Can you provide an example of an MLE estimator of the mean that is biased?
I am not looking for an example that breaks MLE estimators in general by violating regularity conditions.
All examples I can see on the internet refer to the variance, and I can't seem to find anything related to the mean.
EDIT
@MichaelHardy provided an example where we get a biased estimate of the mean of uniform distribution using MLE under a certain proposed model.
However
https://en.wikipedia.org/wiki/Uniform_distribution_(continuous)#Estimation_of_midpoint
suggests that MLE is a uniformly minimum unbiased estimator of the mean, clearly under another proposed model.
At this point it is still not very clear to me what's meant by MLE estimation if it is very hypothesized model dependent as opposed to say a sample mean estimator which is model neutral. At the end I am interested in estimating something about the population and don't really care about the estimation of a parameter of a hypothesized model.
EDIT 2
As @ChristophHanck showed the model with additional information introduced bias but did not manage to reduce the MSE.
We also have additional results:
http://www.maths.manchester.ac.uk/~peterf/CSI_ch4_part1.pdf (p61)
http://www.cs.tut.fi/~hehu/SSP/lecture6.pdf (slide 2)
http://www.stats.ox.ac.uk/~marchini/bs2a/lecture4_4up.pdf (slide 5)
"If a most efficient unbiased estimator ˆθ of θ exists (i.e. ˆθ is
unbiased and its variance is equal to the CRLB) then the maximum
likelihood method of estimation will produce it.""Moreover, if an efficient estimator exists, it is the ML estimator."
Since the MLE with free model parameters is unbiased and efficient, by definition is this "the" Maximum Likelihood Estimator?
EDIT 3
@AlecosPapadopoulos has an example with Half Normal distribution on math forum.
It is not anchoring any of its parameters like in the uniform case. I would say that settles it, though he hasn't demonstrated the bias of the mean estimator.
Best Answer
Christoph Hanck has not posted the details of his proposed example. I take it he means the uniform distribution on the interval $[0,\theta],$ based on an i.i.d. sample $X_1,\ldots,X_n$ of size more than $n=1.$
The mean is $\theta/2$.
The MLE of the mean is $\max\{X_1,\ldots,X_n\}/2.$
That is biased since $\Pr(\max < \theta) = 1,$ so $\operatorname{E}({\max}/2)<\theta/2.$
PS: Perhaps we should note that the best unbiased estimator of the mean $\theta/2$ is not the sample mean, but rather is $$\frac{n+1} {2n} \cdot \max\{X_1,\ldots,X_n\}.$$ The sample mean is a lousy estimator of $\theta/2$ because for some samples, the sample mean is less than $\dfrac 1 2 \max\{X_1,\ldots,X_n\},$ and it is clearly impossible for $\theta/2$ to be less than ${\max}/2.$
end of PS
I suspect the Pareto distribution is another such case. Here's the probability measure: $$ \alpha\left( \frac \kappa x \right)^\alpha\ \frac{dx} x \text{ for } x >\kappa. $$ The expected value is $\dfrac \alpha {\alpha -1 } \kappa.$ The MLE of the expected value is $$ \frac n {n - \sum_{i=1}^n \big((\log X_i) - \log(\min)\big)} \cdot \min $$ where $\min = \min\{X_1,\ldots,X_n\}.$
I haven't worked out the expected value of the MLE for the mean, so I don't know what its bias is.