# Show that an estimator has a lower variance than another estimator

Let $X$ be observed data. Let $\hat{\theta}(X)$ be an unbiased estimate of $\theta$ and let T be a sufficient statistic for $\theta$. Define the new estimator $\hat\theta^{*}$ of $\theta$,

$$\hat\theta^{*}(X) =E(\hat\theta(X)| T)$$

Then, show that:

• $\hat\theta^{*}(X)$ has a variance that is lower than (or equal to) that of $\hat\theta$

Hint: for any two random variables $X$ and $Y$, $\operatorname{VAR}(X)= E(\operatorname{VAR}[X|Y]) +VAR[E(X|Y)]$ and $E(E(X|Y)=E(X)$

• Strings like VAR are interpreted as concatenated variables and thus get italicized. To get the right font and spacing for such function names, you can either use predefined commands like \sin, or generally \operatorname{name} to produce $\operatorname{name}$. – joriki Sep 30 '12 at 18:00

We have $\mathrm{Var}(E(\hat{\theta}|T))$ $=\mathrm{Var}(\hat{\theta})-E(\mathrm{Var}(\hat{\theta}|T))$. Since $\mathrm{Var}(\hat{\theta}|T)$ is a non-negative random variable, its expected value is also non-negative. So, we have that $\mathrm{Var}(\hat{\theta}^{*})=\mathrm{Var}(E(\hat{\theta}|T))$ $\leq \mathrm{Var}(\hat{\theta})$
• Also, see my comment under the question regarding the formatting of $\operatorname{Var}$. – joriki Sep 30 '12 at 22:53