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 $F_t$.


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 $\mu$.

  • $\begingroup$ 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

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*}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.