Suppose $X$ and $Y$ are two real-valued random variables with a specified joint probability distribution $P_{X,Y}.$ I wish to determine if there is a $\sigma$-finite measure $\mu$ on the real line such that $P_{Y|X=x} << \mu$ for $P_X$-almost all $x\in\mathbb{R}$. Call this property $Q.$ Property $Q$ does not always hold e.g. if $X$ is a standard normal variable and $Y=X.$

Is there an equivalent description of property $Q$ that is easy to check for a given probability distribution $P_{X,Y}$?

  • $\begingroup$ One sufficient condition (even though trivial, perhaps) is that if $P \in Q$ and $P' \ll P$ then $P' \in Q$ $\endgroup$
    – Ilya
    Apr 21, 2016 at 15:53
  • $\begingroup$ It seems to me that property $Q$ is equivalent to this: $P_{X,Y}$ is absolutely continuous with respect to the product $P_X\otimes P_Y$ of its marginals. $\endgroup$ Apr 21, 2016 at 21:17
  • $\begingroup$ That is intriguing John. Any suggestions for how I may prove this? $\endgroup$
    – Hedonist
    Apr 21, 2016 at 21:19

1 Answer 1


In one direction it's clear: If $P_{X,Y}\ll P_X\otimes P_Y$, then there is a jointly measurable density $f$, and (a version of) the conditional distribution $P_{Y|X=x}$ is given by $f(x,y)P_Y(dy)$, so the choice $\mu=P_Y$ suffices.

Conversely, suppose that $Q$ holds. Then there is a jointly measurable function $g$ such that $P_{Y|X=x}(dy) = g(x,y)\mu(dy)$, for $P_X$-a.e. $x$. Define $h(y)=\int_R g(x,y)\,P_X(dx)$, and observe that if $h(y)=0$ then $g(x,y)=0$ for $P_X$-a.e. $x$. Define $B=\{y: h(y)=0\}$, and $f(x,y) = 1_{B^c}(y)g(x,y)/h(y)$. I claim that $f\cdot P_X\otimes P_Y=P_{X,Y}$. Indeed, $$ \eqalign{ P_{X,Y}(C) &=\int\int 1_C(x,y)P_X(dx)P_{Y|X=x}(dy)\cr &=\int\int 1_C(x,y)P_X(dx)g(x,y)\mu(dy)\cr &=\int\int 1_C(x,y)1_{B^c}(y)P_X(dx)g(x,y)\mu(dy).\cr } $$ In particular, taking $C=R\times A$, for $A$ a Borel subset of $R$, $$ \eqalign{ P_Y(A)=P_{X,Y}(R\times A) &= \int\int 1_A(y)1_{B^c}(y)P_X(dx)g(x,y)\mu(dy)\cr &= \int 1_A(y)1_{B^c}(y)h(y)\mu(dy)\cr &= \int 1_A(y)h(y)\mu(dy).\cr } $$ That is, $P_Y=h\cdot\mu$. Using this to continue the computation in the last-but-one display (multiply and divide by $h(y)$) we have $$ \eqalign{ P_{X,Y}(C) &=\int\int 1_C(x,y)1_{B^c}(y)P_X(dx)g(x,y)\mu(dy).\cr &=\int\int 1_C(x,y)P_X(dx)f(x,y)P_Y(dy),\cr } $$ and the asserted absolute continuity.


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