# Absolute continuity and conditioning

I'm trying to solve the following problem, which appears in a course about HMM's (problem 6.2).

Let $P,Q$ be probability measures on $(\Omega,\mathcal{G})$ such that $P$ is absolutely continuous with respect to $Q$ and let $\mathcal{G}'\subset\mathcal{G}$ a sub-$\sigma$-field. Show that $$\frac{dP_{\mid\mathcal{G}'}}{dQ_{\mid\mathcal{G}'}}=E_Q\left[\frac{dP}{dQ}\mid\mathcal{G}'\right]$$

I'm not even sure I understand the question.

1) Is $P_{\mid\mathcal{G}'}$ defined by $P_{\mid\mathcal{G}'}(A)=P[A\mid\mathcal{G}']=E_P[1_A\mid\mathcal{G}']$ (and idem for $Q$)?

2) Can we say that $$E_Q\left[\frac{dP}{dQ}\mid\mathcal{G}'\right]=\int_{\Omega}\frac{dP}{dQ}dQ_{\mid\mathcal{G}'}$$ (using the fact that conditional expectation is the expectation under the conditional ditribution)?

3) Using the definition of conditional expectation, I should prove that for any $A\in\mathcal{G}'$, we have $$E_Q\left[\frac{dP}{dQ}1_A\right]=E\left[\frac{dP_{\mid\mathcal{G}'}}{dQ_{\mid\mathcal{G}'}}1_A\right]$$ But under what measure is the expectation on the RHS? And how do I know that $\frac{dP_{\mid\mathcal{G}'}}{dQ_{\mid\mathcal{G}'}}$ even exists? I should show that first right?

Anyway, it seems that I'm a bit lost on this very simple measure theory problem, so I'd appreciate any help.

• I suppose that you refer to $\frac{d P_{|\mathcal{G}'}}{d Q_{|\mathcal{G}'}}$ instead of $\frac{d P_{|\mathcal{G}'}}{d P_{|\mathcal{G}'}}$. – user178826 Oct 18 '16 at 17:26
• Yes of course. I corrected, thanks. – Augustin Oct 18 '16 at 17:49
• $P_{|\mathcal G'}$ is the restriction of $P$ to $\mathcal G'$. – John Dawkins Oct 18 '16 at 19:11
• Isn't it the conditional probability? – Augustin Oct 18 '16 at 19:34
• Because this property is used page 84 for the EM algorithm, and it seems to me that in this context, it is a conditional probability, given the observations. – Augustin Oct 18 '16 at 19:43

## 1 Answer

You are given two probability measures $P,Q$ on the measurable space $\left(\Omega,\mathcal{G}\right)$ s.t. $P\ll Q$ and a $\sigma$-algebra $\mathcal{G'}$ s.t. $\mathcal{G'}\subset\mathcal{G}$.

$P_{|\mathcal{G'}}$ is simply the restriction of $P$ to $\mathcal{G'}$, that is $$P_{|\mathcal{G'}}:\mathcal{G'}\rightarrow [0,1]$$ $$A\mapsto P(A)$$

According to the Radon-Nikodým-theorem, the density of $P$ with respect to $Q$ exists, is unique and denoted by $\frac{dP}{dQ}$. Now, how can one compute the density $\frac{dP_{|\mathcal{G'}}}{dQ_{|\mathcal{G'}}}=\frac{dP}{dQ}_{|\mathcal{G'}}$?

We have $\frac{dP}{dQ}_{|\mathcal{G'}}=E_Q[\varphi|\mathcal{G'}]$, where $\varphi$ is the density of $P$ with respect to $Q$: Let $A\in\mathcal{G'}$. Then $Q(A)=\int_A \varphi dQ=\int_A E[\varphi|\mathcal{G'}] dQ$.

It follows that $E_Q[\frac{dP}{dQ}_{|\mathcal{G'}} 1_A]=E_Q[E_Q[\varphi|\mathcal{G'}]1_A]=E_Q[\varphi 1_A]$