1
$\begingroup$

Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space, $\mathcal E\subseteq\mathcal F$ a sub-$\sigma$-algebra. Assume that $$\mathbb P=\sum_{i=1}^\infty a_i\delta_{\omega_i}$$ where $a_i\in (0,1),$ $\sum_{i=1}^\infty a_i=1$ and $\delta_{\omega_i}$ denotes the point measure at $\omega_i\in\Omega.$ Then, the (unconditional) expectation can be written as $\mathbb E_\mathbb P[X]=\sum_{i=1} a_i X(\omega_i).$ Is there a expression of the conditional expectation $\mathbb E_\mathbb P[X\mid\mathcal E]$ in terms of the $a_i$'s?

$\endgroup$
1
$\begingroup$

Consider the equivalence relation $R$ on $S=\{\omega_i\mid i\in\mathbb N\}$ defined by $\omega_iR\omega_j$ if there exists no $A$ in $\mathcal E$ such that $\omega_i\in A$ and $\omega_j\notin A$. Then $R$ defines a partition $S/R$ of $S$ into equivalence classes.

For every equivalence class $C$ in $S/R$, choose some $\omega_i$ in $C$. For every $\omega_j$ not in $C$, there exists some $A_j(C)$ in $\mathcal E$ such that $\omega_i$ is in $A_j(C)$ but not $\omega_j$. Let $A(C)$ denote the intersection of the sets $A_C(j)$ over every $j$ such that $\omega_j$ is not in $C$. For every equivalence class $C$ in $S/R$, $A(C)$ is in $\mathcal E$ and, for every $i$, $\omega_i$ is in $A(C)$ if and only if $\omega_i$ is in $C$. Let $B(C)$ denote $A(C)$ minus the union of $A(C')$ on every equivalence class $C'\ne C$. For every equivalence class $C$ in $S/R$, $B(C)$ is in $\mathcal E$, for every $i$, $\omega_i$ is in $B(C)$ if and only if $\omega_i$ is in $C$, and the events $B(C)$ are disjoint.

Then, $$ E[X\mid\mathcal E]=\sum\limits_{C}P[B(C)]^{-1}E[X;B(C)]\mathbf 1_{B(C)}, $$ where $E[X;B(C)]=\sum\limits_{i\in\mathbb N}a_iX(\omega_i)\mathbf 1_{\omega_i\in C}$ and $P[B(C)]=\sum\limits_{i\in\mathbb N}a_i\mathbf 1_{\omega_i\in C}$ for every equivalence class $C$.

$\endgroup$
  • $\begingroup$ Indeed, there was a measurability issue. See revised version. $\endgroup$ – Did Nov 13 '13 at 14:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.