# Conditional expectation for a random variable with infinite expectation

Let $(\Omega,\mathcal{F},P)$ be a probability space, $X:\Omega\to\mathbb{R^+}$ be a non-negative $\mathcal{F}-$measurable random variable with $E[X]=\infty$ and $\mathcal{H}\subset\mathcal{F}$ be a sub-$\sigma-$algebra. How can we show the existence of $E[X|\mathcal{H}]$?

For a real valued random variable with finite expectation I can understand the proof from wikipedia. It defines a measure $\mu^X:F\in\mathcal{F}\mapsto\int_F XdP$ and restricts it on $\mathcal{H}$. Then use the Radon–Nikodym derivative to show the existence. In case $E[X]=\infty$ the measure $\mu^X$ is not finite anymore. However, if we impose it to be $\sigma-$finite we can still use the Radon–Nikodym derivative.

I do not think it is true in general that such $\mu^X$ is $\sigma-$finite. Is there a way to show existence for such a random variable without this assumption?

• Perhaps you could look at $X_n=\min(n,X)$? Sep 11, 2018 at 23:26
• When you say the existence of E[X|H], does E[X|H]=∞ count as existing? I would have thought no. And if the expectation is infinite then surely the conditional expectation could also be infinite, for some choices of H, in which case the thing you're trying to show isn't true. Sep 11, 2018 at 23:34
• If it helps, you can show in general that $\mu^X$ is $\sigma$-finite. Sep 11, 2018 at 23:39
• @StevenIrrgang I think working on extended Real line, conditional expectation being $\infty$ still counts as existence, similar to Lebesgue integral for $[0,+\infty]$ valued functions. Also, it could be a case where expectation is infinite while its conditional is not, [1] is an example. Grimmett text, referenced there, derives conditional expectation without finite assumption for continuous r.v.s but not using measure theory. [1]: math.stackexchange.com/questions/725323/… Sep 12, 2018 at 4:18
• @MikeEarnest Would you please explain how we can show it is $\sigma-$finite for a generic probability space and $X$? I guess we can use $X_n$ as suggested in the first comment to find the finite measure sequence, right? Sep 12, 2018 at 4:22

To prove $\mu^X$ is $\sigma$-finite, note $\Omega=\bigcup_{n\in \mathbb N}\{|X|\le n\}$, and $\mu^X(|X|\le n)\le n$.