Say I toss two distinct coins.

Let A be event: there are two heads, $\{HH\}$

Let $\sigma$ be a sigma algebra $\big\{\emptyset,\Omega,\{HH,HT\},\{TT,TH\}\big\}$

How does one understand $P(A|\sigma)$, $E(A|\sigma)$? How does the meaning differ if sigma algebra is chosen differently, say trivial, or full algebra?


The following are the definitions of conditional expectation and conditional probability, could you fit your case in?

Definition 1: If $\mathcal F \subseteq \mathcal G$ are two $\sigma$-fields, and $X$ a $\mathcal G$-measurable integrable random variable, then $\mathbb E[X | \mathcal F]$ is defined as any $\mathcal F$-measurable random variable $Y$, such that $\mathbb E[Y;A]=\mathbb E[X;A]$ for every $A \in \mathcal F$. Here $\mathbb E[X;A]$ is a notation for $\int_AX\,d\mathbb P$.

Definition 2: We define conditional probability as $\mathbb P(A | \mathcal F)= \mathbb E[1_A|\mathcal F]$.

From above definition, such r.v. of $Y$ is guaranteed to exist, and is unique up to a.s. equivalence - this is guaranteed by one version of Radon-Nikodym Theorem (i.e. for finite positive $\mathbb P$, and finite signed $\mathbb Q(A) = \mathbb E[X;A] \ll \mathbb P$, the conditional expectation as a Radon-Nikodym Derivative exits and is unique up to a.s.)

  • $\begingroup$ When the random variable $X$ is a mapping from the probability space $ (\Omega, \mathcal{G}, \mu) $ to the measurable space $ ( \Omega', \mathcal{G}')$ the random variable is $Y$ understood as a mapping between the same spaces? $\endgroup$ – Hamilcar Feb 1 at 14:01
  • 1
    $\begingroup$ Usually the definition of r.v. maps from $\Omega$ to $\mathbb R$. If r.v. $X$ is $\mathcal G$-measurable, then for any set $A \in \mathcal B(\mathbb R)$, $X^{-1}(A) \in \mathcal G$. And according to Definition 1, $Y$ should be another r.v. maps from $\Omega$ to $\mathbb R$, s.t. $\forall A \in \mathcal B(\mathbb R)$, we have $Y^{-1}(A) \in \mathcal F \subset \mathcal G$ $\endgroup$ – Yujie Zha Feb 2 at 3:36

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