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This is the infinite-dimensional sequel to my question, Conditional probabilities are measurable functions - when are they continuous?.

Let $\Omega = \Omega_1 \times \Omega_2$ be a probability space which is Banach, $\mathcal F$ the Borel $\sigma$-algebra on $\Omega$, and $\mathbb P$ a probability measure which need not be the product measure. e.g., $\Omega = C(U_1) \times C(U_2)$ for disjoint, compact $U_1, ~U_2 \subseteq {\mathbb R}$. Let $\mathcal F_2 = \sigma(\Omega_2^\*)$, where $\Omega_2^\*$ is the dual space to $\Omega_2$. In the case of continuous functions, this is the $\sigma$-algebra generated by the evaulation maps $\pi_x$ for $x \in U_2$. Note that these evaluations $\pi_x$ are random variables on $\Omega$.

Goal: I would like an explicit expression for conditional expectations with respect to $\mathcal F_2$. Namely, I want a "reasonable" linear operator $P : \Omega^\* \to \Omega^\*$ such that $$(\*) \qquad \mathbb E(\pi_x | \mathcal F_2) = P\pi_x.$$Ideally, "reasonable" will mean continuous.

I can do this in the case that $\Omega$ is a Gaussian Hilbert space. Decompose the covariance operator of $\mathbb P$ as $$K = \binom{K_{11} ~ K_{12}}{K_{21} ~ K_{22}}.$$(This should be matrix formatted but that doesn't seem to work here). Let $$P = \binom{~~~0 ~~~~~~~~ 0}{K_{22}^{-1} K_{21} ~~ I_2},$$ where $I_2$ is the identity operator on $\Omega_2^\*$. Then using the Gaussian structure, I can show that $(\*)$ holds; using the technology in Anderson & Trapp's Shorted Operators, II I can show that $P$ is continuous.

This is overkill! I think I can adapt my Gaussian calculation without too much trouble to the generic case (since it only deals with covariance operators). On the other hand, I don't know how to show that such an operator $P$ is continuous without relying on heavy-duty functional analysis. Surely this has been studied before, but I can't seem to find a good reference.

Note: Brownian motion is a very special case of this, where $\Omega_2 = C(${$0$}$) = \mathbb R$, the starting value $B_0$. There, one skirts the issue of this conditioning by the Markov property: the "future" $\Omega_1 = C((0,\infty))$ is independent of the "present" $\Omega_2$ up to the starting value $B_0$.

Note: The variance of $\pi_x$ for $x \in U_1$ is the Schur complement $\pi_x^\* (K_{11} - K_{12}^\* K_{22}^{-1} K_{21})\pi_x$.

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Tom--I find using \binom is helpful for small matrices; there is also a meta thread about larger matrices: tea.mathoverflow.net/discussion/164/writing-matrices/#Item_0 –  Steve Huntsman Feb 2 '10 at 20:59
    
Also, have you considered the case where $\mathbb{P}$ is a finite convex combination of simple measures, each with rectangular support? –  Steve Huntsman Feb 2 '10 at 21:06
    
@Steve, thanks for the \binom tip. It's pretty ad hoc (I had to use lots of tildes to make the spacing line up) but it works! Could you explain further your special case? I'm afraid I'm not well-versed enough in Banach space theory to follow. –  Tom LaGatta Feb 2 '10 at 22:38
    
[Disclaimer: I haven't really looked at your question carefully, so I might be overlooking something trivial.] In the simplest case, suppose your measure is uniform on some rectangle. Then everything's a product and it looks pretty straightforward. If now the measure is a finite convex combination of such uniform measures (what I meant by a "simple measure") you should be able to take the summations inside and outside expectations. Then it's a question of convergence to get a reasonably general case. –  Steve Huntsman Feb 2 '10 at 23:46
    
Oh I see. I don't think that'll work here. For example, there is no analogue of Lebesgue/uniform measure on function spaces. The best simple alternative is Gaussian measure. Like I said, I can do it in the case of Gaussian measure on a Hilbert space, but I think it should be doable in a greater generality as is. –  Tom LaGatta Feb 3 '10 at 2:40

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