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Let me reformulate my question. Let $(X_0,X_T)$ be a martingale on $\mathbb R$, then it is known that one has a SDE:

$$Z_t=Z_0+\int_0^t\sigma(s,Z_s)dB_s, \mbox{ for all } t\in [0,T]~~~~~~~~~~~~~~(\ast)$$

s.t. $Z_0\sim X_0$ and $Z_T\sim X_T$. My question is the following: Could we construct a similar SDE above s.t.

$$(Z_0,Z_T)~~\sim~~(X_0,X_T)?$$

PS: Some references: https://www.maths.univ-evry.fr/prepubli/334.pdf

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  • $\begingroup$ What are the quantifiers on $\gamma, \mu,$ and $\nu$? $\endgroup$
    – Pat Devlin
    Nov 29, 2016 at 13:20
  • $\begingroup$ $Z_0,Z_T$ should form a martingale, so not in all case, as e.g. when the variance of $Z_T$ is smaller. $\endgroup$
    – user83457
    Nov 29, 2016 at 13:50
  • $\begingroup$ @michael That's implied by the definition of $\gamma$. Indeed, for any pair $(\mu,\nu)$ satisfying the above condition, one always has a martingale with the marginals $\mu$ and $\nu$. See the paper maths.univ-evry.fr/prepubli/334.pdf $\endgroup$
    – CodeGolf
    Nov 29, 2016 at 14:02
  • $\begingroup$ @PatDevlin Actually. We may understand the problem in another way: Given a martingale $(X_0,X_T)$ with marginal laws $\mu$ and $\nu$, could we embed this martingale to a solution $(Z_t)_{0\le t\le T}$ given by the SDE, i.e. $(X_0,X_T)\sim (Z_0,Z_T)$? $\endgroup$
    – CodeGolf
    Nov 29, 2016 at 14:05

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