Given a a random sample $X_1, X_2, X_3, X_4$ and family of densities $\mathcal{P} = \left\{ f_\theta: \theta \in \Theta \right\}$, where $f_\theta(x) = \frac{1}{2}\mathbb{I}_{[\theta-1, \theta + 1]}$, we can estimate $\theta$ by $\hat{\theta} = X_{(2)} + \frac{1}{5}$. Now we also know that $(S, T)= (X_{(1)}, X_{(4)})$ is a sufficient statistic for $\theta$ so we may Rao-Blackwellize our estimator: $$ \hat{\theta}^*= \mathbb{E}(\hat{\theta}|(X_{(1)},X_{(4)})) = \mathbb{E}(X_{(2)}|(X_{(1)},X_{(4)})) + \frac{1}{5}.$$ Unfortunately I haven't been able to compute the last conditional expectation. Any input would be greatly appreciated! :)
$\begingroup$
$\endgroup$
2
-
1$\begingroup$ I am assuming this is homework or self-study, so will give a hint: given $X_{(1)}$ and $X_{(4)}$, what distribution do each of the remaining two variates have, unconditional upon order? Given that, the next step is to work out the distribution of the smaller of the two variates, and from that get to the expected value. $\endgroup$– jbowmanCommented Dec 13, 2017 at 23:00
-
$\begingroup$ Thanks! I suspected as much. The two remaining variates are uniformly distributed on the interval staked out by the minimum and maximum. Then all that is necessary is to look at the density of minimal order statistic, say $V$, on said interval and compute $\mathbb{E}(V)$. Thanks again! $\endgroup$– Mitch BakerCommented Dec 14, 2017 at 0:20
Add a comment
|