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I have Borel measures $\mu$, $\nu$ on $\mathbb R^2$ with densities $g, h$ respectively with respect to the Lebesgue measure. Now I assume that $\mu(A \times \mathbb R) = \nu(A \times \mathbb R)$ for every Borel set $A$. Now I want to find a Borel measure $\gamma$ on $\mathbb R^3$ such that:

$$\mu(A \times B) = \gamma(A \times B \times \mathbb R)$$


$$\nu(A \times B) = \gamma(A \times \mathbb R \times B)$$

Fine. I try to write the measure $\gamma$ as

$$\gamma(Z) = \int 1_Z(x,y,z) \, d \nu(x,y) \, dz$$ and the same with $\mu$ instead of $\nu$, then I use the density wrt Lebesgue measure but this quickly becomes really messy. Is there an elegant way to do this?

Further, what would be the point of this exercise? Why would someone want this?

This is homework, so only hints.

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After you solve the problem, you may be interested in looking at Swart's paper "A conditional product measure theorem" from Statistics and Probability Letters. He gives assumptions under which such a measure exists (weaker than in your problem), but perhaps more interestingly, a pathological example where such a measure does not exist. –  Noah Stein Mar 11 '11 at 23:58

2 Answers 2

up vote 1 down vote accepted

You are trying to find a random $(X,Y,Z)$ whose marginal distributions $(X,Y)$ and $(X,Z)$ are given. The simplest possible solution is to let $Y$ and $Z$ be conditionally independent given $X$. The existence of densities will help you to make this explicit.

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Wow. Les grands esprits se rencontrent, as they say... –  Did Mar 11 '11 at 23:18
@Didier Ha! It's funny that we both find it easier to think in terms of random vectors than in terms of their image measures. –  Byron Schmuland Mar 11 '11 at 23:25
@Didier, @Byron: Thanks. Seeing this in a probability setting makes it much clearer. Thanks for the hints! –  Jonas Teuwen Mar 11 '11 at 23:34
More Breimanian than Doobian, are we? Joe Doob: "Probability is simply a branch of measure theory, with its own special emphasis and field of application." Leo Breiman: "Probability theory has a right and a left hand. On the right is the rigorous foundational work using the tools of measure theory. The left hand thinks probabilistically, reduces problems to gambling situations, coin-tossing, motions of a physical particle." –  Did Mar 12 '11 at 23:31

In the parlance of probability, when $\mu$ and $\nu$ have mass $1$, you look for the distribution $\gamma$ of a triple $(X,Y,Z)$ of random variables such that $\mu$ is the distribution of $(X,Y)$ and $\nu$ is the distribution of $(X,Z)$. Now we understand the condition that the first marginals of $\mu$ and $\nu$ coincide. And, since we have some latitude on the coupling of $(X,Y,Z)$, we could choose $\gamma$ such that, for example, $Y$ and $Z$ are independent conditionally on $X$.

Since you asked for a hint and not a solution, I stop here.

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