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

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)?

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 $$\frac{1}{\mathbb{E}(\varrho_0)} \varrho_t$$.

• Thank you, it's clear now. One more question related with this that it should be straigthforward but I can't see it. Assume we have a square-integrable martingale M. Define $T:= \inf \{t>0 : |M_t|>\frac{1}{2} \}.$ Then I can define prob measures $Q$ and $R$ such that $E[\frac{dQ}{dP}|\mathscr{F}_t]=1-N^T$ and $E[\frac{dR}{dP}|\mathscr{F}_t]=1+N^T.$ Why is $P= \frac{1}{2}Q + \frac{1}{2}R$?
– UBM
Commented Dec 13, 2019 at 9:40
• It should be $1-M^T_t$ and $1+M^T_t.$
– UBM
Commented Dec 13, 2019 at 9:54
• shouldn't it be $1-M_t^T$ and $1+M_t^T$? If not, what is N?
– Gono
Commented Dec 13, 2019 at 9:55
• @Gono: yes yes, I was editing it... Do you know the answer?
– UBM
Commented Dec 13, 2019 at 9:57
• First of all you need some additional assumptions, e.g. it has to hold $1-M_t^T \ge 0$ and $1+M_t^T \ge 0$ for all $t$. Then it follows for all $t\ge 0$ that $$E[\frac{1}{2}\frac{dQ}{dP} + \frac{1}{2}\frac{dR}{dP}|\mathcal{F}_t] = \frac{1}{2}(1 - M_t^T + 1 + M_t^T) = 1$$ so we get $$P = \frac{1}{2}Q + \frac{1}{2}R$$ on $$\mathcal{F}_\infty = \sigma\left(\bigcup_{n\in\mathbb{N}} \mathcal{F}_n\right)$$
– Gono
Commented Dec 13, 2019 at 10:04