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This is a follow up of another post: Construct a martingale with a given distribution?

Given two distributions $f_1(\cdot)$ and $f_2(\cdot)$ on $\mathbb R$, under what condition can we construct a martingale $X_t$, such that $X_1$ has distribution $f_1$, and $X_2$ has distribution $f_2$?

From martingale property, we know that $E(X_1)=E(X_2)$ and $Var(X_1)\le Var(X_2)$, so we know $f_1(\cdot)$ and $f_2(\cdot)$ can not be arbitrary.

Jay.H
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2 Answers2

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The martingale property means that $\Bbb E[X_2|X_1]=X_1$. Jensen's inequality now implies that if $\varphi:\Bbb R\to\Bbb R$ is a convex function then $$ \Bbb E[\varphi(X_1)]=\Bbb E[\varphi(\Bbb E[X_2|X_1])]\le \Bbb E[\Bbb E[\varphi(X_2)|X_1]]=\Bbb E[\varphi(X_2)]. $$ This "monotonicity along convex functions" property is not only necessary but also sufficient for two distributions to be the marginals at two times of a martingale. This result is due to V. Strassen, in The existence of probability measures with given marginals, at http://projecteuclid.org/download/pdf_1/euclid.aoms/1177700153 . Thus, you need to check that $\int_{\Bbb R}\varphi(x)f_1(x)\,dx\le\int_{\Bbb R}\varphi(x)f_2(x)\,dx $ for all convex $\varphi$.

John Dawkins
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  • Thanks. I need sometime to digest the article, but good to know it is not trivial. Also, it seems that the article only shows existence of discrete martingales. An additional question is: given that all such inequalities hold, is it possible to find a continuous time martingale, with continuous sample paths that "interpolates" the two given distributions? – Jay.H Apr 11 '16 at 20:56
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Modulo technical details (on which I am no expert) the answer is Yes. The result is due to J.L. Doob ["Generalized sweeping-out and probability", J. Functional Analysis vol. 2, (1968) 207–225], who showed the analogous result for submartingales. Strassen's work was applied to the problem by H. G. Kellerer "Markov-Komposition und eine Anwendung auf Martingale" dating to 1972, in Mathematische Annalen. The topic is of some current research interest, under the name "peacock" (for the french acronym PCOC = Processus Croissants pour l'Ordre Convexe).

John Dawkins
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  • Thanks a lot. It seems that there is even a book dedicated to "peacock" : https://books.google.com/books?id=hix-wkrST00C&pg=PA262&lpg=PA262&dq=peacock+stochastic&source=bl&ots=N14SJ1GQBQ&sig=YW2mZtucaBITTSeAKRowY6box4U&hl=en&sa=X&ved=0ahUKEwid8ZK1nonMAhXDZCYKHYg2CoEQ6AEIJTAC#v=onepage&q=peacock%20stochastic&f=false – Jay.H Apr 12 '16 at 13:54