Understanding measures on the space of measures (via examples)

Let $X$ be a Polish space. If it allows for an interesting answer, you may assume $X$ is compact or even $X=[0,1]$. The space $\mathcal{P}(X)$ of Borel probability measures on $X$ is also Polish (via the Prokhorov metric). Measures on $\mathcal{P}(X)$ (i.e. elements of $\mathcal{P}(\mathcal{P}(X))$) arise, for example in the ergodic decomposition. I'm looking to understand $\mathcal{P}(\mathcal{P}(X))$ better, especially through examples.

Q1. Is there a natural example of an element $\mathcal{P}(\mathcal{P}(X))$? Q.1.5 What about when $X=[0,1]$?

Q2. What results in mathematics use or refer to an element of $\mathcal{P}(\mathcal{P}(X))$? I'm aware of the ergodic decomposition and its special case, de Finetti's theorem.

Q3. Where is $\mathcal{P}(\mathcal{P}(X))$ studied? I'm aware of Billingsley's Convergence of Probability Measures and Parthasarathy's Probability Measures on Metric Spaces.

EDIT Regarding Q2, I'm most interested in classical results. The answers given by @NateEldredge and @MichaelGreinecker, while interesting and helpful, seem to regard more modern (i.e. not classical) results. I realize that 'classical' is vague, and I'll try to make what I'm after more precise if necessary. The ergodic decomposition is something I consider 'classical'.

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One example arises in the study of superprocesses. I know only a little bit about these, so what follows may be a bit sketchy and possibly wrong. Allison Etheridge's book An Introduction to Superprocesses looks like a promising place to read more.

For this example, let's take $X$ to be some nice compact manifold, say for instance a sphere $S^d$ or torus $(S^1)^d$, on which we know how to run Brownian motion. ($X = \mathbb{R}^d$ might be more natural but its non-compactness complicates things slightly.) Also, to start, I actually would rather consider positive finite measures which need not have total mass 1. So let $\mathcal{M}(X)$ denote the space of all such measures on $X$. Under the weak topology, this is still a nice space; it is Polish (and if I am not mistaken, even locally compact), and so its Borel $\sigma$-field gives it a nice measurable structure as well.

A typical example is branching Brownian motion. Imagine that you start with $n$ particles in $X$, each moving independently according to Brownian motion. However, after an exponentially distributed amount of time, a particle splits into a random number of new "offspring" particles. Each new particle starts at the time and place where the split occurred, and evolves according to its own (conditionally) independent branching Brownian motion (and may itself split). (One could allow the number of new particles to be zero, in which case the particle can be seen as having died.)

If we want to consider branching Brownian motion as a stochastic process $Y_t$, we need to decide in what state space it should take its values: it has to be able to represent the location of all the particles alive at a given moment, however many that happens to be. Our first guess might be to represent the state of the process as a (finite) subset of $X$, so that our state space becomes the set of all finite subsets of $X$; let me call it $\mathcal{F}(X)$. However, this is a bit awkward:

• First, we would have to decide how to define an appropriate topology and measurable structure on $\mathcal{F}(X)$, which inherits enough of the topology of $X$ that we can keep track of the fact that each particle moves continuously.

• Second, it doesn't account for the possibility of two particles at the same location, as happens when a particle has just split, or if two particles have collided (though in this model they just pass through each other without interacting). So maybe we need multisets or something.

• Third and most seriously, $\mathcal{F}(X)$ is not a vector space. We cannot make sense of the expectation of $Y_t$. Also, if we want to study the behavior as the initial number of particles goes to infinity, we will want to be able to rescale $Y_t$ somehow.

The solution to these problems is as follows: instead of representing the state of the process by a subset of $X$, we represent it by a measure, where the location of the particles are marked by unit point masses. So the state where there are particles at locations $\{x_1, \dots, x_m\}$ corresponds to the measure $\sum_{i=1}^m \delta_{x_i}$. We can thus think of $Y_t$ as a stochastic process taking its values in $\mathcal{M}(X)$. This solves all the above problems:

• $\mathcal{M}(X)$ has a nice natural topology and measurable structure, as discussed above. We can also show, for example, that the process $Y_t$ is càdlàg; the jumps correspond to times when a particle splits (so if a single particle at location $x$ splits into two offspring, the process jumps from $\delta_x$ to $2 \delta_x$.

• This model naturally counts locations with multiplicity, by just putting more mass at a point if it's occupied by several particles.

• $\mathcal{M}(X)$ is (almost) a vector space (okay, it's a cone in the vector space of signed measures); we can scale and take positive linear combinations, and that's enough to handle expectations and scaling limits.

