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Let $\mu_t, t \geq 0,$ be a family of probability measures on the real line. One can assume whatever one wishes about them, although typically they will be continuous in some topology (usually at least the topology of weak convergence of measures), and they will be absolutely continuous with respect to Lebesgue measure. The basic question is as follows:

Is there a Markov process $X_t$ such that its marginal distribution at each time is $\mu_t$?

An obvious example is when $$d \mu_t = \frac{e^{-x^2/2t}}{\sqrt{2 \pi t}} dx$$ and $\mu_0 = \delta_0$, in which case we know that Brownian motion is such a Markov process. I am curious to know if there is any general theory along these lines.

Edit

As per Byron's comment below, I would like the Markov process to be continuous. Ideally I would like to have an SDE description of the process.

The SDE description actually suggests one possible answer: simply compute and play with the time and space derivatives of the density function to see if they satisfy some sort of parabolic equation (like the heat equation), use this to get the adjoint of the generator, and then compute the generator itself. This is a very plausible option, but I was hoping that there might be something more systematic.

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3 Answers 3

The following result was proved in

Kellerer, H.G. (1972) Markov-Komposition und eine Anwendung auf Martingale. Math. Ann., 198, 99–122.

Let $p(y, t)$ be a family of marginal densities, with finite first moment, such that for $s , t$ the density at time $t$ dominates the density at time $s$ in the convex order. Then there exists a Markov process $X(t)$ with these marginal densities under which $X(t)$ is a submartingale. Furthermore, if the means are independent of $t$ then $X(t)$ is a martingale.

Constructive versions have then been studied by Madan and Yor in

http://projecteuclid.org/download/pdf_1/euclid.bj/1078681382

The SDE description you mention is actually discussed in Section 2.3 of the paper.

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If you are willing to drop continuity in the parameter $t$, then you could let $(X_t)$ be independent with distribution $\mu_t$.

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Good point. Yeah I meant to say that the process should be continuous (or at least enjoy some kind of regularity property). But I'll stick with continuous and edit my post. Thanks! –  Tom Alberts Feb 9 '12 at 4:34
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Hi,

This is not the general answer you are looking for but it might be sufficient for your needs, here are my two cents.

If you are given a Semimartingale then there exists conditions that ensures that there exists a Markov process such that this markov process has the same marginal distributions that the marginal distributions of your semimartingale.

Take a look here, where it is motivated by application to mathematical finance and where the first version of this theorem is known as "Gyongy's Lemma" I believe.

Best Regards

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Thank you! You are right that it's not exactly what I'm looking for, but the references in here are invaluable. It's already exposed me to a lot of material which I didn't know about before. I'm going to leave the question open for a little bit to see if anyone else has some input. –  Tom Alberts Feb 10 '12 at 1:01
    
@Tom alberts : No problems, I'm glad if this reference is of any help to you. Best regards –  The Bridge Feb 10 '12 at 17:35
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