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?


1 Answer 1


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, 2019 at 14:01
  • 2
    $\begingroup$ 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 $Y^{-1}(A) \in \mathcal F \subset \mathcal G$ $\endgroup$
    – Jay Zha
    Feb 2, 2019 at 3:36

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