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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 $F_t$.

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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 $\mu$.

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  • $\begingroup$ Do you know of a proof, which relies on the $L^2$-definition (and minimizer properties)? $\endgroup$
    – AIM_BLB
    Dec 23 '18 at 15:20
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Alternatively, note that, for ant $A \in \mathcal{F}_t$, \begin{align*} \int_A \frac{dQ_X}{d\pi}S(T) d\pi &=\int_A S(T) dQ_X\\ &=\int_A E_{Q_X}(S(T)\mid \mathcal{F}_t) dQ_X\\ &=\int_A E_{Q_X}(S(T)\mid \mathcal{F}_t) \frac{dQ_X}{d\pi} d\pi\\ &=\int_A E_{\pi}\left(E_{Q_X}(S(T)\mid \mathcal{F}_t) \frac{dQ_X}{d\pi} \mid \mathcal{F}_t \right)d\pi\\ &=\int_A E_{Q_X}(S(T)\mid \mathcal{F}_t) E_{\pi}\left(\frac{dQ_X}{d\pi} \mid \mathcal{F}_t \right)d\pi. \end{align*} Therefore, \begin{align*} E_{\pi}\left(\frac{dQ_X}{d\pi}S(T) \mid \mathcal{F}_t\right) &= E_{Q_X}(S(T)\mid \mathcal{F}_t) E_{\pi}\left(\frac{dQ_X}{d\pi} \mid \mathcal{F}_t \right). \end{align*}

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