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


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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

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