Conditional Change of Measure and Bayes’ Formula

Summary

The document asks how the conditional-expectation change-of-measure identity relates to the unconditional formula. For equivalent measures P and Q with Radon–Nikodym density dQ/dP, it recalls that an expectation under Q can be written as a P expectation weighted by that density. It then presents the conditional version, which includes the conditional expectation of the density as a denominator.

The questioner's proposed explanation—that this extra term equals one because the conditional Q expectation of one is one—is mistaken in general. The density's conditional expectation under P need not equal one; it is the ratio that normalizes the weighted conditional expectation. The discussion gives no worked resolution or financial application, so its value is conceptual: it identifies a measure-change detail relevant to conditional pricing and risk-neutral expectations. The assumptions and integrability conditions needed for the identities are not developed.

Key ideas

• Changing from P to Q weights expectations under P by the Radon–Nikodym density.
• The conditional change-of-measure identity includes a normalization by the density's conditional expectation under P.
• That conditional normalization is not generally equal to one.
• The document raises the issue but does not provide a complete derivation or application.

Tags

# Understanding Bayes Rule of conditional expectation


# Understanding Bayes Rule of conditional expectation












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!

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.