# Calculating mean squared error of estimators

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 $$E[(T-\theta)^2]$$?

Thanks

• The estimators are not same. Notice the range of $\theta$. Sep 28, 2020 at 13:34
• @StubbornAtom : thanks for the hint Sep 29, 2020 at 9:19
• @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? Sep 30, 2020 at 1:58
• A comparison of the MSEs is shown here: math.stackexchange.com/q/2804396/321264. Oct 2, 2020 at 4:19

EDIT

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

EDIT2:

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

$$\mathbb{V}[\overline{X}_n]=\frac{1}{n^2}n\mathbb{V}[X_1]=\frac{\theta(1-\theta)}{n}$$

• 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 $$MSE=(\frac{1}{2}-\theta)^2$$
• How do you know the estimators are unbiased? Sep 28, 2020 at 22:05
• @Viv4660 : made some edits Sep 29, 2020 at 9:19
• @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? Sep 30, 2020 at 1:57
• @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 Sep 30, 2020 at 7:14
• @Tikak : I update my answer with another edit.... Oct 1, 2020 at 8:09