2
$\begingroup$

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 $E(E(X|Y)=E(X)$

  • 2
    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 $\operatorname{name}$.2012-09-30

1 Answers 1

2

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 $\mathrm{Var}(\hat{\theta}^{*})=\mathrm{Var}(E(\hat{\theta}|T))$ $\leq \mathrm{Var}(\hat{\theta})$

  • 0
    Also, see my comment under the question regarding the formatting of $\operatorname{Var}$.2012-09-30