22

I need to prove that if $E \int_a^b |X_u|\,du = \int_a^b E|X_u|\,du$ is finite then:

$$E\left[\left.\int_a^b X_u\,du \;\right|\; \mathcal{G}\right] = \int_a^b E[X_u \mid \mathcal{G}]\,du.$$

I just dont have any idea how to approach this problem.

saz
  • 124,107
luka5z
  • 6,601
  • 7
  • 53
  • 103

1 Answers1

24

This is a direct consequence of Fubini's theorem:

Set $X := \int_a^b X_r \, dr$ and $Y_r:=\mathbb{E}(X_r \mid \mathcal{G})$. Then $\int_G Y_r \, d\mathbb{P}= \int_G X_r \, d\mathbb{P}$ for all $G \in \mathcal{G}$; thus $$\int_a^b \int_G Y_r \, d\mathbb{P} \, dr = \int_a^b \int_G X_r \, d\mathbb{P} \, dr. \tag{1}$$ By Fubini's theorem, $$\int_G \left(\int_a^b Y_r \, dr\right) \, d\mathbb{P} \stackrel{\text{Fub.},(1)}{=} \int_G \left(\int_a^b X_r \, dr\right) \, d\mathbb{P}=\int_G X \, d\mathbb{P} \tag{2}$$ for all $G \in \mathcal{G}$. Therefore $$\mathbb{E} \left( \int_a^b X_r \, dr \mid \mathcal{G} \right) \stackrel{\text{def}}{=} \mathbb{E}(X \mid \mathcal{G}) \stackrel{(2)}{=} \int_a^b Y_r \, dr \stackrel{\text{def}}{=} \int_a^b \mathbb{E}(X_r \mid \mathcal{G}) \, dr.$$

Reference: Measures, Integrals and Martingales by René Schilling, 2nd edition, Theorem 27.17.

saz
  • 124,107
  • 1
    Sorry to ask this 10 years later...how do we know that $Y(\omega, r)$ is product measurable, to apply Fubini in the first place? – qp212223 Apr 21 '25 at 22:14
  • @qp212223: although you most likely have solved this by now, you can do this using a monotone class argument. You can easily show this for $X(\omega, r)$ equal to an indicator of $A \times B$ where $A \in \mathcal{A}$ is in the $\sigma$-algebra of the probability space and $B \in \mathcal{B}([a,b])$. Then you can show that the sets for which this hold form a monotone class of functions and hence the property holds for all measurable $X(\omega, r)$. More details are described in the reference mentioned in the answer. – Finn Heijink Dec 20 '25 at 23:45