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.


@MichaelHardy provided an example where we get a biased estimate of the mean of uniform distribution using MLE under a certain proposed model.



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.


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?


@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.

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    $\begingroup$ The mean of a uniform on zero and theta. $\endgroup$ Dec 17, 2016 at 20:02
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    $\begingroup$ I cannot follow your distinction between "estimating something about the population" and "a parameter of a hypothesized model". In all of parametric statistics, we parameterize a population by some parameters. Of course, we may thus run into issues of misspecification, but that does not seem to be relevant to the issue at hand. $\endgroup$ Dec 18, 2016 at 9:30
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    $\begingroup$ For example, that a population can be characterized by its parameters/moments, like the mean and variance (which would be sufficient for a normal population, for example). And: I do not think that people are any more or less pedantic with you than with anybody else on this forum. $\endgroup$ Dec 18, 2016 at 9:38
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    $\begingroup$ If you are feeling unhappy about the apparent sleight of hand of switching between "parameter" and "mean", let me define a certain non-negative distribution in terms of its mean $\mu$, with density $\frac{1}{2\mu}$ on its support of $[0, 2\mu]$... $\endgroup$
    – Silverfish
    Dec 18, 2016 at 12:50
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    $\begingroup$ Regarding your edit 2, many of these results are derived under regularity conditions which are not satisfied for the uniform example discussed in this thread, for which the sample space depends on the parameter. $\endgroup$ Dec 18, 2016 at 16:56

6 Answers 6


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.

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    $\begingroup$ Cagdas, It's not legitimate to ask for a countexample and then deny that you would propose something else! It's like asking for an example of a fruit that is not red, being shown a blueberry, and then saying it doesn't count because you don't like blueberries. $\endgroup$
    – whuber
    Dec 17, 2016 at 21:45
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    $\begingroup$ That's not relevant to the question you asked. $\endgroup$
    – whuber
    Dec 17, 2016 at 21:49
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    $\begingroup$ @CagdasOzgenc : Whether the MLE is biased or not depends on the model. There's no such thing as an MLE without a model. And if you alter the model, you alter the MLE. $\endgroup$ Dec 17, 2016 at 21:52
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    $\begingroup$ @CagdasOzgenc Here's a socratic question: the sample mean is an unbiased estimator of what? You need a model to have a parameter to be estimating. $\endgroup$ Dec 17, 2016 at 22:32
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    $\begingroup$ The mean of an i.i.d. sample is an unbiased estimator of the population mean, but one cannot speak of a maximum-likelihood estimator of anything without more structure than what is needed to speak of an unbiased estimator of something. $\endgroup$ Dec 17, 2016 at 23:14

Here's an example that I think some may find surprising:

In logistic regression, for any finite sample size with non-deterministic outcomes (i.e. $0 < p_{i} < 1$), any estimated regression coefficient is not only biased, the mean of the regression coefficient is actually undefined.

This is because for any finite sample size, there is a positive probability (albeit very small if the number of samples is large compared with the number of regression parameters) of getting perfect separation of outcomes. When this happens, estimated regression coefficients will be either $-\infty$ or $\infty$. Having positive probability of being either $-\infty$ or $\infty$ implies the expected value is undefined.

For more on this particular issue, see the Hauck-Donner-effect.

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    $\begingroup$ This is quite clever. I wonder if the MLE of logistic regression coefficients is unbiased conditional on the nonoccurence of the Hauck-Donner effect? $\endgroup$ Dec 28, 2016 at 1:48
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    $\begingroup$ @gung: Short answer: ignoring the Hauck-Donner effect, there is still upward bias in absolute regression coefficients (i.e. negative coefficients have downward bias, positive have upward bias). Interestingly, there appears to be a bias toward 0.5 in estimated probabilities. I've started writing up about it on this post, but haven't put up my results on the biases of the estimated probabilities. $\endgroup$
    – Cliff AB
    Dec 28, 2016 at 18:36
  • $\begingroup$ David Firth has some related papers on using jeffrey priors as penalty for logistic regression $\endgroup$
    – nan hu
    Dec 16, 2021 at 3:10

Although @MichaelHardy has made the point, here is a more detailed argument as to why the MLE of the maximum (and hence, that of the mean $\theta/2$, by invariance) is not unbiased, although it is in a different model (see the edit below).

