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In this paper, there is a proof of Hoeffding identity for covariance (see page 541 or page 5 in pdf file, Theorem 1.11). A part of the proof is the following equality:

$$ 2\text{cov}(X,Y) = 2(E(XY)-E(X)E(Y))=E((X_1-X_2)(Y_1 - Y_2))$$

As far as I understand this transition is done by taking into account that$\ (X_1,Y_1)$ is iid copy of$\ (X_2,Y_2)$. It should be pretty simple, but I'm having hard time writing this explicitly. Can you please give me a hint?

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up vote 2 down vote accepted


Since $\mathbb{E}[X]=\mathbb{E}[X_1]=\mathbb{E}[X_2]$ (and the same for $Y$ and $XY$), we can write:

$$ \begin{align} 2\text{cov}(X,Y)&=2(\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y])\\ &= \mathbb{E}[X_1Y_1]-\mathbb{E}[X_2]\mathbb{E}[Y_1]+\mathbb{E}[X_2Y_2]-\mathbb{E}[X_1]\mathbb{E}[Y_2] \end{align} $$

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