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.