2
$\begingroup$

Let $\mathcal{F}$ be a $\sigma$-algebra, $P$ and $Q$ be equivalent martingale measures and $\frac{dQ}{dP}$ the Radon Nikodym Derivative.
I learned that $\Bbb{E}_Q[X]=\Bbb{E}_P[\frac{dQ}{dP}X] $, which makes sense if one looks at the following: $$\Bbb{E}_Q[X]=\int_\Omega X dQ=\int_\Omega X \frac{dQ}{dP} dP=\Bbb{E}_P[\frac{dQ}{dP}X]$$Recently I was introduced to the Bayes Formula for conditional expectation, which states that $$\Bbb{E}_Q[X|\mathcal{F}]\;\; \Bbb{E}_P[\frac{dQ}{dP}|\mathcal{F}]=\Bbb{E}_P[\frac{dQ}{dP}X|\mathcal{F}] $$Comparing this with the version that I learned first, the term $\Bbb{E}_P[\frac{dQ}{dP}|\mathcal{F}]$ has been included and the only explanation I have is that $$\Bbb{E}_P[\frac{dQ}{dP}|\mathcal{F}]=\Bbb{E}_Q[1|\mathcal{F}]=1$$Is this understanding correct? And if so, why bother including the term on some occasions and excluding it on other occasions? Thank you for clearing up my confusion!

$\endgroup$
1
  • 2
    $\begingroup$ Have a look of here. $\endgroup$ – Gordon Jan 13 at 18:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.