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

$\endgroup$
  • 2
    $\begingroup$ 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 $\operatorname{name}$. $\endgroup$ – joriki Sep 30 '12 at 18:00
2
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ There's a preview window underneath the answer field. If it's not working for you, it might be worthwhile to post a bug report at meta.math.stackexchange.com (stating your browser version). $\endgroup$ – joriki Sep 30 '12 at 21:21
  • $\begingroup$ Also, see my comment under the question regarding the formatting of $\operatorname{Var}$. $\endgroup$ – joriki Sep 30 '12 at 22:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.