showing that $E[(\hat\theta -\theta)^2] \lt Var(\bar X)=\dfrac{1}{n}$.

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Suppose $X_1, X_2, \dots, X_n$ are i.i.d $N(\theta, 1),\theta_0 \lt\theta$ , 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 $$\hat\theta = \begin{cases} \theta_0, &\bar X \lt \theta_0\\ \bar X, &\theta_0 \le \bar X \\ \end{cases}$$

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