Baye’s rule

Let $\mu$ and $\nu$ be two probability measures on a measurable space $(\Omega,\mathcal{G})$ such that

$d\nu(\omega)=f(\omega)d\mu(\omega)$

for some $f\in L^{1}(\mu)$ . Let $X$ be a random variable on $(\Omega,\mathcal{G})$ such that

$E_{\nu}[|X|]=\int_{\Omega}|X(\omega)|f(\omega)d\mu(\omega)<\infty$

( $\nu$ -integrable)

Let $\mathcal{H}$ be a $\sigma$ -algebra, $\mathcal{H}\subset\mathcal{G}$ . Then,

$E_{\nu}[X|\mathcal{H}].E_{\mu}[f|\mathcal{H}]=E_{\mu}[fX|\mathcal{H}]\quad a.s.$

$\displaystyle E_{\nu}[X|\mathcal{H}]=\frac{E_{\mu}[fX|\mathcal{H}]}{E_{\mu}[f|% \mathcal{H}]}\quad a.s.$