I have not really seen any probability books calculate conditional expectation, except for $\sigma$-algebras generated by a discrete random variable. They simply state the existence of conditional expectation, along with its properties, and leave it at that. I find this a little upsetting and am trying to find a method to compute it. This is what I think it "ought to be".

Let $(\Omega, \mathscr{F},\mu)$ be a probability space with $\mathscr{G}\subseteq \mathscr{F}$ a $\sigma$-algebra. Let $\xi:\Omega\to \mathbb{R}$ be a random variable. Our goal is to compute $E[\xi|\mathscr{G}]$.

Fix $\omega\in \Omega$, we need to compute $E[\xi|\mathscr{G}](\omega)$. Let $A\in \mathscr{G}$ be such $\omega\in A$. Intuition says that $E[\xi|A] = \frac1{\mu(A)}\int_A \xi$ is an approximation to the value of $E[\xi|\mathscr{G}](\omega)$, provided of course that $\mu(A) \not = 0$ which we now assume.

Intuition also says that, if we can find a smaller event $B\subseteq A$, with $\omega\in B$, and $\mu(B) \not = 0$, then $E[\xi|B]$ is a better approximation of $E[\xi|\mathscr{G}](\omega)$ than $E[\xi|A]$.

Hence the optimal such approximation of $E[\xi|\mathscr{G}](\omega)$ should be $E[\xi|M]$ where $M\in \mathscr{G}$, with $\omega\in M$, and with the minimum property. The minimum property here is simply if $A\in \mathscr{G}$ with $\omega\in A$, then $M\subseteq A$.

But there are two issues:

(i) Does such an $M$ even exist? If $\mathscr{G}$ is at most countable this is trivially true. Thus, let us assume that $\mathscr{G}$ is indeed countable.

(ii) What if $\mu(M) = 0$, then $E[\xi|M]$ is undefined! In this case we will assume that we can produce a sequence of events $M_n\in \mathscr{G}$, such that $M_n \downarrow M$ and $\mu(M_n) > 0$.

Intuition says that, $$E[\xi|\mathscr{G}](\omega) = \lim_{n\to \infty} \frac{1}{\mu(M_n)}\int_{M_n}\xi =\lim_{n\to \infty} \frac{1}{\mu(M_n)}\int_{\Omega} \xi.\mathbf{1}_{M_n} $$

As a reality-check, the Monotone Convergence Theorem implies, $$ \int_{\Omega} \xi.\mathbf{1}_{M_n} \to \int_{\Omega} \xi.\mathbf{1}_{M} = \int_{\Omega} 0 = 0 $$ Continuity in measure implies, $$\mu(M_n) \to \mu(M) = 0$$ Thus, our limit is of the indeterminate form "$\frac{0}{0}$", which is what we want.

1) Will this computation correctly compute conditional expectation?

2) What are some assumptions on the probability space for this to hold?

  • 2
    $\begingroup$ Aside: It is a well-known theorem that no sigma algebras are countable, so your (i) needs some revision, as it's basically assuming finiteness of $|\mathcal{G}|$. $\endgroup$
    – cardinal
    Commented Sep 4, 2016 at 4:37
  • $\begingroup$ @cardinal The $\sigma$-algebra generated by a simple random variable will be countable. $\endgroup$ Commented Sep 5, 2016 at 7:19
  • 2
    $\begingroup$ The $\sigma$-algebra of a simple random variable will be finite, which in concert with the result I mentioned above, significantly simplifies your (i). $\endgroup$
    – cardinal
    Commented Sep 5, 2016 at 22:52
  • 1
    $\begingroup$ You should look into the borel paradox $\endgroup$ Commented Mar 24, 2017 at 0:12

1 Answer 1


This does not answer the question but it does provide a sort of "counter-example". Not quite, but it does address a potential issue that can take place when using your intuition to approximate the conditional approximation.

The book by Brezniak, "Basic Stochastic Processes", computes the following conditional expectation exercise via the formal definition. I redid his example using the 'method of approximations' as asked in the original post.

Consider the following example. $\Omega = [0,1]$ with $\mu$ the standard Lebesgue measure.

Define the random variables, $\xi(\omega) = 2\omega^2$ and $\eta(\omega) = 1 - |2\omega - 1|$. We will calculate $E[\xi|\eta]$. Given $\omega\in \Omega$, the conditional expectation $E[\xi|\eta](\omega)$ ought to be equal to $E[\xi|\eta = \eta(\omega)]$. However, the event $(\eta = \eta(\omega))$ is the set $\{ \omega,1-\omega\}$, which is of measure zero, and so $[\xi|\eta = \eta(\omega)]$ is undefined.

So we will approximate the event $A=\{\omega,1-\omega\}$. Choose a small $\varepsilon > 0$, and construct the event $A_{\varepsilon} = [\omega - \varepsilon,\omega + \varepsilon] \cup \{ 1 - \omega \}$. The events $A_{\varepsilon}$ approximate $A$, and approach $A$ in the limit as we shrink the $\varepsilon$. Furthermore, $\mu(A_{\varepsilon}) = 2\varepsilon$.

We calculate, in the limit, $$ E[\xi|A_{\varepsilon}] = \frac{1}{2\varepsilon} \int_{\omega - \varepsilon}^{\omega + \varepsilon} 2t^2 ~ dt \to 2\omega^2 $$ But this is the wrong answer!

However, if we approximate by $B_{\varepsilon} = [\omega - \varepsilon,\omega + \varepsilon] \cup [1- \omega - \varepsilon,1-\omega + \varepsilon] $ then, $$ E[\xi|B_{\varepsilon}] = \frac{1}{4\varepsilon} \left\{ \int_{\omega - \varepsilon}^{\omega + \varepsilon} 2t^2 ~ dt + \int_{1-\omega - \varepsilon}^{1-\omega + \varepsilon} 2t^2 ~ dt \right\} \to \omega^2 + (1-\omega)^2 $$ Which is the right answer!

Why does one approach work and the other does not? Clearly, in the first approximation, the approximating sets $A_{\omega}$ did not belong to the $\sigma$-algebra generated by $\xi$. In the second approximation, the approximating sets $B_{\omega}$ did belong to $\sigma(\xi)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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