Conditional probability is defined in terms of conditional expectation as $$ P(A \mid \mathcal{N}) := E(I_A \mid \mathcal{N}) $$ where $(\Omega, \mathcal{F}, P)$ is a probability space, $\mathcal{N}$ is a sub $\sigma$-algebra of $\mathcal{F}$, and $A \in \mathcal{F}$.

Conversely, can conditional expectation be represented in terms of conditional probability?

Following is one way I think might be possible, which is inspired by the case conditional on a discrete random variable, where conditional expectation can be represented as expectation with respect to corresponding conditional probability.

If the conditional probability is regular, i.e. $\forall \omega \in \Omega, P(\cdot \mid \mathcal{N})(\omega)$ is a probability measure on $(\Omega, \mathcal{F})$, define $Y: \Omega \rightarrow \mathbb{R}$ for any random variable $X$ as $$ Y(\omega):=\int_\Omega X \, dP(\cdot \mid \mathcal{N})(\omega). $$ Is $Y$ a version of the conditional expectation $E(X \mid \mathcal{N})$, i.e. $$ Y = E(X \mid \mathcal{N}) \text{ a.e. }? $$

Thanks and regards!


1 Answer 1


the answer for your question is yes. Note that, all you need to prove is the following equality $$ \int_{E} X\ dP = \int_{E} \left[\int_{\Omega} X\ dP(\cdot|\mathcal{N})(\omega) \right] dP|_{\mathcal{N}} \qquad \qquad (1). $$ for all $E\in\mathcal{N}$.

The simplest way to do this is start from characteristic functions and then use linearity and finally the powerfull approximation theorems of the measure theory.

Let us start with the case $X=1_{A}$ where $A\in \mathcal{F}$. In this case the above equality holds as long as $$ P(A\cap E)=\int_{\Omega} \left[\int_{\Omega} 1_{A\cap E}\ dP(\cdot|\mathcal{N})(\omega) \right] dP|_{\mathcal{N}}. $$ But this is certainly true and we can prove this manipulating the rhs. Indeed $$ \int_{\Omega} \left[\int_{\Omega} 1_{A\cap E}\ dP(\cdot|\mathcal{N})(\omega) \right] dP|_{\mathcal{N}} = \int_{\Omega} \mathbb{E}(A\cap E|\mathcal{N}) dP|_{\mathcal{N}} = \int_{E} \mathbb{E}(A|\mathcal{N}) dP|_{\mathcal{N}}, $$ where we used in the last equality that $E$ is $\mathcal{N}$ measurable. But the right hand side above is by the properties of conditional expectation equals to $P(A\cap E)$. So the identity (1) holds for any characteristic function and for all $E\in\mathcal{N}$.

By linearity of the integral (1) is certainly true if $X$ is any simple function. Let $X$ be any positive $\mathcal{F}-$measurable function. We know that there is a sequence of simple functions $\varphi_n \uparrow X$, here the up arrow means monotone convergence. Since at this point is clear for us that $$ \begin{eqnarray} \int_{E} \varphi_n\ dP &=& \int_{E} \left[\int_{\Omega} \varphi_n\ dP(\cdot|\mathcal{N})(\omega) \right] dP|_{\mathcal{N}} \\ &=& \int_{E} \mathbb{E}(\varphi_n|\mathcal{N}) dP|_{\mathcal{N}}, \end{eqnarray} $$ we can finish the proof by invoking the Monotone Convergence Theorem.

Edition. I corrected the error pointed out by Didier Piau.

Addition. Theorem. Let $(\Omega,\mathcal{F},\mu)$ be a $\sigma$-finite measure space, $\mathcal{A}$ a sub-$\sigma$-algebra of $\mathcal{B}$, and $\nu=\mu|_{\mathcal{A}}$. If $f\in L^1(\mu)$, there exist $g\in L^1(\nu)$ (thus $g$ is $\mathcal{A}$-measurable) such that $$ \int_{E} f\ d\mu =\int_{E} g\ d\nu $$ for all $E\in A$; if $g'$ is another such function then $g=g'$ $\nu$-a.e. (Note that $g$ is the conditional expectation of $f$ on $\mathcal{A}$. (Folland)

  • $\begingroup$ Thanks! I was confused about $(\omega)$ in $dP(\cdot|\mathcal{N})(\omega)$ and $dP|_{\mathcal{N}}(\omega)$, but now I understand it means different in the two. $\endgroup$
    – Tim
    Nov 2, 2011 at 0:04
  • $\begingroup$ @Tim, you are welcome. You absolutely right, I made terrible choice for the name of the dummy variable, sorry about that. $\endgroup$
    – Leandro
    Nov 2, 2011 at 0:25
  • $\begingroup$ None of your fault. After all it is me who started using $P(\cdot|\mathcal{N})(\omega)$, and $\int \ldots dP(\omega)$ is one of standards used in integral (I often just write $\int \ldots dP$). $\endgroup$
    – Tim
    Nov 2, 2011 at 0:32
  • $\begingroup$ Hope you don't mind I change the notations. $\endgroup$
    – Tim
    Nov 2, 2011 at 0:37
  • $\begingroup$ @Tim Good idea. $\endgroup$
    – Leandro
    Nov 2, 2011 at 1:43

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