Context: 2nd year university statistics course textbook question

So I had to find two estimators (using method-of-moments and maximum likelihood estimation) of $\theta$ for a random sample $X_1, ..., X_n$ from a population with pmf $f(X=x)=\theta^x(1-\theta)^{1-x}$ for $x=0$ or $x=1$ where $\theta \in [0, 0.5]$ is a model parameter. I recognise this is a Bernoulli distribution.

I found that both methods gave the same estimator $T=\frac{1}{n} \sum^n_{i=1}X_i$ (the sample mean). The next part of the question required me to find the mean squared error of the two estimators. I have a couple questions:

  1. Since the estimators are the same, does this mean their mean squared erors will be too?
  2. How should I go about calculating the mean squared error? I know $MSE(T) = Var(T)+[Bias(T)]^2$, but for the $Var(T)$ component I don't know how to calculate $E(T^2)$. Or would it be better to calculate it via $E[(T-\theta)^2]$?


  • $\begingroup$ The estimators are not same. Notice the range of $\theta$. $\endgroup$ Commented Sep 28, 2020 at 13:34
  • $\begingroup$ @StubbornAtom : thanks for the hint $\endgroup$
    – tommik
    Commented Sep 29, 2020 at 9:19
  • $\begingroup$ @StubbornAtom How do you implement this range of $\theta$ into the solution for the MLE estimator? And why is it not part of the solution for the MoM estimator? $\endgroup$
    – Tikak
    Commented Sep 30, 2020 at 1:58
  • 1
    $\begingroup$ A comparison of the MSEs is shown here: math.stackexchange.com/q/2804396/321264. $\endgroup$ Commented Oct 2, 2020 at 4:19

1 Answer 1



  1. The two estimator are not the same.

    • $\hat{\theta}_{MM}=\overline{X}_n$
    • $\hat{\theta}_{ML}=min[\overline{X}_n;\frac{1}{2}]$
  2. I do not know if the exercise asks you to find analytically the two MSE's, but if $\overline{X}_n\leq\frac{1}{2}$ the two MSE's are the same, and equal to the sample means' variance: $\frac{\theta(1-\theta)}{n}$. On the contrary, if $\overline{X}_n>\frac{1}{2}$ the first estimator does not make sense.

Restricted MLE

in this example, Likelihood's domain is restricted in $\theta \in[0;0.5]$ so it is self evident that if $\overline{X}>0.5$ the likelihood is strictly increasing and its argmax is on the border: $\hat{\theta}_{ML}=0.5$

Let's look at the following example:

Let's draw an unfair coin 10 times. Suppose we have the two following cases

  1. 3 Successes on 10 Draws

  2. 7 Successed on 10 Draws

the two likelihoods are the following

enter image description here


Let's have a focus on the MSE(ML)

This changes if the estimator "sample mean" is greater than 0.5 or not.

  • If $\overline{X}_n\leq 0.5$ we have $\hat{\theta}_{ML}=\overline{X}_n$ so it is an unbiased estimator and thus its MSE=VAR(Sample mean) that is $\frac{\theta(1-\theta)}{n}$ as well known and easy proved below


  • If $\overline{X}_n> 0.5$ we have $\hat{\theta}_{ML}=\frac{1}{2}$ thus its $ MSE= (Bias)^2$ given that its variance is zero. (The estimator is constant). In other words $MSE=(\frac{1}{2}-\theta)^2$
  • $\begingroup$ How do you know the estimators are unbiased? $\endgroup$
    – Viv4660
    Commented Sep 28, 2020 at 22:05
  • $\begingroup$ @Viv4660 : made some edits $\endgroup$
    – tommik
    Commented Sep 29, 2020 at 9:19
  • $\begingroup$ @tommik I am confused as to how you got the estimator using MLE to be the minimum of the sample mean and $0.5$. Is this something you calculate when finding the critical point of the log-likelihood function? $\endgroup$
    – Tikak
    Commented Sep 30, 2020 at 1:57
  • $\begingroup$ @Tikak : In your example there is a restriction over $\theta$ domain so, if $\overline{X}>0.5$ the likelihood is strictly increasing. I did you an example in the edit of my answer $\endgroup$
    – tommik
    Commented Sep 30, 2020 at 7:14
  • 1
    $\begingroup$ @Tikak : I update my answer with another edit.... $\endgroup$
    – tommik
    Commented Oct 1, 2020 at 8:09

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