Suppose $X_1, X_2, \dots, X_n$ are i.i.d $N(\theta, 1),\theta_0 \le \theta \le \theta_1$, where $\theta_0 \lt \theta_1$ are two specified numbers. Find the MLE of $\theta$ and show that it is better than the sample mean $\bar X$ in the sense of having smaller mean squared error.

Trial: Here MLE of $\theta$ is $$\hat\theta = \begin{cases} \theta_0, &\bar X \lt \theta_0\\ \bar X, &\theta_0 \le \bar X \le \theta_1 \\ \theta_1, &\bar X \gt \theta_1 \end{cases}$$

I can't show that $E[(\hat\theta -\theta)^2] \lt Var(\bar X)=\dfrac{1}{n}$.

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    $\begingroup$ First of all, get your notation correct. You are asked to show that $E[(\hat{\theta}-\theta)^2] < \operatorname{var}(\bar{X})$ and not $E[\hat\theta -\theta]^2$ as you have it. Your expression would be taken to mean $\left(E[\hat\theta -\theta]\right)^2$ the square of the mean, whereas what is wanted is the mean of the square. Then argue that the difference between the distributions of $\bar{X}$ and $\hat{\theta}$ is that all the probability mass to the left of $\theta_0$ (to the right of $\theta_1$) has been replaced by a point mass at $\theta_0$ (at $\theta_1$) and this reduces the variance. $\endgroup$ Commented May 10, 2013 at 11:58
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    $\begingroup$ @DilipSarwate unfortunately some authors actually use such an ambiguous notation. I applaud your encouragement of its avoidance. $\endgroup$
    – Glen_b
    Commented May 11, 2013 at 6:36

1 Answer 1


I assume that you already know that $\operatorname{var}(\bar{X}) = \frac{1}{n}$ or can prove this result if it is also asked for, and also that you know that $E[\bar{X}] = \theta$.

Expanding on my comment, let $g(x)$ denote the probability density function of $\bar{X}$. Then, $$\begin{align} \operatorname{var}(\bar{X}) &= \int_{-\infty}^\infty (x-\theta)^2g(x)\,\mathrm dx\\ &= \int_{-\infty}^{\theta_0} (x-\theta)^2g(x)\,\mathrm dx + \int_{\theta_0}^{\theta_1}(x-\theta)^2g(x)\,\mathrm dx + \int_{\theta_1}^\infty (x-\theta)^2g(x)\,\mathrm dx\\ &> (\theta_0-\theta)^2\int_{-\infty}^{\theta_0}g(x)\,\mathrm dx + \int_{\theta_0}^{\theta_1}(x-\theta)^2g(x)\,\mathrm dx + (\theta_1-\theta)^2\int_{\theta_1}^\infty g(x)\,\mathrm dx\\ &= (\theta_0-\theta)^2P\{\hat{\theta}=\theta_0\} + \int_{\theta_0}^{\theta_1}(x-\theta)^2g(x)\,\mathrm dx + (\theta_0-\theta)^2P\{\hat{\theta}=\theta_1\}\\ &= E[(\hat{\theta}-\theta)^2]. \end{align}$$ I will leave it to you to figure out how the inequality in the third line comes about, how two integrals become probabilities involving $\hat{\theta}$ in the fourth line, and why the sum on the fourth line works out to be $E[(\hat{\theta}-\theta)^2]$ as claimed on the last line.

Incidentally, though you have tagged this with normal-distribution, the result holds as long as the i.i.d. random variables have finite variance.


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