Conditional probability given a $\sigma$-algebra

I've been studying some statistics by the ways of measure theory, and came up with a problem in understanding conditional probability. The book gives the following definitions:

Let $$(\Omega,\mathcal{F},\mathbb{P})$$ be a probability space, $$\mathcal{G} \subset \mathcal{F}$$ be a $$\sigma$$-algebra and $$A \in \mathcal{F}$$. Then, we define a probability measure on $$\mathcal{G}$$ by the law $$\nu(G) = \mathbb{P}(A \cap G).$$ If $$\mathbb{P}(G) = 0$$, then $$\nu(G) = \mathbb{P}(G \cap A) \leq \mathbb{P}(G) = 0,$$ hence $$\nu \ll \mathbb{P}$$. We define $$\mathbb{P}(A \mid \mathcal{G})$$ as $$d\nu/d\mathbb{P}$$. This way, the conditional probability is kind of well-defined, since Radon-Nikodym derivatives are equal almost everywhere.

My problem here is: I don't understand what is the intuition behind this definition. For some help in understanding my problem, there is an important result

The function $$\mathbb{P}(A \mid \mathcal{G})$$ has the following properties almost everywhere:

• $$0 \leq \mathbb{P}(A \mid \mathcal{G}) \leq 1$$;
• $$\mathbb{P}(\emptyset \mid \mathcal{G}) = 0$$;
• If $$\{A_n\}_{n = 1}^\infty$$ is a sequence of two-by-two disjoint sets in $$\mathcal{F}$$, then $$\mathbb{P}\left(\bigcup_{n=1}^\infty A_n \mid \mathcal{G}\right) = \sum_{n = 1}^\infty \mathbb{P}(A_n \mid \mathcal{G}).$$

This lemma says that, for almost every point, $$\mathbb{P}(A \mid \mathcal{G})$$ is a probability measure in $$\mathcal{F}$$ (as a function of $$A$$). Now, I don't have a single clue about what a "field" of probability measures mean and how it is related in any way with concepts of elementary probability. Even more, is there any "real-world" applications of this concept? With real-world applications I mean like some kind of elementary statistics problem that can be solved using this, some applicable situation that might appear in a non-math context.

• @jerkajerka thanks! Commented Dec 27, 2022 at 14:11

A key scenario is when we have a random variable $$X:\Omega\rightarrow\mathbb{R}$$ and we have $$\mathcal{G}=\sigma(X)$$, the sigma algebra generated by the random variable $$X$$, so that: $$P[A|\mathcal{G}]=P[A|X] \quad \forall A \in \mathcal{F}$$

Example 1: Suppose we play one of two teams equally likely. Let $$X \in \{1,2\}$$ be the random variable specifying which team we play. Suppose:

• Given we play team 1: We win with probability 1/3
• Given we play team 2: We win with probability 1/4

In this example we have $$\sigma(X) = \{\phi, \Omega, \{X=1\}, \{X=2\}\}$$. Let $$Win$$ be the event that we win. Then $$P[Win|X] = \left\{\begin{array}{cc} 1/3 & \mbox{ if X=1} \\ 1/4 & \mbox{ if X=2} \end{array}\right.$$ Also, for all $$A \in \mathcal{F}$$ we have $$P[A|X] = \left\{\begin{array}{cc} P[A|X=1] & \mbox{ if X=1} \\ P[A|X=2] & \mbox{ if X=2} \end{array}\right.$$ So you can see we select one of two probability measures depending on the value of $$X$$. Since $$X$$ is really $$X(\omega)$$ (a function of $$\omega \in \Omega$$) you can view $$P[A|X]$$ as a function of $$\omega \in \Omega$$. It is a $$\sigma(X)$$-measurable function from $$\Omega$$ to $$\mathbb{R}$$.

Example 2: Suppose $$X:\Omega\rightarrow\mathbb{R}$$ takes values in a finite or countably infinite set $$S_X$$ and $$P[X=x]>0$$ for all $$x \in S_X$$. Then $$\sigma(X) = \sigma(\{\omega\in \Omega : X(\omega)=x\}_{x\in S_X})$$ and $$\boxed{P[A|X] = \sum_{x\in S_X}P[A|X=x]1_{\{X=x\}} \quad \forall A \in \mathcal{F}}$$ where $$1_{\{X=x\}}$$ is an indicator function that is 1 if $$X=x$$ and 0 else. So which probability measure we use depends on the value of $$X(\omega)$$. If $$\omega_1 \in \Omega$$ satisfies $$X(\omega_1)=x_1$$ then $$P[A|X](\omega_1)=P[A|X=x_1] \quad \forall A \in \mathcal{F}$$ You can show that this definition of $$P[A|X]$$ satisfies the requirements of a conditional expectation of $$1_A$$ given $$\sigma(X)$$: It is a $$\sigma(X)$$-measurable function from $$\Omega\rightarrow\mathbb{R}$$ (meaning it is a pure function of $$X(\omega)$$); $$E[P[A|X]1_B]=E[1_{A\cap B}]$$ for all $$B \in \sigma(X)$$.

• Man, this is INSANELY cool, thanks a lot for the answer, now I can finally grasp what is happening here. Commented Dec 28, 2022 at 8:00

For the purpose of intuition, I suggest you peak at a much more basic textbook introduction like Hogg, McKean and Craig's "Introduction to Mathematical Statistics" in which conditional probabilities are introduced as "revised" probabilities after being granted some "information" (a reduced sample space). That book will immediately give you down-to-earth applications from playing cards, lotteries, etc.

The "important result" you highlight shows that the measure theory definition of a conditional probability measure via Radon-Nikodym actually delivers something with the properties of a probability measure. Hence, it justifies the name.