We estimate the upper bound of the uniform distribution $U[0,\theta]$. Here, $y_{(n)}$ is the MLE, for a random sample $y$. We show that $y_{(n)}$ is not unbiased. Its cdf is \begin{eqnarray*} F_{y_{(n)}}(x)&=&\Pr\{Y_1\leqslant x,\ldots,Y_n\leqslant x\}\\ &=&\Pr\{Y_1\leqslant x\}^n\\ &=&\begin{cases} 0&\qquad\text{for}\quad x<0\\ \left(\frac{x}{\theta}\right)^n&\qquad\text{for}\quad 0\leqslant x\leqslant\theta\\ 1&\qquad\text{for}\quad x>\theta \end{cases} \end{eqnarray*} Thus, its density is $$f_{y_{(n)}}(x)= \begin{cases} \frac{n}{\theta}\left(\frac{x}{\theta}\right)^{n-1}&\qquad\text{for}\quad 0\leqslant x\leqslant\theta\\ 0&\qquad\text{else} \end{cases} $$ Hence, \begin{eqnarray*} E[Y_{(n)}]&=&\int_0^\theta x\frac{n}{\theta}\left(\frac{x}{\theta}\right)^{n-1}dx\\ &=&\int_0^\theta n\left(\frac{x}{\theta}\right)^{n}dx\\ &=&\frac{n}{n+1}\theta \end{eqnarray*}

EDIT: It is indeed the case that (see the discussion in the comments) the MLE is unbiased for the mean in the case in which both the lower bound $a$ and upper bound $b$ are unknown. Then, the minimum $Y_{(1)}$ is the MLE for $a$, with (details omitted) expected value $$ E(Y_{(1)})=\frac{na+b}{n+1} $$ while $$ E(Y_{(n)})=\frac{nb+a}{n+1} $$ so that the MLE for $(a+b)/2$ is $$ \frac{Y_{(1)}+Y_{(n)}}{2} $$ with expected value $$ E\left(\frac{Y_{(1)}+Y_{(n)}}{2}\right)=\frac{na+b+nb+a}{2(n+1)}=\frac{a+b}{2} $$

EDIT 2: To elaborate on Henry's point, here is a little simulation for the MSE of the estimators of the mean, showing that while the MLE if we do not know the lower bound is zero is unbiased, the MSEs for the two variants are identical, suggesting that the estimator which incorporates knowledge of the lower bound reduces variability.

theta <- 1
mean <- theta/2
reps <- 500000
n <- 5
mse <- bias <- matrix(NA, nrow = reps, ncol = 2)

for (i in 1:reps){
  x <- runif(n, min = 0, max = theta)
  mle.knownlowerbound <- max(x)/2
  mle.unknownlowerbound <- (max(x)+min(x))/2
  mse[i,1] <- (mle.knownlowerbound-mean)^2
  mse[i,2] <- (mle.unknownlowerbound-mean)^2
  bias[i,1] <- mle.knownlowerbound-mean
  bias[i,2] <- mle.unknownlowerbound-mean


> colMeans(mse)
[1] 0.01194837 0.01194413

> colMeans(bias)
[1] -0.083464968 -0.000121968
  • $\begingroup$ Because Wikipedia is proposing a different model to begin with. That's where my confusion lies. $\endgroup$ Dec 18, 2016 at 9:31
  • $\begingroup$ Yes, but once we adjust to the special case discussed here, namely $a=0$, we are back at square 1. In that case, we do not need the sample minimum for estimation anymore, as we know that the lower bound is zero, so that the MLE of the midpoint (=median=mean) simply becomes $(max+0)/2$ again. $\endgroup$ Dec 18, 2016 at 9:34
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    $\begingroup$ I have not worked out the details, but the MLE in that model could be unbiased if the minimum overestimates the lower bound by the same amount as the maximum underestimates the maximum, so that the midpoint is being estimated without bias. $\endgroup$ Dec 18, 2016 at 9:41
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    $\begingroup$ @CagdasOzgenc: unbiasedness is not the only or even the most important measure of better. By knowing one end of the support precisely, you may lose the balance between errors in estimating the mean, but you end up with (for example) a better estimate of the range $\endgroup$
    – Henry
    Dec 18, 2016 at 10:12
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    $\begingroup$ Maximum likelihood estimators are not always "best" across all criteria for small sample sizes. So what? They don't pretend to be, either. If you want to use a different estimator for your problem that has better properties according to some criterion for sample sizes that are in the neighborhood of your actual sample size, you're free to do so. I do so, and so do other people. No one is claiming that using MLE is justified in all situations just because it's MLE. $\endgroup$
    – jbowman
    Dec 18, 2016 at 15:10