So if $Y_t$ is a process valued in $\mathcal{M}(X)$, then the law of $Y_t$ at any fixed time $t$ is a probability measure on $\mathcal{M}(X)$, i.e. an element of $\mathcal{P}(\mathcal{M}(X))$. (The law of the entire process is a probability measure on the Skorohod space $\mathcal{D}([0, \infty), \mathcal{M}(X))$ of càdlàg paths in $\mathcal{M}(X)$, which is rather more complicated, but still a Polish space.) We get other nice properties of $Y_t$ as well; for instance, thanks to the exponential reproduction times and independence of the offspring, $Y_t$ is Markov.

There are other models where the number of particles remains constant; in this case, we can renormalize our measures to have total mass 1, and get a process valued in $\mathcal{P}(X)$, whose one-dimensional distributions are elements of $\mathcal{P}(\mathcal{P}(X))$. Or, we could say each particle starts with a certain amount of mass, and when it splits, that mass is divided among its offspring so that the total amount of mass in the system is conserved. Then we can represent the state with particles at $x_1, \dots, x_m$ with respective masses $c_1, \dots, c_m$ by the measure $\sum c_i \delta_{x_i}$; if we take the total mass of the system to be 1 we again have a $\mathcal{P}(X)$ valued process. (Here, to preserve the Markov property I guess we have to assume that if two particles collide they coalesce.)

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Sometimes, you want to randomly pick a point by a randomly chosen measure. The $\sigma$-algebra on $\mathcal{P}(X)$ is generated by sets of the form $$\big\{\mu\in\mathcal{P}(X):\mu(A)\leq r\big\},$$ with $A\subseteq X$ measurable and $r\in[0,1]$. Now let $\pi:X\to\mathcal{P}(Y)$ be a measurable function. Let $\mathcal{Y}$ be the $\sigma$-algebra on $Y$. Then we can define a function $$\kappa:X\times\mathcal{Y}\to [0,1]$$ by letting $\kappa(x,B)=\pi(x)(B)$ for all $x\in X$ and $B\in\mathcal{Y}$. One can rather easily show that $\kappa(x,\cdot)$ is a probability measure for all $x\in X$ and $\kappa(\cdot,B)$ is measurable for all $B\in\mathcal{Y}$. Moreover, every such function can be obtained this way. They are known as Markov kernels, probability kernels, transition probabilities, conditional probabilities, random mappings, probabilistic mappings, Young measures... they occur quite often in probability theory. Now if $\nu$ is a probability measure on $X$, we can construct a probability measure on $Y$ that corresponds to the intuitive notion of picking a point $x\in X$ according to $\nu$ and then picking a point $y\in Y$ according to $\pi(x)$. We write this measure $\nu\otimes\pi$ and have $$\nu\otimes\pi(B)=\int_X \pi(x)(B)~d\nu(x)$$ for all $B\in\mathcal{Y}$. But this can actually be seen as using $\mathcal{P}(\mathcal{P}(X))$. Using a slightly more complicated construction, we have a composition of kernels and there is actually a category with measurable spaces as objects and kernels as morphisms. If we have a whole sequence of kernels, we can use them to construct discrete time stochastic processes by the so called Ionescu-Tulcea theorem. This is the foundation of discrete time Markov processes. We can identify probability measures on $Y$ with kernels constant in the first coordinate and measurable functions $f:X\to Y$ with the kernel such that $\pi(x)$ puts probability $1$ on $f(x)$. So kernels are a very useful building block for probability theory. A good ressource on Kernels is "Foundations of Modern Probability" by Kallenberg. A book that goes into a lot of depth is "Statistical Decision Rules and Optimal Inference" by Cencov. A good starting point for the categorial viewpoint can be found here. You seem to be interested in the more topological approach. Erik Balder's material on Young measures might be useful for this (but it is hard to read).

An even more direct place where probabilities over probabilities matter is in game theory. There one studies sometimes beliefs over beliefs over beliefs over.. infinite hierarchies of beliefs. The seminal paper on the topic is this, but it is known to be really hard to read. To get the basic idea, read this. For the case of Polish spaces, this paper by Heifetz might be a good starting point (paywall warning!).

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Why did you state the form of the $\sigma$-algebra on $\mathcal{P}(X)$? –  Quinn Culver Aug 7 '12 at 14:39
Would you please elaborate on "But this can actually be seen as using $\mathcal{P}(\mathcal{P}(X))$."? –  Quinn Culver Aug 7 '12 at 14:39
@QuinnCulver: I stated the form of the $\sigma$-algebra because it might not be immediately obvious to everyone that it coincides with the Borel $\sigma$-algebra of the weak*-topology on a space of Borel measures on a Polish space. They do, but the formulation makes sense even in nontopological settings, and most of what I wrote holds for arbitrary measurable spaces. –  Michael Greinecker Aug 7 '12 at 17:06
@QuinnCulver: It is using $\mathcal{P}(\mathcal{P}(X))$ the same way it is used in de Finetti's theorem, which you mentioned in your post. –  Michael Greinecker Aug 7 '12 at 17:07
Okay, that's what I thought (regarding the use of $\mathcal{P}(\mathcal{P}(X))$). –  Quinn Culver Aug 7 '12 at 20:40