# Expected Value of Bootstrap Standard Error Estimator

• May 21st 2013, 06:32 PM
Expected Value of Bootstrap Standard Error Estimator
Let $\displaystyle x_1 , x_2 , \ldot , x_n$ be a random sample iid to an unknown distribution $\displaystyle F$, and suppose that $\displaystyle \hat F$ is the empirical distribution of the $\displaystyle x's$.

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

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

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

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

Proof.

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

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

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

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

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

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

Is this right? Thank you.