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Let $\{p_t\}_{t \geq 0}$ be a family of densities. Is there any result concerning the existence of a semi-martingale $\{X_t\}_{t \geq 0}$ such that for all $t\geq 0$, the density of $X_t$ is $p_t$ ?

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It really only relies on the fact that if $p$ is a density wrt. the Lebesgue measure, then there exists a random variable $X$ with density $p$. – Stefan Hansen Jul 2 '12 at 15:43
My bad, it is indeed a stupid question. The only interesting case would be if the covariance structure was specified, e.g. finite dimensional laws. Kolmogorov extension theorem would be the answer in this case. – vanna Jul 2 '12 at 15:54
@StefanHansen I changed stochastic process to semi-martingale, which is what I am actually looking at. – vanna Jul 2 '12 at 15:59
up vote 2 down vote accepted

After some investigations I found this theorem due to Kellerer.

Theorem (Kellerer, 1972) $-$ Let $(\mu_t)_{t\in[0,T]}$ be a family of probability measures of $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ with first moments, such that for $s<t$, $\mu_t$ dominates $\mu_s$ in the convex order, i.e. for each convex function $\phi : \mathbb{R} \rightarrow \mathbb{R}$ $\mu_t$-integrable for each $t\in[0,T]$, we have $$ \int_\mathbb{R} \phi d\mu_t \ge \int_\mathbb{R} \phi d\mu_s $$ Then there exists a Markov process $(M_t)_{t\in [0,T]}$ with these marginals for which it is a submartingale. Furthermore if the means are independant of $t$ then $(M_t)_{t\in[0,T]}$ is a martingale.

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