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Let $(\nu_t)_{t \in [0,1]}$ be Borel probability measures on a stochastic basis $(\Omega,\mathcal{F},(\mathcal{F}_{t \in [0,1]})_t,\mathbb{P})$.

Does there exist a semi-martingale $(X_t)_{t\in[0,1]}$ on this stochastic basis, such that $X_t\sim \nu_t$ for everty $t \in [0,1]$? If not, what conditions are needed on the measures $\nu_{\cdot}$ for this to be possible?

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  • $\begingroup$ No. For any $\nu$ there are many $X$ such that $X \sim \nu$. Starting with a martingale $X_i$, we can change one or more $X_i$ to $Y_i$ with $Y_i \sim X_i$ and destroy the martingale property. A more sensible question may be: for which systems $(\nu_i)$ does there exist a martingale $(X_i)$ with $X_i \sim \nu_i\;\forall i$? There is a certain "order" relation $\preceq$ on probability measures, and the condition you want is $\nu_i \preceq \nu_j$ for $i \le j$. That is a different qustion, so I do not say more here. $\endgroup$
    – Gerald Edgar
    Jan 14, 2021 at 11:57
  • $\begingroup$ @GeraldEdgar Good point, I reframed the question with your suggestion. However, I'm wondering; what is this (partial?) ordering? $\endgroup$
    – ABIM
    Jan 14, 2021 at 13:01
  • $\begingroup$ Certainly not without extra assumptions: just pick a path $x(t)$ of unbounded variation, and set $\nu_t = \delta_{x(t)}$ to be a point-mass at $x(t)$. Then $X_t = x(t)$ almost surely, and hence $X_t$ is clearly not a semi-martingale. (I assume $\nu_t$ are Borel measures on $\mathbb R$ rather than on $\Omega$.) $\endgroup$
    – Mateusz Kwaśnicki
    Jan 14, 2021 at 13:46
  • $\begingroup$ Hehe, that's an very clever degenerat example. I guess part of my question (albehit maybe too implicit) is what conditions would such a system of measures require? $\endgroup$
    – ABIM
    Jan 14, 2021 at 16:44
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    $\begingroup$ It being Friday I'll confine myself to continuous $X$, and seek out reasonable necessary conditions. As a first approximation further suppose $X_t = M_t + V_t$, in which $M$ s a continuous martingale and $V$ is adapted, continuous, and of integrable variation. Then for a bounded $C^1$ function $f$ (with bounded derivative) one has $$ \Bbb E[f(X_t)]=\Bbb E[f(X_0)]+\Bbb E\left[\int_0^t f'(X_s)\,dV_s\right], $$ implying (I think) that $t\mapsto \Bbb E[f(X_t)]$ is of finite variation on $[0,1]$. How far a condition like this is from being sufficient is not clear to me. $\endgroup$
    – John Dawkins
    Jan 16, 2021 at 0:00

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