Conditional Expectation and the Time-Zero Measure Density
Summary
The document clarifies a common error about the process formed by conditioning a Radon-Nikodym derivative on the information available at time t. Its unconditional expectation is one at every time, by the tower property, but that does not mean the conditional expectation itself equals one at every time. A conditional expectation is an information-dependent random variable, generally measurable with respect to the current filtration; integrating it over the whole sample space gives its unconditional expectation.
At time zero, the density process equals one if the initial sigma-algebra is trivial, since conditioning then agrees with taking the ordinary expectation of the likelihood ratio. The reply emphasizes that integrating over all outcomes defines unconditional expectation, rather than conditional expectation at a general time. The result assumes equivalent probability measures and a probability density dQ/dP; if the initial information is nontrivial, the time-zero conditional density need not be identically one, even though its expectation is one.
Key ideas
- The conditional expectation of a likelihood ratio is a random variable tied to the information available at time t.
- Its unconditional expectation remains one at every time by the tower property.
- A time-zero density equal to one requires the initial sigma-algebra to be trivial.
- Integration across the entire sample space computes unconditional expectation, not conditional expectation at a general time.
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# Radon-Nikodim Derivative at time 0
# Radon-Nikodim Derivative at time 0
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!
## Answer by Alex (score 2)
https://quant.stackexchange.com/a/53417
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 $\{\emptyset,\Omega\}$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.