Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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: $$ \mathrm{E}[f (X)(f (X) − f (X'))] = \sum_{i=1}^n \mathrm{E} [f (X) (f (X^{[i−1]} ) − f (X^{[i]} ))] $$

Thanks!

share|improve this question
1  
Concatenation + definition of $X^{[0]}$ + definition of $X^{[n]}$. –  Did Oct 18 '12 at 17:31
    
@did: Thanks, Didier! –  Tim Oct 18 '12 at 17:56

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.