I have $E(X) < \infty$. Under which conditions follows that $E(X|A)<\infty$ ?

(A is an event of the form {$Y=y$} if it should matter)

If I can use the formula $E(X|A)=\frac{E(X 1_A)}{P(A)}$ ($1_A$ the indicator function of the set A) then it would follow at least for $P(A)\neq0$, but I am not sure if this formula is applyable as I am working with the general definition (A random variable Y is called conditional expectation value and we write $Y=E(X|\mathcal{A})$ if Y is $\mathcal{A}$-measurable and for all $A \in \mathcal{A}$: $E(X1_A)=E(Y1_A)$.

How can I assert that $E(X|A) <\infty$? (Or if my way seems valid, how can I deduce the formula from the general definition?)


1 Answer 1


(Since your $A$ and $\mathcal A$ are suspiciously close (and I wonder whether you do not use one for the other at least once in your question), let me use $G$ to denote the sigma-algebra.)

You recall the definition of $Y=E(X|G)$. An important point is that this random variable $Y$ is defined only if $X$ is integrable (that is, when $E(|X|)$ is finite), and that $Y$ is always integrable (otherwise $E(Y1_A)$ might not be defined for some $A$ in $G$).

Hence yes, $E(X|G)$ (which is a random variable) is (almost surely) finite, since it is integrable by definition.

For a reference, see the little blue book by David Williams, more commonly called Probability with martingales.

  • $\begingroup$ okay :) thanks a lot, now it's clear. $\endgroup$
    – Johannes L
    Aug 2, 2011 at 10:28

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