Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $p_t$ ?

share|improve this question
1  
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

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.