# Expected Value of Bootstrap Standard Error Estimator

• May 21st 2013, 07:32 PM
Expected Value of Bootstrap Standard Error Estimator
Let $x_1 , x_2 , \ldot , x_n$ be a random sample iid to an unknown distribution $F$, and suppose that $\hat F$ is the empirical distribution of the $x's$.

Define $\hat { \theta }^*(b)$ to be the Bootstrap estimator of a statistic $\theta$ and define $\hat { \theta }^*( \cdot ) = \sum _{b=1}^B \frac { \hat { \theta }^*(b) }{B}$

Define the standard error $\hat {se}_B= \sqrt { \frac { \sum ^B_{b=1}[ \hat { \theta }^*(b) - \hat { \theta }^*( \cdot )]^2 }{B-1}}$

and let $\hat {se}_ \infty = se _{ \hat F } = \sum ^m_{j=1} \omega _j [ \hat \theta ^*(b) - \hat \theta ^* ( \cdot )]^2$ where weights $\sum ^m_{j=1} \omega _j =1$

Show that $E_ { \hat F } ( \hat {se}_B^2)= \hat {se}^2_ \infty$

Proof.

Now, $E_ { \hat F } ( \hat {se}_B^2)= E_ { \hat F } \{ \frac { \sum ^B_{b=1}[ \hat { \theta }^*(b) - \hat { \theta }^*( \cdot )]^2 }{B-1} } \}$

$= \frac {1}{B-1} \sum ^B_{b=1} E_ { \hat F } [ \hat { \theta }^*(b) - \hat { \theta }^*( \cdot )]^2$

since $E_ { \hat F } [ \hat { \theta }^*(i) - \hat { \theta }^*( \cdot )]^2 = E_ { \hat F } [ \hat { \theta }^*(j) - \hat { \theta }^*( \cdot )]^2 \ \ \ \ \ \forall i \neq j$ because they are iid, we then have:

$= E_ { \hat F } [ \hat { \theta }^*(b) - \hat { \theta }^*( \cdot )]^2$

$= \sum ^m_{j=1} \omega _j [ \hat \theta ^*(b) - \hat \theta ^* ( \cdot )]^2$

$= \hat {se}_ \infty ^2$

Is this right? Thank you.