# A problem about conditional probablity

Consider probability space $\left( \Omega\mathcal{,F,}\mathbb{P} \right)$. Given an event $A\mathcal{\in F}$ and a $\sigma$-algebra $\mathcal{G \subseteq F}$, we define the conditional probability given $\sigma$-algebra $\mathcal{G}$, denoted by $\mathbb{P}\left( A|\mathcal{G} \right)$, as a non-negative $\left( \Omega\mathcal{,G} \right) \rightarrow \left( \mathbb{R}\mathcal{,B}\left( \mathbb{R} \right) \right)$ random variable s.t. $\forall G \in \mathcal{G,}\mathbb{P}\left( A \cap G \right) = \int_{G}^{}{\mathbb{P}\left( A \middle| \mathcal{G} \right)d\mathbb{P}}$. Existence and uniqueness is proved below.

For any $A \in \mathcal{F}$ and $\mathcal{G \subseteq F}$, the conditional probability exists and is a.s. unique. Notice $\mathbb{P}_{A}\left( \cdot \right):=\mathbb{P}\left( A\bigcap_{}^{}\left( \cdot \right) \right)$ is a measure for $\left( \Omega\mathcal{,G} \right)$ by checking those axioms, and we can also verify $\mathbb{P}_{A}\mathbb{\ll P}$. Also, $\mathbb{P}$ is also a valid measure for $\left( \Omega\mathcal{,G} \right)$, then by Randon-Nikodym Theorem we have a non-negative $\mathcal{G}$-measurable function $\mathbb{P}\left( A \middle| \mathcal{G} \right) = \frac{d\mathbb{P}_{A}}{d\mathbb{P}}$ exists and is a.s. unique.

We can further show given any $\omega \in \Omega$, then $\mathbb{P}_{\mathcal{G}}^{\left( \omega \right)}\left( \cdot \right):=\mathbb{P}\left( \cdot \middle| \mathcal{G} \right)\left( \omega \right)$ is a.s. a valid measure on $\left( \Omega\mathcal{,F} \right)$.

First, $\mathbb{P}_{\mathcal{G}}^{\left( \omega \right)} \geq 0$ by definition. Secondly, $\mathbb{P}_{\mathcal{G}}^{\left( \omega \right)}\left( \varnothing \right) = 0$ because by definition $0 = \mathbb{P}\left( \varnothing \cap G \right) = \int_{G}^{}{\mathbb{P}\left( \varnothing \middle| \mathcal{G} \right)d\mathbb{P,\forall}G \in \mathcal{G \Rightarrow}\mathbb{P}\left( \varnothing \middle| \mathcal{G} \right) \equiv 0}$ a.s. Thirdly, for any $A_{1},A_{2}\mathcal{\in F,}A_{1}\bigcap A_{2} = \varnothing$, then $A_{1}\bigcap G$ is disjoint form $A_{2}\bigcap G$ and

$$\int_{G}^{}{\mathbb{P}\left( A_{1}\bigcup A_{2} \middle| \mathcal{G} \right)d\mathbb{P}}\mathbb{= P}\left( \left( A_{1}\bigcup A_{2} \right) \cap G \right)\mathbb{= P}\left( \left( A_{1}\bigcap G \right)\bigcup\left( A_{2}\bigcap G \right) \right)\mathbb{= P}\left( A_{1}\bigcap G \right)\mathbb{+ P}\left( A_{2}\bigcap G \right) = \int_{G}^{}{\mathbb{P}\left( A_{1} \middle| \mathcal{G} \right)d\mathbb{P}} + \int_{G}^{}{\mathbb{P}\left( A_{2} \middle| \mathcal{G} \right)d\mathbb{P}} = \int_{G}^{}{\left( \mathbb{P}\left( A_{1} \middle| \mathcal{G} \right)\mathbb{+ P}\left( A_{2} \middle| \mathcal{G} \right) \right)d\mathbb{P}}$$

Then by uniqueness we have a.s. $\mathbb{P}\left( A_{1}\bigcup A_{2} \middle| \mathcal{G} \right) = \mathbb{P}\left( A_{1} \middle| \mathcal{G} \right)\mathbb{+ P}\left( A_{2} \middle| \mathcal{G} \right)$.

I need help with showing the following,

Given any random variable $Y:\left( \Omega\mathcal{,F} \right) \rightarrow \left( S,\mathcal{E} \right)$, let $\sigma(Y)$ be its generated $\sigma$-algebra, then $\mathbb{P}_{\sigma\left( Y \right)}^{\left( \omega_{1} \right)} = \mathbb{P}_{\sigma\left( Y \right)}^{\left( \omega_{2} \right)}$ a.s. if $Y\left( \omega_{1} \right) = Y\left( \omega_{2} \right)$

• The proofs you consider complete are actually lacking tons of "almost surely". For example, $$P(\varnothing\mid\mathcal G)(\omega)=0$$ is not guaranteed for every $\omega$ in $\Omega$ but only almost surely with respect to $P$, in the sense that for every version of the random variable $P(\varnothing\mid\mathcal G)$, one has $$P(P(\varnothing\mid\mathcal G)\ne0)=0$$ Likewise, the statement you are asking about does not hold. – Did Jul 25 '18 at 6:17
• I think there are many wrong sattements here. $P^{(\omega )}_\mathcal G$ is defined only almost everywhere so we cannot say that for fixed $\omega$ this gives a measure. I think you should take a look at Wikipedia entries for 'regular conditional probabilities'. – Kabo Murphy Jul 25 '18 at 6:19
• Unrelated: you might want to replace every \bigcap and \bigcup in your post by \cap and \cup. – Did Jul 25 '18 at 6:19

What you can show is that for any $A\in\mathcal{F}$,
$$\mathbb{P}(A\mid Y)(\omega_1)=\mathbb{P}(A\mid Y)(\omega_2)$$ when $Y(\omega_1)=Y(\omega_2)$.
Since $\mathbb{P}(A\mid Y)=\mathsf{E}[1_A\mid Y]$ and the latter is $\sigma(Y)$-measuralbe, there exists a Borel function $\phi$ such that
$$\mathbb{P}(A\mid Y)(\omega)=\phi(Y(\omega))$$ for all $\omega\in\Omega$.