Conditional Expectation under Change of Measure

Theorem

Let \mathbb{P} and \mathbb{Q} be measures on (\Omega, {\mathcal{F}}) with \mathbb{Q} \ll \mathbb{P}, {\mathcal{G}} \subset {\mathcal{F}} a sub \sigma-algebra and X an integrable random variable (\mathbb{P}\lvert{X}\rvert < \infty) then

\displaystyle   \mathbb{P}(X\vert {\mathcal{G}}) =  \frac  {\mathbb{Q}\bigg(X\frac{\mathrm{d}\mathbb{P}}{\mathrm{d}\mathbb{Q}}\bigg\vert {\mathcal{G}}\bigg)}  {\mathbb{Q}\bigg(\frac{\mathrm{d}\mathbb{P}}{\mathrm{d}\mathbb{Q}}\bigg\vert {\mathcal{G}}\bigg)}

Proof

\displaystyle   \begin{aligned}  \mathbb{Q}\bigg(\mathbb{I}_A \mathbb{Q}\bigg(X \frac{\mathrm{d}\mathbb{P}}{\mathrm{d}\mathbb{P}}\bigg\vert {\mathcal{G}}\bigg)\bigg)  &=  \mathbb{Q}\bigg(\mathbb{I}_A X \frac{\mathrm{d}\mathbb{P}}{\mathrm{d}\mathbb{P}}\bigg) \\  &=  \mathbb{P}\big(\mathbb{I}_A X \big) \\  &=  \mathbb{P}\big(\mathbb{I}_A \mathbb{P}(X \vert {\mathcal{G}})\big) \\  &=  \mathbb{Q}\bigg(\mathbb{I}_A \frac{\mathrm{d}\mathbb{P}}{\mathrm{d}\mathbb{Q}}\mathbb{P}(X \vert {\mathcal{G}})\bigg) \\  &=  \mathbb{Q}\bigg(\mathbb{I}_A \mathbb{Q}\bigg(\frac{\mathrm{d}\mathbb{P}}{\mathrm{d}\mathbb{Q}}\bigg\vert {\mathcal{G}}\bigg)\mathbb{P}(X \vert {\mathcal{G}})\bigg) \\  \end{aligned}

Thus

\displaystyle   \mathbb{Q}\bigg(\mathbb{I}_A\mathbb{P}(X\vert {\mathcal{G}}){\mathbb{Q}\bigg(\frac{\mathrm{d}\mathbb{P}}{\mathrm{d}\mathbb{Q}}\bigg\vert {\mathcal{G}}\bigg)}\bigg) =  \mathbb{Q}\bigg(\mathbb{I}_A \mathbb{Q}\bigg(X\frac{\mathrm{d}\mathbb{P}}{\mathrm{d}\mathbb{Q}}\bigg\vert {\mathcal{G}}\bigg)\bigg)\quad \mathrm{for\,all}\, A \in {\mathcal{G}}

Hence

\displaystyle   \mathbb{P}(X\vert {\mathcal{G}}){\mathbb{Q}\bigg(\frac{\mathrm{d}\mathbb{P}}{\mathrm{d}\mathbb{Q}}\bigg\vert {\mathcal{G}}\bigg)} =  \mathbb{Q}\bigg(X\frac{\mathrm{d}\mathbb{P}}{\mathrm{d}\mathbb{Q}}\bigg\vert {\mathcal{G}}\bigg)\quad {\mathbb{Q}-\mathrm{a.s.}}

We note that

\displaystyle   A = \bigg\{\omega \,\bigg\vert\, \mathbb{Q}\bigg(\frac{\mathrm{d}\mathbb{P}}{\mathrm{d}\mathbb{Q}}\bigg)\bigg\}

is {\mathcal{G}}-measurable (it is the result of a projection) and that

\displaystyle   0  =  \mathbb{Q}\bigg(\mathbb{I}_A\mathbb{Q}\bigg(  \frac{\mathrm{d}\mathbb{P}}{\mathrm{d}\mathbb{Q}}  \bigg\vert {\mathcal{G}}\bigg)\bigg)  =  \mathbb{Q}\bigg(\mathbb{I}_A  \frac{\mathrm{d}\mathbb{P}}{\mathrm{d}\mathbb{Q}}  \bigg)  =  \mathbb{P}(\mathbb{I}_A)

Hence

\displaystyle   \mathbb{P}(X\vert {\mathcal{G}}) =  \frac  {\mathbb{Q}\bigg(X\frac{\mathrm{d}\mathbb{P}}{\mathrm{d}\mathbb{Q}}\bigg\vert {\mathcal{G}}\bigg)}  {\mathbb{Q}\bigg(\frac{\mathrm{d}\mathbb{P}}{\mathrm{d}\mathbb{Q}}\bigg\vert {\mathcal{G}}\bigg)}\quad {\mathbb{P}-\mathrm{a.s.}}

as required.

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One thought on “Conditional Expectation under Change of Measure

  1. Pingback: Girsanov’s Theorem | Maths, Stats & Functional Programming

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