= Conditional expectation under a martingale density
{title2=$\mathbb E_Q[Y\mid\mathcal F_s]=\mathbb E_P[YZ_T\mid\mathcal F_s]/Z_s$}
Let $Z_s>0$ be a <martingale> with $Z_0=1$ and let $dQ=Z_T\,dP$ on a finite horizon. Its restriction to $\mathcal F_s$ has <Radon-Nikodym derivative> $Z_s$. If $Y$ is Q-integrable and $U=\mathbb E_Q[Y\mid\mathcal F_s]$, then for every $A\in\mathcal F_s$,
$$
\mathbb E_P[\mathbf1_AZ_sU]=\mathbb E_Q[\mathbf1_AU]
=\mathbb E_Q[\mathbf1_AY]=\mathbb E_P[\mathbf1_AYZ_T].
$$
Also $Z_sU$ is P-integrable because $\mathbb E_P[Z_s|U|]=\mathbb E_Q|U|<\infty$. The defining uniqueness of <conditional expectation> gives the displayed formula. In particular, if $N_sZ_s$ is a P-martingale and $N_s$ is Q-integrable, conditioning first at an intermediate time proves $N_s$ is a Q-martingale.
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