Let $\Omega$ be a Banach space; for the sake of this post, we will take $\Omega = {\mathbb R}^2$, but I am more interested in the infinite dimensional setting. Take $\mathcal F$ to be the Borel $\sigma$-algebra, and let $\mathbb P$ be a probability measure on $(\Omega, \mathcal F)$.

Denote by $\vec x = (x,y)$ a point in $\Omega$, and let $\mathcal F_1$ be the $\sigma$-algebra generated by the first coordinate. Fix $y_0 \in \mathbb R$ and $\eta > 0$, and consider $$f(\vec x) = \mathbb P( ~|y - y_0| \le \eta~ |\mathcal F_1).$$ (More generally, one can consider $f(\vec x) = \mathbb E( \varphi(\vec x) | \mathcal F_1)$ for some suitable $\varphi : {\mathbb R}^2 \to \mathbb R$.)

The function $f$ is measurable; that comes from the definition of conditional expectations. I would like to find some reasonable sufficient conditions such that $f$ is continuous and positive.

I feel like this should be relatively elementary material, but unfortunately I'm having trouble finding any references. How should I approach this?

I've included the [fa.functional-analysis] tag because in general I want to consider $\Omega$ to be a space of smooth functions. I'm guessing that to give some additional structure, I'll need to assume that $\mathbb P$ is absolutely continuous with respect to a Gaussian measure, because I don't know any other reasonable measures on function space.

  • $\begingroup$ I'll have a think about this once I've converted from probability to analysis (it's been too long since I thought in the former mode). Small quibble/question: shouldn't it really be $f(x)$, since you are "integrating out" over the second variable? Also, since $f$ is a conditional probability, it has to be positive as a function. (Or did you mean positive semi-definite?) $\endgroup$
    – Yemon Choi
    Jan 28, 2010 at 22:34
  • $\begingroup$ @Yemon, the conditional probability is defined as a measurable function on the probability space. Thus f really is a function of both variables. A good way to think about it is that when you condition on the first coordinate x, the distribution of x collapses onto a delta function. Thus the probability space remains R², but effectively there is only one dimension of randomness left. On your second point, you're right that as a conditional probability, the function f is automatically non-negative. I want conditions such that it's strictly positive on the whole space. $\endgroup$ Jan 28, 2010 at 23:36
  • 1
    $\begingroup$ I don't think I quite answered your question about integrating out the variable. Suppose Omega is an abstract probability space, and we consider a conditional probability P(A|F) for some sigma-algebra F and event A. Then this is still a random variable / measurable function on Omega: the probability space doesn't change. My example is the same, it's just that I've specified Omega explicitly. I think the function f(x,y) is constant in the second coordinate, for the intuitive reason you said, that we integrate out the variable y. But formally it still depends on it. $\endgroup$ Jan 28, 2010 at 23:46
  • 1
    $\begingroup$ When everything is reasonably nice, $\mathbb{E}(\psi|\mathcal{F})$ is the projection of $\psi$ onto $L^2(\Omega, \mathcal{F},\mathbb{P})$. $\endgroup$ Jan 29, 2010 at 0:30
  • $\begingroup$ (deleted previous version of this comment, as it was a bit snippy/waspish) @Tom: OK, my question was more about your notation (different from what I'm used to) than about the defn of conditional probability. FWIW, I know the definition. $\endgroup$
    – Yemon Choi
    Jan 29, 2010 at 2:46

5 Answers 5


Even if your probability measure is absolutely continuous with respect to Lebesgue measure on $\Omega={\mathbb R}^2$, I don't think this suffices for the function $f$ you have defined to be continuous (just take $X$ to be independent from $Y$, i.e. your probability measure is just the product of two probability measures, "one on each axis", and choose the one for $X$ to be something in $L^1({\mathbb R}, {\mathcal B}, dx)$ which is discontinuous.

On the other hand, if ${\mathbb P}$ is not just absolutely continuous with respect to Lebesgue measure on $\Omega$, but has a continuous density function wrt said measure, then your function $f$ will be continuous -- just because integrating over a ball of radius $\eta$ with centre $\eta$ can only smooth things out, so that continuity of the original density function goes over to continuity of your conditional probability. [This is a fairly straightforward observation using basic properties of usual integration in the plane.]

So in your example, the Gaussian structure isn't really relevant as far as I can see. Also, if the original density function is strictly positive on the cylinder $\{(x,y) : |y-y_0|<\eta \}$, then the conditional probability you've defined will also be strictly positive; this condition is evidently not necessary, but I suspect in the examples you're interested in something like it should hold.

