# Coupling of two random variables

Let $X_1,X_2$ be two random variables on $\Bbb R$ with distributions $P_1$ and $P_2$, and let $\Bbb P$ be their coupling, $$\Bbb P\circ X_1^{-1} = P_1,\quad \Bbb P\circ X_2^{-1} = P_2.$$ Is that true that for the diagonal $\{X_1 = X_2\}$ we have $$\Bbb P(X_1 = X_2) = \int_\Bbb R \Bbb P(X_1 = X_2,X_2 = y)P_2(\mathrm dy) \\\tag{1} = \int_\Bbb R \Bbb P(X_1 = y)P_2(\mathrm dy)?$$ I guess, no - otherwise we would have $$\Bbb P(X_1 = X_2) = \int_\Bbb R P_1(y)P_2(\mathrm dy)$$ which does not depend on $\Bbb P$ and furthermore, we get $0$ in case $X_1$ is a continuous random variable.

Q. Is the formula $(1)$ true, and if not is there exists a similar formula?

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Let $\mu$ be their joint distribution on $\mathbb{R}^2$. By a standard result for the existence of regular conditional probabilities, there exists a probability kernel $\kappa:\mathbb{R}\times\mathcal{B}\to[0,1]$ (the conditional distribution of $X_2$ given the value of $X_1$)such that for every Borel set $B\subseteq\mathbb{R}^2$, we have $$\mu(B)=\int_\mathbb{R} \int_{\mathbb{R}} 1_B ~d\kappa(x,\cdot) dP_1.$$ Let $D$ be the diagonal. Then we have $$\mu(D)=\int_\mathbb{R}\kappa\big(x,\{x\}\big)~dP_1.$$ The difference to your formula is that you use the distribution instead of the conditional distribution and that is only justified when the two random variables are independent.
Oh, indeed - I just lost the conditioning. Could you refer me to any source of the existence of such kernel on spaces more general than $\Bbb R$? Also, in case $\Bbb P$ is the $\gamma$-coupling does $\kappa$ has a known form? –  Ilya Jan 15 '13 at 14:22