0
$\begingroup$

Problem:

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.

$\endgroup$
1
$\begingroup$

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

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

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.