Change of measure of conditional expectation

Question

How can I prove that: $E_π [ (dQ_X/dπ) S (T)| F_t ]= E_{Q_X} [S(T) | F_t]E_π [ dQ_X/dπ | F_t ]$. Obviously $E_π [(dQ_X/dπ) S(T) ]= E_{Q_X} [S(T)]$ I know that much, but how to prove when it is conditioned on $F_t$.

Answer 1

I guess the $Q_X$, $S(T)$ and $\mathcal F_t$ come from a particular context, so we can simplify the notation and prove that if $\mu,\nu$ are two probability measures such that $\nu\ll\mu$ and the random variables $X$ and $\frac{\mathrm d\nu}{\mathrm d\mu}$ are integrable, then $$\mathbb E_\mu\left(\frac{\mathrm d\nu}{\mathrm d\mu}\cdot X\mid \mathcal F\right)=\mathbb E_{\nu}(X\mid \mathcal F)\cdot\mathbb E_\mu\left(\frac{\mathrm d\nu}{\mathrm d\mu}\cdot\mid\mathcal F\right).$$ To see that, we go back to the definition of conditional expectation. Let $F\in\mathcal F$. Then $$\int_F\mathbb E_\mu\left(\frac{\mathrm d\nu}{\mathrm d\mu}\cdot X\mid \mathcal F\right)\mathrm d\mu=\int_F\frac{\mathrm d\nu}{\mathrm d\mu} X\mathrm d\mu=\int_FX\mathrm d\nu.$$ Now, the trick is to use $\mathcal F$-measurability of $\mathbb E_\nu(X\mid\mathcal F)$ to write $$\mathbb E_{\nu}(X\mid \mathcal F)\cdot\mathbb E_\mu\left(\frac{\mathrm d\nu}{\mathrm d\mu}\cdot\mid\mathcal F\right)=\mathbb E_\mu\left(\mathbb E_{\nu}(X\mid \mathcal F)\cdot\frac{\mathrm d\nu}{\mathrm d\mu}\cdot\mid\mathcal F\right),$$ then integrate over $F$ with respect to $\mu$.