7
$\begingroup$

Let $(\Omega_i, \mathcal{F}_i), i=1,2$ be measurable spaces. Their product measurable space is $(\Omega, \mathcal{F})$.

Let $\mu_1$ be a probability measure on $(\Omega_1, \mathcal{F}_1)$, and let $(\mu_{ω_1})_{ω_1∈\Omega_1}$ be a transition probability from $\Omega_1$ to $\Omega_2$. Then there exists a probability measure $\mu$ , defined by $$\mu(A)= \int_{\Omega_1} \mu_{ω_1}(A_{ω_1})\mu_1(dω_1), \quad \forall A \in \mathcal{F}$$ where $A_{ω_1}:= \{\omega_2 \in \Omega_2: (\omega_1, \omega_2) \in A\}$.

My questions are:

  1. Conversely, given any probability measure $\mu$ on $(\Omega, \mathcal{F})$, do there exist a probability measure $\mu_1$ on $(\Omega_1, \mathcal{F}_1)$, and a transition probability $(\mu_{ω_1})_{ω_1∈\Omega_1}$ from $\Omega_1$ to $\Omega_2$, such that $$\mu(A)= \int_{\Omega_1} \mu_{ω_1}(A_{ω_1})\mu_1(dω_1), \quad \forall A \in \mathcal{F} ?$$ Can they be explicitly or implicitly determined?

  2. Are such probability measure $\mu_1$, and transition probability $(\mu_{ω_1})_{ω_1∈\Omega_1}$ unique?

  3. What if considering general measures instead of probability measures? Are the answers yes only up to scaling of measures?

Thanks and regards!

$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

  1. The answer is in general no. The first such an example was constructed by Dieudonne and can be found in many advanced books on probability. Existence is guaranteed under various assumptions that are mostly topological or emulate topological regularity conditions. The strongest results obtained this way are found in a paper by Pachl. The simplest case where existence is guaranteed is when one deal with (the product of) the real line. An accessible proof for that case can be found, for example in the book by Lehmann and Romano (I like their exposition, but the result can be found in many books). Computation of these probabilites is in general not possible. Rao has written a book that carefully looks at the challenges of calculating these conditional probabilities- and pretty much everything else about the topic. The book makes for hard reading.

  2. They are unique up to a measure zero subset of $\Omega_1$ for countable generated probability or measure spaces $\Omega$ (10.4.3. in Bogachev). Otherwise, not necessarily (10.10.44 in Bogachev).

  3. One can certainly do this for finite measures and to some degree also for infinite measures. The simple proof in Lehmann and Romano alluded to above does not work for general spaces, however (at least not without major adaptions).

Generally, the whole area, known as regular conditional probabilities and, relatedly, disintegrations, is fairly technical and advanced. A good guide is given by Chapter 10 in Bogachev.

$\endgroup$
6
  • $\begingroup$ +1 Thanks! So in 1, for a given joint probability measure, are you saying that my question about existence of the transition probability is equivalent to the question of whether the conditional probability is regular? $\endgroup$
    – Tim
    Jan 22, 2012 at 0:59
  • $\begingroup$ Mostly. But it is a special case, where the conditioning field comes from the product structure. But this is not enough for existence. A high-level answer for why is given here by Blackwell: If regular conditional probabilities for product spaces would exist, wee could use the Ionescu-Tulcea theorem to show that a probability measure on an infinite product exists extending a given family of finite dimensional distributions- but there is a counterexample by Andersen and Jessen. $\endgroup$
    – Michael Greinecker
    Jan 22, 2012 at 7:07
  • $\begingroup$ Thanks! In disintegration theorem $\int_{Y}f(y)\,\mathrm{d}\mu (y)=\int_{X}\int_{\pi^{-1}(x)}f(y)\,\mathrm{d}\mu_{x}(y)\mathrm{d}\nu(x).$ I was wondering if it is equivalent to $ \int_{Y}f(y)\,\mathrm{d}\mu(y)= \int_{X} \int_X f(y) \, \mathrm{d} \mu_{x} (y) \mathrm{d} \nu (x)$? The difference lies in the integral region of the inner integral. I think yes, because " $μ_x$ 'lives on' the fiber $π^{-1}(x)$: for $ν$-almost all $x ∈ X$,$\mu_{x}\left( Y \setminus \pi^{-1}(x)\right)=0$, and so $μ_x(E)=μ_x(E∩π^{-1}(x))$"? $\endgroup$
    – Tim
    Jan 24, 2012 at 3:53
  • $\begingroup$ In the version given in wikipedia, yes. The two concepts are equivalent when one conditions on a countably generated $\sigma$-algbera. But there exists more general versions of disintegrations that do not give you a regular conditional probability. In the wikipedia article, it essentially tells you that the function given $K(x,B)=\mu_x(B)$ is a probability kernel. You can always change the function on a null set so as to actually obtain oe, so that $\mu_(x)$ is supported on $\pi^{-1}(x)$ for al $x$. $\endgroup$
    – Michael Greinecker
    Jan 24, 2012 at 7:23
  • $\begingroup$ Interesting question. I don't know much about measure theory, but is it correct to say that the question is equivalent to the following? given a "starting" dsitribution and an "ending" distribution and a transition matrix, can we derive a continuous generator matrix that will describe this transition at all points in time between the start and finish? The answer reminds me of the fact that the solution to my problem only exists and is unique when we are dealing with a homogenous markov chain, which strikes me as sounding similar to the topological conditions you mentioned. Am I way off here? $\endgroup$
    – Paul
    Feb 3, 2017 at 20:39

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.