Given a prob measure P and a P-UI martingale $\{ \rho_t \}$ define $Q \sim P$ s.t. $E[\frac{dQ}{dP}|\mathscr{F}_t]=\rho_t$

Question

I know that if we have a filtered probability space $(\Omega, \mathscr{F},\{ \mathscr{F}_t \},P)$ and we have a probability measure Q equivalent to P, then there exists the Radon-Nikodym $\frac{dQ}{dP}:= \rho$ and we can define a UI martingale via $\rho_t:=E[\rho|\mathscr{F}_t].$ I would like to prove some type of converse result. If I'm given a probability measure $P$ and a UI martingale $\{ \rho_t \},$ how can I define a prob measure Q equivalent to P such that $\rho_t=E[\frac{dQ}{dP}|\mathscr{F}_t]$ (or prove that there exists such a Q)?

Answer 1

First of all, note that we need to assume that $\varrho_t \geq 0$ and $\mathbb{E}(\varrho_t)=1$ for all $t \geq 0$ (because otherwise $\varrho_t$ cannot be a conditional density of a probability measure).

Since $(\varrho_t)_{t \geq 0}$ is a uniformly integrable martingale, it follows from the martingale convergence theorem that the limit $\varrho_{\infty} := \lim_{t \to \infty} \varrho_t$ exists in $L^1$ and almost surely. In particular, $\varrho_{\infty} \geq 0$ and $\mathbb{E}(\varrho_{\infty}) = \lim_{t \to \infty} \mathbb{E}(\varrho_t)=1$. This means that

$$\mathbb{Q}(A) := \int_A \varrho_{\infty} \, d\mathbb{P}$$

defines a probability measure. Moreover, $(\varrho_t)_{t \in [0,\infty]}$ is a martingale and so $$\mathbb{E} \left( \frac{d\mathbb{Q}}{d\mathbb{P}} \mid \mathcal{F}_t \right) = \mathbb{E}(\varrho_{\infty} \mid \mathcal{F}_t) = \varrho_t.$$

Remark: The assumption $\mathbb{E}(\varrho_t)=1$ is not very restrictive. Let $(\varrho_t)_{t \geq 0}$ be a non-negative, uniformly integrable martingale. If $\mathbb{E}(\varrho_t)=0$, then $\varrho_t=0$ and so $\mathbb{Q}:=0$ satisfies $\mathbb{E}(d\mathbb{Q}/d\mathbb{P} \mid \mathcal{F}_t)=\varrho_t$. On the other hand, if $\mathbb{E}(\varrho_t)=\mathbb{E}(\varrho_0)$ is strictly positive (note that the expectation does not depend on $t$ because of the martingale property), then we can apply the above reasoning to the rescaled process $\frac{1}{\mathbb{E}(\varrho_0)} \varrho_t$.