Completing here the omission in my answer over at math.se referenced by the OP,

assume that we have an i.i.d. sample of size $n$ of random variables following the Half Normal distribution. The density and moments of this distribution are

$$f_H(x) = \sqrt{2/\pi}\cdot \frac 1{v^{1/2}}\cdot \exp\big\{-\frac {x^2}{2v} \big\} \\ E(X) = \sqrt{2/\pi}\cdot v^{1/2}\equiv \mu,\;\; \operatorname{Var}(X) = \left(1-\frac 2 \pi \right)v$$

The log-likelihood of the sample is

$$L(v\mid \mathbf x) = n\ln\sqrt{2/\pi}-\frac n2\ln v -\frac 1 {2v} \sum_{i=1}^n x_i^2$$

The first derivative with respect to $v$ is

$$\frac {\partial}{\partial v}L(v\mid\mathbf x) = -\frac n{2v} + \frac 1 {2v^2} \sum_{i=1}^n x_i^2,\implies \hat v_\text{MLE} = \frac 1n \sum_{i=1}^nx_i^2$$

so it is a method of moments estimator. It is unbiased since,

$$E(\hat v_\text{MLE}) = E(X^2) = \operatorname{Var}(X) + [E(X)])^2 = \left(1-\frac 2 \pi \right)v + \frac 2 \pi v = v$$

But, the resulting estimator for the mean is downward biased due to Jensen's inequality

\begin{align} \hat \mu_\text{MLE} = \sqrt{2/\pi}\cdot \sqrt {\hat v_\text{MLE}} \implies & E\left(\hat \mu_\text{MLE}\right) = \sqrt{2/\pi}\cdot E\left(\sqrt {\hat v_\text{MLE}}\,\right) \\[6pt] & < \sqrt{2/\pi}\cdot \left[\sqrt {E(\hat v_\text{MLE})}\,\right] = \sqrt{2/\pi}\cdot \sqrt v = \mu \end{align}


The famous Neyman Scott problem has an inconsistent MLE in that it never even converges to the right thing. Motivates the use of conditional likelihood.

Take $(X_i, Y_i) \sim \mathcal{N}\left(\mu_i, \sigma^2 \right)$. The MLE of $\mu_i$ is $(X_i + Y_i)/2$ and of $\sigma^2$ is $\hat{\sigma}^2 = \sum_{i=1}^n \frac{1}{n} s_i^2$ with $s_i^2 = (X_i - \hat{\mu}_i)^2/2 + (Y_i - \hat{\mu}_i)^2/2 = (X_i - Y_i)^2 / 4$ which has expected value $\sigma^2/4$ and so biased by a factor of 2.

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    $\begingroup$ While this example holds true, this actually defies one of the basic regularity conditions for asymptotic results of MLE's: that $k / n \rightarrow 0$, where $k$ is the number of parameters estimated and $n$ is the sample size. $\endgroup$
    – Cliff AB
    Dec 18, 2016 at 18:38
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    $\begingroup$ @CliffAB the assumption violation is that the parametric dimension is not fixed. The dimension of $\Theta$ goes to $\infty$ as $n \rightarrow \infty$. I think that's what you're saying, but don't know what $k$ means. The practical illustration of this example of course is that these results would be biased even in small samples and you have to use conditional likelihood, like a mixed effects model, to estimate $\sigma$ in this case. $\endgroup$
    – AdamO
    Dec 19, 2016 at 15:06

There is an infinite range of examples for this phenomenon since

  1. the maximum likelihood estimator of a bijective transform $\Psi(\theta)$ of a parameter $\theta$ is the bijective transform of the maximum likelihood estimator of $\theta$, $\Psi(\hat{\theta}_\text{MLE})$;
  2. the expectation of the bijective transform of the maximum likelihood estimator of $\theta$, $\Psi(\hat{\theta}_\text{MLE})$, $\mathbb{E}[\Psi(\hat{\theta}_\text{MLE})]$ is not the bijective transform of the expectation of the maximum likelihood estimator, $\Psi(\mathbb{E}[\hat{\theta}_\text{MLE}])$;
  3. most transforms $\Psi(\theta)$ are expectations of some transform of the data, $\mathfrak{h}(X)$, at least for exponential families, provided an inverse Laplace transform can be applied to them.

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