# Is it possible to obtain the Radon-Nikodym derivative $\left. \frac{dQ}{dP} \right|_{\mathscr{F}}$ from a UI martigale?

I have the following theorem:

Theorem. Let P be a probability measure on $$(\Omega, \{ \mathscr{F} \}, \mathscr{F}).$$ Given an a.s. strictly positive UI $$(\{ \mathscr{F} \},P)$$-martingale $$\rho$$ with $$E_P[\rho_{\infty}]=1$$, $$\left. \frac{dQ}{dP} \right|_{\mathscr{F}_{\infty}}:= \rho_{\infty}$$ defines via $$Q(A) = \int_A \rho_{\infty} dP \quad \text{for every } A \in \mathscr{F}_{\infty}$$ a measure Q equivalent to P with respect to $$\mathscr{F}_{\infty}.$$

Remark. Note that if $$\mathscr{F}$$ is richer than $$\mathscr{F}_{\infty},$$ we can not recover $$\left. \frac{dQ}{dP} \right|_{\mathscr{F}}$$ from the martingale $$\rho.$$

My question: I don't understand the remark. Why can't we get Q equivalent to P with respect to the "richer" $$\mathscr{F}$$? Why the "best" we can do is to obtain $$\left. \frac{dQ}{dP} \right|_{\mathscr{F}_{\infty}}$$?

We know that $$\rho_{\infty} \in L^1(\Omega, \mathscr{F}_{\infty}, P)$$ but I think it is also true that $$\rho_{\infty} \in L^1(\Omega, \mathscr{F}, P).$$ Hence, I could have defined $$Q(A) = \int_A \rho_{\infty} dP \quad \text{for every } A \in \mathscr{F},$$ and then we would have $$Q \ll P$$ with respect $$\mathscr{F}.$$ Also, since $$\rho$$ is an a.s. strictly positive UI martingale, we have $$P \ll Q.$$ Is there anything wrong with this argument?

• @GeorgeDewhirst Measurability w.r.t. a smaller sigma algebra implies measurability w.r.t. a larger one. So measurability is not the issue. Commented Dec 22, 2019 at 11:44

If we define $$Q(A) = \int_A \rho_{\infty} dP \quad \text{for all } A \in \mathscr{F},$$ then $$\rho_{\infty}$$ is the Radon-Nikodym derivative of Q with respect to P on $$\mathscr{F}.$$ Since $$\rho_{\infty}$$ is $$\mathscr{F}$$-measurable, $$E[\rho_{\infty} | \mathscr{F}] = \rho_{\infty}.$$ But if $$\mathscr{F}$$ is richer than $$\mathscr{F}_{\infty},$$ it could happen that $$E[\rho_{\infty} | \mathscr{F}] = \rho_{\infty} \neq E[\rho_{\infty} | \mathscr{F}_{\infty}],$$ so the UI martingale wouldn't be $$\{ \mathscr{F}_t \}$$-adapted, which would be a contradiction. Thus, we are forced to define $$Q$$ only for all $$A \in \mathscr{F}_{\infty}.$$