MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is a sequel to another question I have asked.

The notion of disintegration is a refinement of conditional probability to spaces which have more structure than abstract probability spaces; sometimes this is called regular conditional probability. Let $Y$ and $X$ be two nice metric spaces, let $\mathbb P$ be a probability measure on $Y$, and let $\pi : Y \to X$ be a measurable function. Let $\mathbb P_X(B) = \mathbb P(\pi^{-1} B)$ denote the push-forward measure of $\mathbb P$ on $X$. The disintegration theorem says that for $\mathbb P_X$-almost every $x \in X$, there exists a nice measure $\mathbb P^x$ on $Y$ such that $\mathbb P$ "disintegrates":$$\int_Y f(y) ~d\mathbb P(y) = \int_X \int_{\pi^{-1}(x)} f(y) ~d\mathbb P^x(y) d\mathbb P_X(x)$$ for every measurable $f$ on $Y$.

This is a beautiful theorem, but it's not strong enough for my needs. Fix a Borel set $B \subseteq X$, and let $p(x) = \mathbb P^x(B)$. Part of the theorem is that $p$ is a measurable function of $x$. Suppose that the map $\pi : Y \to X$ is continuous instead of simply measurable. My question: What is a general sufficient condition for $p(x)$ to be continuous?

To me, this is an obvious question to ask, since if $x$ and $x'$ are two close realizations of a random $x \in X$, then the measures $\mathbb P^x$ and $\mathbb P^{x'}$ should be close too, at least in many natural situations. However, in my combing through the literature, I haven't been able to find an answer to this question. My guess is that most people are content to integrate over $x$ when they use the theorem. For my purposes, I need some estimates which I get by continuity.

At this point, I've managed to prove and write down a pretty good sufficient condition for the case I care about (Banach spaces), using an abstract Wiener space-type construction. However, I am hoping that an expert can point me toward a good reference that does this in wider generality.

share|cite|improve this question
B must be a subset of Y? – Andrey Gogolev Feb 20 '10 at 1:09
Also, sufficient condition on what, on B, on pi, both? – Andrey Gogolev Feb 20 '10 at 1:24
Andrey: B above is a Borel set in Y. The function pi need only be measurable for the general disintegration theorem to apply. However, in my case, pi is continuous function. – Tom LaGatta Feb 20 '10 at 15:21

Tue Tjur studied the existence of continuous disintegrations in a 1975 preprint "A Constructive Definition of Conditional Distributions," Issue 13, Copenhagen Universitet. He gives necessary and sufficient conditions for their existence, at least in the setting of Radon measures.

He also discusses sufficient structure, and there considers a basic probability space that is an open subset of a finite-dimensional Euclidean space and the problem of conditioning on a random variable taking values in an open subsets of a Euclidean space $\mathbb R^k$ such that, when the random variable is considered as a (measurable) function, it is surjective and continuously differentiable with differential of maximal rank. This particular special case may be too narrow for you, but perhaps the general case can give you some guidance.

The article is a bit hard to track down, so let me know if you need help finding it. The existence of continuous disintegrations arises also in the study of the computability of conditional probability, which is my interest.

share|cite|improve this answer

Probably is not general as you want, but if you don't think before about that can be a begining...

Proposition: If $\pi:Y\to X$ is bijective function such that $\pi^{-1}$ is continuous then $\mathbb{P}^{x_n}\to\mathbb{P}^{x}$ (weak topology) whenever $x_n\to x$.

Proof: it follows from the Disintegration Theorem that for all $B\in\mathcal{B}(Y)$ we have

$$ \begin{array}{rcl} \mathbb{P}(B)&=&\displaystyle\int_X\int_{\pi^{-1}(x)}\chi_B(y)\ d\mathbb{P}^x(y)\ d\mathbb{P}_X(x) &=&\displaystyle\int_X \mathbb{P}^x(B\cap\pi^{-1}(x)) \ d\mathbb{P}_X(x) \end{array} $$ For the other hand we have that $$ \mathbb{P}(B)=\displaystyle\int_X \chi_B(\pi^{-1}(x))\ d\mathbb{P}_X(x) =\displaystyle\int_X \chi_B(\pi^{-1}(x))\delta_{\pi^{-1}(x)}(\pi^{-1}(x)) \ d\mathbb{P}_X(x) $$ so $$ \displaystyle\int_X \mathbb{P}^x(B\cap\pi^{-1}(x)) \ d\mathbb{P}_X(x)=\mathbb{P}(B)=\displaystyle\int_X \delta_{\pi^{-1}(x)}(B\cap\pi^{-1}(x)) \ d\mathbb{P}_X(x) $$

Since $\mathbb{P}^x$ is a probability measure and $B\cap\pi^{-1}(x)$ is a singleton or empty set, we have $$ \delta_{\pi^{-1}(x)}(B\cap\pi^{-1}(x))\geq \mathbb{P}^x(B\cap\pi^{-1}(x)) $$ and from previous integral equality almost surely we have $$ \delta_{\pi^{-1}(x)}(B\cap\pi^{-1}(x))=\mathbb{P}^x(B\cap\pi^{-1}(x)). $$ Fix $x$ and take $B=\pi^{-1}(x)$, then from the above equality, follows that $\mathbb{P}^x=\delta_{\pi^{-1}(x)}$.

If $x_n\to x$ then $p(x_n)=\mathbb{P}^{x_n}$ converge to $p(x)$ in the weak topology. In fact, by the continuity of $\pi^{-1}$ we get that $\int f\ p(x_n) \to\int f\ p(x)$ for all bounded uniformly continuous functions $f$.

share|cite|improve this answer

I think you need some hypothesis on the measure to be pushed, at least in the very common case where $\pi$ is the projection to a factor in a product.

Take any family $\mathbb{P}^x$ of measures in a space $X'$, where $x$ runs over $X$, and let $Y=X'\times X$, $\pi$ be the projection on $X$, and $\mathbb{P}=\int_X \mathbb{P}^x dx$ where $dx$ is any measure on $X$. Then $\pi$ is very regular (smooth if $X'$ and $X$ are smooth manifolds for example) but yet, any kind of lack of regularity can appear in $\mathbb{P}^x$ (which are by construction the disintegration measures, since they are unique up to a negligible set).

I guess that in this setting, assuming $\mathbb{P}$ to be absolutly continuous with continuous density would be sufficient.

Edit: My guess seems wrong, as is shown by the restriction of Lebesgue measure to a L shaped polygon. You will probably need strong restrictions on $\mathbb{P}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.