0
$\begingroup$

From a note, in the proof for Theorem 1 Efron Stein-Inequality:

Suppose that $X_1 , \dots, X_n , X_1' , \dots, X_n'$ are independent with $X_i$ and $X_i'$ have the same distribution for all $i$.

Let $X = (X_1, \dots, X_n)$, $X' = (X_1', \dots, X_n')$, $X^{(i)} = (X_1, \dots, X_{i-1}, X_i', X_{i+1}, \dots, X_n)$, and $X^{[i]} = (X_1', \dots, X_i', X_{i+1}, \dots, X_n)$.

$f: \mathbb{R}^n \to \mathbb{R}$ is some measurable function.

Why is this true: $ \mathrm{E}[f (X)(f (X) āˆ’ f (X'))] = \sum_{i=1}^n \mathrm{E} [f (X) (f (X^{[iāˆ’1]} ) āˆ’ f (X^{[i]} ))] $

Thanks!

  • 0
    @did: Thanks, Didier! – 2012-10-18

0 Answers 0