Let $(\Omega, \mathcal{F}, P)$ be a probability measure space. Let $Q$ be another (finite) measure on $(\Omega, \mathcal{F})$. Let $Q \ll P$ so by the Radon-Nikodym derivative there exists a $Y \in \mathcal{L}^{1}(\Omega, \mathcal{F}, P)$ such that $\frac{dQ}{dP} = Y$ with $Q(A) = \int_A Y~dP$. What is $E(Y)$ ?

My attempt:

Intuitively, we should have that $E(Y) = E(\frac{dQ}{dP}) = \int_{\Omega} \frac{dQ}{dP}~dP$ so the $dP$s should "cancel out" so that we get $Q(\Omega)$ as the final answer. This makes sense: if $Q$ is finite, then $Y \in \mathcal{L}^{1}$.

I know from Williams 1991, chapter 14, that $Q(F) = \int_F Y~dP, \forall F \in \mathcal{F}$, and I think this makes it very likely that my guess is correct, but I couldn't figure out how to use this to prove it. Any help would be appreciated.


Just put $F=\Omega$; you can do this since $\Omega$ is an element of the sigma-algebra by definition. Then you have $Q(\Omega)=\int_{\Omega} YdP=E(Y)$.

  • $\begingroup$ thanks, now I can't believe I didn't see this, but it fits perfectly! $\endgroup$ – baibo Oct 12 '19 at 15:45

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