Let $P$ and $Q$ be probabilities on $(\Omega, \mathcal{A})$, and let $\mathcal{F}$ be a sigma-subalgebra of $\mathcal{A}$. Assume $P \ll Q$. Assume that $P(\cdot \mid \mathcal{F})$ and $Q(\cdot \mid \mathcal{F})$ are regular.

Is it true that $$P\big(\big\{\omega: P(\cdot \mid \mathcal{F})(\omega) \ll Q(\cdot \mid \mathcal{F})(\omega)\big\}\big)=1?$$ Is the set in question even measurable in general?

This seems like a very natural question to me, but I have not been able to find a single reference addressing it.


My first attempt was as follows. Suppose for contradiction that there were some $A \in \mathcal{A}$ such that $$0< P\big(\big\{\omega: 0=Q(A \mid \mathcal{F})(\omega) < P(A \mid \mathcal{F})(\omega)\big\}\big).$$ Since $F=\{\omega: 0=Q(A \mid \mathcal{F})(\omega) < P(A \mid \mathcal{F})(\omega)\} \in \mathcal{F}$, we can conclude that $Q(A \cap F) = 0 < P(A \cap F)$, which contradicts $P \ll Q$.

But that observation doesn't use regularity at all, and, indeed, all we have shown is that for each $A$, there's a $P$-probability $1$ set on which $Q(A \mid \mathcal{F})=0 \implies P(A \mid \mathcal{F})=0$. But I am asking about something stronger.

I should note that the claim I ask about seems plausible to me because it holds for elementary conditional probabilities: If $P(F)>0$ and $P \ll Q$, then $P(\cdot \mid F) \ll Q(\cdot \mid F)$. Using this basic fact, one can show that the claim is true when $\mathcal{F}$ is generated by a countable partition. So all that is left really is to account for general $\mathcal{F}$.


Here is an attempt to prove the claim for the case where $\mathcal{A}$ is generated by a countable algebra $\mathcal{B}$. For arbitrary probability measure $\mu_1$, $\mu_2$ defined on an algebra $\mathcal{G}$, write $\mu_1 \ll_s \mu_2$ iff $$\forall m \in \mathbb{N}, \ \exists r \in \mathbb{N}, \ \forall G \in \mathcal{G}: \mu_2(G) < 1/r \implies \mu_1(G)<1/m.$$

I make use of the following

Claim: If $\mu_1$ and $\mu_2$ are probability measures on $(\Omega, \mathcal{A})$, a measurable space, and $\mathcal{G}$ is an algebra that generates $\mathcal{A}$, then $\mu_1 \ll \mu_2$ if and only if $\mu_1 \ll_s \mu_2$.

So, in our case $$\{\omega: P(\cdot \mid \mathcal{F})(\omega) \ll Q(\cdot \mid \mathcal{F})(\omega)\} = \{\omega: P^{\mathcal{B}}(\cdot \mid \mathcal{F})(\omega) \ll_s Q^{\mathcal{B}}(\cdot \mid \mathcal{F})(\omega)\},\tag{1}$$ where $P^{\mathcal{B}}(\cdot \mid \mathcal{F})(\omega)$ is the restriction of $P(\cdot \mid \mathcal{F})(\omega)$ to $\mathcal{B}$, and similarly for $Q^{\mathcal{B}}(\cdot \mid \mathcal{F})(\omega)$. So it suffices to show that the set on the right-hand side of (1) has $P$-probability $1$.

First, note that \begin{align} \{\omega: P^{\mathcal{B}}(\cdot \mid \mathcal{F})(\omega) \ll_s Q^{\mathcal{B}}(\cdot \mid \mathcal{F})(\omega)\} &= \bigcap_{m \in \mathbb{N}} \bigcup_{r \in \mathbb{N}} \bigcap_{B \in \mathcal{B}} \{Q(B \mid \mathcal{F}) < 1/r \implies P(B \mid \mathcal{F}) < 1/m\}\\ &=:C. \end{align} So, $P(C) < 1$ implies $$\exists m, \ \forall r, \ \exists B \in \mathcal{B}: P\big(Q(B \mid \mathcal{F}) < 1/r \implies P(B \mid \mathcal{F}) < 1/m \big) < 1$$ iff $$\exists m, \ \forall r, \ \exists B \in \mathcal{B}: P\big(\{Q(B \mid \mathcal{F})<1/r\} \cap \{P(B \mid \mathcal{F}) \geq 1/m\} \big)>0.\tag{2}$$ Now, $$P\big(\{Q(B \mid \mathcal{F})<1/r\} \cap \{P(B \mid \mathcal{F}) \geq 1/m\} \big)>0.$$ implies $$Q\big(\{Q(B \mid \mathcal{F})<1/r\} \cap \{P(B \mid \mathcal{F}) \geq 1/m\} \big)>0$$ because $P \ll Q$, in which case $$P(B \cap \{P(B \mid \mathcal{F}) \geq 1/m\}) = \int_{\{P(B \mid \mathcal{F}) \geq 1/m\}} P(B \mid \mathcal{F})dP \geq 1/m$$ and $$Q(B \cap \{Q(B \mid \mathcal{F}) < 1/r\}) = \int_{\{Q(B \mid \mathcal{F}) < 1/r\}} Q(B \mid \mathcal{F})dQ <1/r.$$ But then, by (2), $$\exists m, \ \forall r, \ \exists B \in \mathcal{B}: Q(B \cap \{Q(B \mid \mathcal{F}) < 1/r \ \text{and} \ P(B \cap \{P(B \mid \mathcal{F}) \geq 1/m,$$ which implies $$\exists m, \ \forall r, \ \exists A \in \mathcal{\mathcal{A}}: Q(A) < 1/r \ \text{and} \ P(A) \geq 1/m.$$ This contradicts $P \ll Q$.

  • Your contradiction is odd: $A$ depends on $\omega.$ – Will M. Nov 8 at 22:33
  • Set $E$ the event $P(\cdot \mid \mathcal{F}) \not\ll Q(\cdot \mid \mathcal{F}).$ Then, if $\omega \in E$ there exists a set $A = A(\omega) \in \mathcal{A}$ such that $P(A\mid \mathcal{F}) > 0$ while $Q(A\mid \mathcal{F}) = 0$. – Will M. Nov 8 at 22:39
  • @WillM. Ah, I see. Thanks. – aduh Nov 8 at 22:42

Here is another approach. Let $\varphi$ denote the Radon-Nikodym derivative of $P$ with respect to $Q$. Define $$ K(\omega, A):={ \int_A \varphi(\omega')Q(d\omega'|\mathcal F)(\omega)\over \int_\Omega\varphi(\omega')Q(d\omega'|\mathcal F)(\omega)},\qquad A\in\mathcal A, $$ with the understanding that the ratio vanishes when the denominator is zero. You can check that (i) $(\omega,A)\mapsto K(\omega,A)$ has the properties required of the regular conditional distribution $P(\cdot|\mathcal F)(\omega)$, and (ii) clearly $K(\omega,\cdot)\ll Q(\cdot|\mathcal F)(\omega)$ for a.e. $\omega$.

  • Ah, very good! I figured there should be some simple proof. But let me think about this for a little bit. Thanks. – aduh Nov 10 at 18:50

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