Restricted Radon–Nikodym Derivatives Across Equivalent Measures

Summary

The document poses an identity for three equivalent probability measures and a sub-sigma-algebra: the density of one measure relative to another, restricted to the smaller information set, factors through a third measure. This kind of relationship is useful in probability arguments that underpin changes of measure in mathematical finance.

The questioner's attempted derivation invokes conditional expectations and a measure-change rule, but the steps shown are not a reliable proof: conditional expectations of a product do not generally factor into a product of conditional expectations, and the displayed density ratios are inconsistent. A correct route uses conditional expectations of the full Radon–Nikodym derivatives and the tower property, with care about which measure each conditional expectation uses. The document itself contains no completed proof, trading application, or empirical evidence, so it serves mainly as a mathematical prompt rather than a worked explanation.

Key ideas

• The problem concerns how Radon–Nikodym densities behave when restricted to a sub-sigma-algebra.
• Equivalent measures allow the relevant density ratios to be defined almost surely.
• A proof must track the measure used for each conditional expectation.
• Conditional expectations of products cannot generally be split into products of conditional expectations.
• The document's proposed derivation is incomplete and does not establish the identity.

Tags

# Let $\mathbb{P} \sim \mathbb{Q} \sim \mathbb{R}$ be equivalent probability measures on some measurable space


# Let $\mathbb{P} \sim \mathbb{Q} \sim \mathbb{R}$ be equivalent probability measures on some measurable space












Let $\mathbb{P} \sim \mathbb{Q} \sim \mathbb{R}$ be equivalent probability measures on some measurable space $(\Omega, \mathcal{F})$, and let $\mathcal{G} \subset \mathcal{F}$ be a sub- $\sigma$-algebra. Show that $$ \left.\frac{d \mathbb{R}}{d \mathbb{P}}\right|_{\mathcal{G}}=\left.\left.\frac{d \mathbb{R}}{d \mathbb{Q}}\right|_{\mathcal{G}} \frac{d \mathbb{Q}}{d \mathbb{P}}\right|_{\mathcal{G}} $$

My attempt:

I am not sure if this is the way I am suppose to answer the question:

By definition, we can take the expectation of $\left.\frac{d \mathbb{R}}{d \mathbb{P}}\right|_{\mathcal{G}}$ which gives:

$$E_{p}[\frac{d \mathbb{R}}{d \mathbb{P}}\mid G_{t}]$$

$$=E_{p}\left[\frac{d \mathbb{R}}{d Q} \cdot \frac{d Q}{d \mathbb{R}} \mid G_{t}\right]$$

Using Bayes theorem we get:

$$E_{p}\left[\frac{d \mathbb{R}}{d Q} \cdot \frac{d Q}{d \mathbb{R}} \mid G_{t}\right] = E_{Q}\left[\frac{d R}{d Q} \mid G_{t}\right] \cdot E_{P}\left[\frac{d Q}{d \mathbb{P}} \mid G_{t}\right]$$

Hence we get the measure change: $$\left.\left.\frac{d \mathbb{R}}{d \mathbb{Q}}\right|_{\mathcal{G}} \frac{d \mathbb{Q}}{d \mathbb{P}}\right|_{\mathcal{G}}$$

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