How is this a MLE?

#1
In all the literature I can find it is stated (and "proven" trivially) that for i.i.d. samples r with Rayleigh distribution \(\sigma\) the MLE is \(\widehat{\sigma} = \frac{\sum r_i^2}{2n}\), and it is an unbiased estimator for \(\sigma\).

But any Monte Carlo test shows that's not true: Only the square root of that MLE is anything like an estimator for \(\sigma\).

Obviously I'm misunderstanding what it means to be a Maximum Likelihood Estimator. Can somebody explain what I'm missing?
 

BGM

TS Contributor
#2
I presume you follow this parametrization for Rayleigh distribution:

http://en.wikipedia.org/wiki/Rayleigh_distribution

Then it is very standard to show that the MLE of \( \sigma^2 \), \( \widehat{\sigma^2} = \frac {1} {2n} \sum_{i=1}^n r_i^2 \)

By functional invariance property of MLE,

\( \hat{\sigma} = \widehat{\sqrt{\sigma^2}} = \sqrt{\widehat{\sigma^2}}
= \sqrt{\frac {1} {2n} \sum_{i=1}^n r_i^2} \)
 
#3
Yes, I was using that parameterization for the distribution. Although I can see it makes just as much sense to use \(\sigma^2\) as the distribution parameter.

I assume then that taking the square root of the MLE is what introduces bias?

In any case, once we sample the parameter we want to be able to say things about the estimated mean and stdev, both of which are a function of the square root. This question now ties into my other about computing correction factors on this distribution. I think I need to plead ignorance of theoretical statistics:

  1. Does the fact that this MLE is "unbiased" mean that one would not apply correction factors to estimates derived from even very small samples?
  2. Having an estimated parameter for a continuous distribution, if we then want to look at sample moments (mean, stdev) and confidence intervals that are nonlinear functions (in this case, sqrt) of the unbiased parameter, is there a better approach than to apply correction factors?