one is the MLE and the other is not. Which is the "right" estimator? One answer is that it depends on whether you really believe in you assumption of a lognormal distribution or not -- the sample mean of the original data is going to be a decent estimator even if the data are not lognormal. Another answer is that the MLE is likely to be more efficient if the lognormal assumption is true.

The best answer is to figure out what assumptions you want to make, and simulate.

Hope this helps.

- Peter Perkins
The MathWorks, Inc.
.

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