Question about Girsanov and Conditional Expectation

Question

could you please help me to prove the following :

$$ E^*\left[X|F_t\right]=\frac{E\left[X\frac{dP^*}{dP}|F_t \right]}{\rho_t} $$

Given that $\rho_t $ a martingale

$$ \rho_t=\frac{dP^*}{dP}|F_t$$

This is what I’ve done, but no success. I think there’s something wrong with my understanding :

$$ \frac{E\left[X\frac{dP^*}{dP}|F_t \right]}{\rho_t} =\frac{E\left[X\rho_t|F_t\right] }{\rho_t} $$ $$=\frac{\rho_t E\left[X |F_t \right]}{\rho_t} $$ $$=E[X |F_t ]\neq E^*[ X |F_t] $$

Answer 1

Note that \begin{align*} \rho_t = E\left(\frac{dP^*}{dP}\mid\mathcal{F}_t \right), \end{align*} instead of \begin{align*} \rho_t = \frac{dP^*}{dP}\big| \mathcal{F}_t. \end{align*} Then, for any $C \in \mathcal{F}_t$, \begin{align*} \int_C E\left(X \frac{dP^*}{dP}\mid\mathcal{F}_t \right) dP &=\int_C X \frac{dP^*}{dP} dP\\ &=\int_C X dP^*\\ &=\int_C E^*(X\mid \mathcal{F}_t) dP^*\\ &=\int_C E^*(X\mid \mathcal{F}_t) \frac{dP^*}{dP} dP\\ &=\int_C E\left(E^*(X\mid \mathcal{F}_t) \frac{dP^*}{dP} \mid \mathcal{F}_t\right)dP\\ &=\int_C E^*(X\mid \mathcal{F}_t)E\left(\frac{dP^*}{dP} \mid \mathcal{F}_t\right)dP\\ &=\int_C E^*(X\mid \mathcal{F}_t) \rho_t\, dP. \end{align*} That is, \begin{align*} E\left(X \frac{dP^*}{dP}\mid\mathcal{F}_t \right) = E^*(X\mid \mathcal{F}_t) \rho_t, \end{align*} or \begin{align*} E^*(X\mid \mathcal{F}_t) = \frac{E\left(X \frac{dP^*}{dP}\mid\mathcal{F}_t \right)}{\rho_t}. \end{align*}