Let $(\Omega,\mathcal F,\mathbf P,\{\mathcal F_t\}_{t\ge 0})$ be a filtered probability space, and let $\mathbf Q$ be a probability measure on $(\Omega,\mathcal F,\{\mathcal F_t\}_{t\ge 0})$ such that $\mathbf Q\ll\mathbf P$. Then it's well known that the 'density process' \begin{equation}\tag{1} Z_t:=\mathbf E^{\mathbf P}\left[\frac{d\mathbf Q}{d\mathbf P}\bigg|\mathcal F_t\right] \end{equation} is a $\mathbf P$-martingale (see, for instance, [Jacod and Shiryaev, 2003, Theorem III.3.4]).

Now we consider the inverse problem, that is,

under what condition a $\mathbf P$-martingale $\{Z_t\}$ can be the density process of some probability measure $\mathbf Q$ which is absolutely continuous w.r.t $\mathbf P$?

I cannot find out the answer to the inverse problem in Jacod and Shiryaev's book. Can anyone give some comments or reference to it? Thank in advance!


1 Answer 1


Essentially, $Z_t$ has to be a.s. positive and uniformly integrable. Then, it's convergent for $t\to\infty$ in $L^1$ to an a.s. positive $Z_\infty$, and $$Z_t=E(Z_\infty\mid\mathcal{F}_t).$$ Obviously, you can use $Z_\infty$ as the density you're looking for.
It's surprisingly difficult to find a reference, though. In the discrete case, it would be Theorem 4.5.6 in this lecture. It's also Theorem T18 in Chapter V, §3 of "Probability and Potentials" by P. A. Meyer, but I own the Russian translation, so...

  • $\begingroup$ Thank you. I still got a question here. If $Z$ is a positive and uniformly integrable martingale on $(\Omega,\mathcal F,\mathbf P,\{\mathcal F_t\}_{t\ge 0})$, then it's easy to get that the measure $\mathbf Q:=Z_\infty.\mathbf P$ is the only measure on $\mathcal F_\infty:=\bigvee_{0\le t\le\infty}\mathcal F_t$ such that the equation (1) holds, by the Caratheodory Extension Theorem. My question is if $\mathbf Q$ is unique on $\mathcal F$ satisfying (1)? $\endgroup$
    – Dreamer
    Commented Feb 5, 2018 at 3:57
  • $\begingroup$ @Q. Huang If $\mathcal{F}$ is bigger than $\mathcal{F}_\infty$, $\mathbf Q$ can't be unique. $\endgroup$
    – user436658
    Commented Feb 5, 2018 at 4:41

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