In between these two extremes, I'm not sure what else one can say. Perhaps, from your point of view, it's more important to go up to infinite-dimensional $\Omega$ but place restrictions on the kind of probability measure which you wish to consider.

  • $\begingroup$ @Yemon: for your first example of independent coordinate variables, wouldn't the function f be constant? If I understood correctly Tom's definition of $f$, it is defined as $f(x') = P(|y-y_0| < \eta | x = x')$, where $(x,y)$ is a vector valued random variable, as opposed to the notation $f(\vec{x})$. Then if $x$ is independent of $y$, then clearly there is a version of conditional probability such that $P(|y-y_0| < \eta | x= x') = P(|y-y_0| < \eta)$. Isn't this a constant function? $\endgroup$
    – John Jiang
    Apr 8, 2010 at 21:20


Here is an idea, I really don't know if there is something to do with it, but this is the best I can think of (if I understood properly your question).

If you consider Homogenous Feller Markov processes (over some Banach Space if you like) then you can express the semi-group of transition with respect to the last known value of the process. And moreover the Feller property constrains the semi-group (or conditional probabilities) to be continuous in this last known value if I remember well. From there you might be able to find some conditions in the same spirit to Markov property and Feller conditions to be extended over an other Index that time.

So the trick is to "convert" (if possible in your case) the conditioning sigma field into a sigma field gereatad by one random variable, then express this conditionnal probability (or semigroup of transition) as a function of the value in this variable, and impose some continuity condition to get your property (this seems "ad hoc" though).

In any case this is only giving some sufficients conditions so the genrality could be questionable.

Hope this helps


  • $\begingroup$ Interesting suggestion, The Bridge. Thanks! $\endgroup$ Feb 3, 2010 at 18:30

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. He also discusses sufficient structure, and there the question of the existence of joint densities arises. 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.

  • $\begingroup$ Hi Daniel. I'm interested in taking a look at the preprint you mention, but cannot find it. I do have access to Tjur's two books; do you know if they contain the relevant material? $\endgroup$
    – R Hahn
    Oct 30, 2013 at 16:31
  • $\begingroup$ No, they are in a tech report. Feel free to contact me. $\endgroup$
    – Daniel Roy
    Oct 30, 2013 at 19:48
  • $\begingroup$ @DanielRoy, that's very exciting, and I'd love to learn more. Do you have a PDF copy? Please email me if so: tlagatta@gmail.com . I'm a big fan of your work, and am looking forward to chatting more with you. $\endgroup$ Nov 5, 2013 at 22:23
  • $\begingroup$ @TomLaGatta: Tjur's article has been made legally available online by the University of Copenhagen. $\endgroup$
    – Alex M.
    May 3, 2018 at 17:48

Since a troll bumped this question to the front page, I might as well answer it. The technology which provides the solution is called regular conditional probability or disintegration.


Here's an answer in the case that $X$ and $Y$ are Gaussians. There's such a rigid algebraic structure there that I don't know whether I'm exploiting that, or a more general property of the distribution. I think this answer also generalizes to the case of Gaussian measures on function space, but I'm not sure if it's a sufficiently general answer.

Let $X \sim N(0, \sigma_X)$ and $Z \sim N(0,1)$ be independent Gaussians. Let $$Y = \frac{\rho\sigma_Y}{\sigma_X} X + \sqrt{1-\rho^2} \sigma_Y Z,$$ so that $Y \sim N(0,\sigma_Y)$. Clearly, $$\mathbb E(Y|X) = \frac{\rho\sigma_Y}{\sigma_X} X \qquad \mathrm{and} \qquad \mathbb E(Y^2|X) - \frac{\rho^2\sigma_Y^2}{\sigma_X^2} X^2 = (1-\rho^2)\sigma_Y^2.$$

Conditioned on $X$, $Y$ again has a Gaussian distribution---this is a special property of Gaussians. Thus $$f(x,y) = \mathbb P( ~|Y - y_0| \le \eta ~| X) = \frac{1}{\sqrt{2\pi(1-\rho^2)} \sigma_Y} \int_{y_0 - \eta}^{y_0 + \eta} \exp \left(-|u-\tfrac{\rho \sigma_Y}{\sigma_X} x|^2 / 2(1-\rho^2)\sigma_Y^2 \right) ~du.$$This does not depend on $y$, as I explained in my comment to Yemon above. Moreover, this is clearly a continuous function on ${\mathbb R}^2$, and $f(x,y) > 0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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