# 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.