I have a very basic question about filtrations and Radon-Nikodym derivatives. I am reading the Andersen-Piterbarg, more in particular Eq. (1.12). They define the process $\zeta(t) = E^P_t[\frac{dQ}{dP}]$, where $Q\sim P$ are equivalent measures. Now, their claim is that obviously $\zeta(0) = 1$. Now, I see that the whole sample space $\Omega$ belongs to $\mathcal{F}_0$, which thus implies $\zeta(0) = 1$ (using the definition of expected value). But why it doesn't hold for every $t$? I mean, aren't the $\mathcal{F}_t$ also sigma-algebras, and thus contain $\Omega$, which would imply, by the definition of conditional expectation, $$ \int_{\Omega} \zeta(t, \omega) dP(\omega) = \int_{\Omega} E^P_t[\frac{dQ}{dP}](\omega)dP(\omega) \stackrel{def \:\&\: \Omega \in \mathcal{F}_t}{=} \int_{\Omega} \frac{dQ}{dP}(\omega)dP(\omega) = \int_{\Omega} dQ(\omega) = 1. $$ What am I doing wrong here? Does $\Omega$ not belong to $\mathcal{F}_t$? Thanks in advance!
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At time $t=0$, you get \begin{align*} \zeta(0)=E^P_0\left[\frac{\mathrm{d}Q}{\mathrm{d}P}\right]=E^P\left[\frac{\mathrm{d}Q}{\mathrm{d}P}\right]=\int_\Omega \frac{\mathrm{d}Q}{\mathrm{d}P}\mathrm{d}P=\int_\Omega \mathrm{d}Q = Q(\Omega)=1, \end{align*} because $Q$ is a probability measure.
But at a general time point $t$, you cannot write $E_t^P[X]=\int_\Omega X \mathrm{d}P$. That integral is the definition of the unconditional expectation! In fact, it only works at time $t=0$ if you assume that the filtration begins with the trivial $\sigma$-algebra $\{\emptyset,\Omega\}$.
