4
$\begingroup$

I'm trying to understand the problem arising when using Kolmogorov's extension theorem to prove the existence of the Dirichlet process on an arbitrary measurable space $\left(\mathcal{X},\mathcal{A}\right)$.

Recalling the definition from Ferguson: A $\left[0,1\right]$-valued process $\left(\mu_{A}\right)_{A\in\mathcal{A}}$ is a Dirichlet process with parameter $\alpha\in\mathcal{M}_{1}\left(\mathcal{X}\right)$ if for any finite measurable partition $A_{1},\ldots,A_{k}$ of $\mathcal{X}$:

$$ \left(\mu_{A_{1}},\ldots,\mu_{A_{k}}\right)\sim\mbox{Dir}\left(\alpha H\left(A_{1}\right),\ldots,\alpha H\left(A_{k}\right)\right) $$

where $\mbox{Dir}$ denotes the Dirichlet distribution.

Since this family of marginals is consistent (follows from the “aggregation property” of the Dirichlet), Kolmogorov's extension theorem tells me that there's a probability space $\left(\Omega,\mathcal{F},\mathbb{P}\right)$ supporting this process and that a can choose $\Omega=\left[0,1\right]^{\mathcal{A}}$, $\mathcal{F}=\mathcal{B}\left(\left[0,1\right]\right)^{\otimes\mathcal{A}}$ and $\mu_{A}\left(\omega\right)=\omega_{A}$.

My understanding is that Kolmogorov's does not give us any properties of $\mu$ besides its existence, and that we need to show that it almost surely is a probability measure. The tricky point seems to be the $\sigma$-additivity: For any countable collection of measurable sets $\mathcal{D}\subset\mathcal{A}$ we need

$$\mu\left(\bigcup_{A\in\mathcal{A}}A\right)=\sum_{A\in\mathcal{A}}\mu\left(A\right)\mbox{a.s.} $$

and if the $\sigma$-algebra $\mathcal{A}$ does not have a countable generator, then there are uncountably many countable collections $\mathcal{D}$.

Now my questions are:

  1. (Probably trivial:) Why does this equality even hold for a single countable collection? Doesn't this require me to know countable marginals where I only know finite ones?

  2. What is exactly the problem with uncountability? Is it that null sets sum up, so that potentially $\mathbb{P}\left[\mu\mbox{ is }\sigma\mbox{-additive}\right]<1$ even if the above equation holds for every collection of measurable sets? Or is the event $\left\{ \mu\mbox{ is }\sigma\mbox{-additive}\right\}$ not even measurable w.r.t the product algebra $\mathcal{F}$?

  3. What gets easier by assuming that $\mathcal{X}$ is polish? Does every Borel-$\sigma$-algebra on a polish space have a countable generator?

My references are:

  • Ferguson (1973), "A Bayesian analysis of some nonparametric problems"
  • Sethuraman (1991), "A constructive definition of Dirichlet priors"
$\endgroup$
1
  • $\begingroup$ I asked this question on math.se before but didn't get any answer, so I deleted the questions there. $\endgroup$ Commented Sep 26, 2014 at 9:45

1 Answer 1

6
$\begingroup$

Constructing the Dirichlet process using Kolmogorov raises several problems, and your question shows that you are already on to some of them. In short, an arbitrary measurable space is not sufficient, you need more topological structure.

I think it is useful to distinguish two cases:

i) Construction of the Dirichlet process, which is a particularly simple random probability measure (since it is discrete, homogeneous, and its weights can be obtained using stick-breaking). Because of its special structure, it can be constructed without resorting to Kolmogorov, and hence can be generalized to more general measurable spaces (provided you can really identify a setting where a Polish space is too restrictive).

ii) Construction of arbitrary random probability measures, for which you do need a projective limit technique in general (though not precisely Kolmogorov). The latter is also true if you want to construct the Dirichlet process from its marginals.

Let me try to give a short answer for each case:

i) In the particular case of the DP, there is arguably a generalization to arbitrary measurable spaces via the stick-breaking construction: If you define the DP on a Polish space, you can proceed to show that it is a discrete, homogeneous random measure almost surely, so a draw $\mu$ is necessarily of the form $$ \mu = \sum_{k\in\mathbb{N}}C_k\delta_{\Theta_k}\qquad\text{ (equality in distribution),} $$ where the variables $\Theta_k$ are iid and independent of the weights, and the weights $C_k$ can be sampled from the stick-breaking construction. That works on an arbitrary measurable space---with measurable singletons, say---and $\mu$ is always countably additive and normalized almost surely, and has Dirichlet marginals on any finite measurable partition. Whether or not this should be called a Dirichlet process if the space is not Polish depends of course on how you define what a Dirichlet process is in the first place.

ii) Regarding projective limit (i.e. "Kolmogorov") constructions of random probability measures, this article should answer your questions (the appendix summarizes the issues with using Kolmogorov on arbitrary measurable spaces):

http://projecteuclid.org/euclid.ejs/1319028571

To come back to your specific questions:

1) That you only need the finite-dimensional marginals is precisely the non-trivial statement of Kolmogorov's theorem. Intuitively (and somewhat imprecisely), you can think of this as a regularity result: A continuous function, for example, is completely determined by its values on a suitably large (namely dense) subset. Roughly speaking, Kolmogorov's theorem says that a probability measure is a set function that is sufficiently regular to be determined by its values on a sufficiently large subsystem of measurable sets, and that the "cylinder sets" determined by the finite-dimensional marginals constitute such a sufficiently large subsystem.

2) The event $\lbrace \mu \text{ is }\sigma\text{-additive }\rbrace$ is indeed not measurable in the product $\sigma$-algebra; that is one reason why Kolmogorov is not quite the right tool here, since it inherently uses a product topology. Even with more suitable tools, there are still an uncountable number of countable collections, so null sets can add up. (For more details, see the paper referenced above.)

3) Yes, any Polish topology has a countable basis: Since the space is separable, it has a countable dense subset; since it is metric, you can choose balls of rational radii around the points in this set and obtain a countable system. You can then show that these generate the topology, and hence the Borel sets. The system of balls can also be used to construct an index set for the stochastic process construction, consisting of measurable finite partitions (see Section 3.1 in the paper above).

$\endgroup$
2
  • $\begingroup$ Thanks for the exhaustive answer! Considering myself a reader of "passing familiarity with measure-theoretic probability", the appendix of your paper is very helpful. If wish it became an academic tradition put some additional background information into appendixes to make papers more accessible to non-specialists. $\endgroup$ Commented Sep 27, 2014 at 16:51
  • $\begingroup$ One small correction: Kolmogorov's theorem for uncountable index sets doesn't use the the Borel $\sigma$-algebra from the product topology, but the Baire $\sigma$-algebra from the product topology, i.e. the coarsest $\sigma$-algebra that makes all real continuous functions measurable. For example, with the uncountable product of nontrivial Hausdorff spaces, singleton sets are closed in the product topology and therefore Borel measurable, but they are not in the $\sigma$-algebra generated by projections, since any set in that $\sigma$-algebra must be generated by countably many projections. $\endgroup$ Commented Sep 22, 2016 at 8:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .