Why is the conditional probability distribution in terms of $\omega$?

174 Views Asked by At

Let $(\Omega, \mathcal{F}, P)$ be a probability space, $\mathcal{G} \subseteq \mathcal{F}$ a $\sigma$-field in $\mathcal{F}$. Given $A \in \mathcal{F}$, the Radon-Nikodym theorem implies that there is ${ }^{[3]}$ a $\mathcal{G}$-measurable random variable $P(A \mid \mathcal{G}): \Omega \rightarrow \mathbb{R}$, called the conditional probability, such that $$ \int_{G} P(A \mid \mathcal{G})(\omega) d P(\omega)=P(A \cap G) $$ for every $G \in \mathcal{G}$, and such a random variable is uniquely defined up to sets of probability zero. A conditional probability is called regular if $\mathrm{P}(\cdot \mid \mathcal{B})(\omega)$ is a probability measure on $(\Omega, \mathcal{F})$ for all $\omega \in \Omega$ a.e. Special cases:

  • For the trivial sigma algebra $\mathcal{G}=\{\emptyset, \Omega\}$, the conditional probability is the constant function $\mathrm{P}(A \mid\{\emptyset, \Omega\})=\mathrm{P}(A)$.
  • If $A \in \mathcal{G}$, then $\mathrm{P}(A \mid \mathcal{G})=1_{A}$, the indicator function (defined below). Let $X: \Omega \rightarrow E$ be a $(E, \mathcal{E})$-valued random variable. For each $B \in \mathcal{E}$, define $$ \mu_{X \mid \mathcal{G}}(B \mid \mathcal{G})=\mathrm{P}\left(X^{-1}(B) \mid \mathcal{G}\right) . $$ For any $\omega \in \Omega$, the function $\mu_{X \mid \mathcal{G}}(\cdot \mid \mathcal{G})(\omega): \mathcal{E} \rightarrow \mathbb{R}$ is called the conditional probability distribution of $X$ given $\mathcal{G}$. If it is a probability measure on $(E, \mathcal{E})$, then it is called regular.

Why does the conditional probability distribution depend on $\epsilon \in \Omega$? I would think that the distribution would simply be a single probability measure over $\mathbb{R}$, given by the pushforward of P(A| \mathcal{G}). It looks like we have a different measure for every $\omega \in \Omega$.

Why is this the case and is it not defined as $$\mu_{X \mid \mathcal{G}}(\cdot \mid \mathcal{G}): \mathcal{E} \rightarrow \mathbb{R}$$

1

There are 1 best solutions below

2
On BEST ANSWER

By definition of $P(\cdot | \cdot)$, the object $\mu_{X|\mathcal{G}}(B | \mathcal{G}) \stackrel{\text{def}}{=} P(X^{-1}(B)|\mathcal{G})$ is an $\mathbb{R}$-valued $\mathcal{G}$-measurable random variable for fixed $B \in \mathcal{E}$ and $\mathcal{G} \subset \mathcal{F}$. In particular, for fixed $\mathcal{G}$, the collection $\{\mu_{X|\mathcal{G}}(B | \mathcal{G})\}_{B \in \mathcal{E}}$ is a collection of random variables, not real numbers whereas if we fix $\omega$ the collection $\{\mu_{X|\mathcal{G}}(B | \mathcal{G})(\omega)\}_{B \in \mathcal{E}}$ is in fact a collection of real numbers.

Thus, we need to fix $\omega$ for $\mu_{X|\mathcal{G}}(\cdot | \mathcal{G})(\omega)$ to be a well-defined map $\mathcal{E} \to \mathbb{R}$