1

Given a filtered probability space $(\Omega, \mathcal{F}, \{\mathcal{F}_t\}_{t \in [0, T]}, \mathbb{P})$, and a $\{\mathcal{F}_t\}$-adapted stochastic process $f(t, \omega)$ such that $$ \int_{0}^{T} f(t) dt < \infty. $$ The above integral is a Lebesgue integral.

Let $c$ be a constant with $0 < c < T$. Does the following integral still make sense? $$ \int_{0}^{T-c} \mathbb{E}[f(t + c)\vert \mathcal{F}_t] dt < \infty. $$

David
  • 135
  • 1
    It depends on whether f has a finite expectation. If it does, then yes because every random variable with a finite expectation is bounded. https://math.stackexchange.com/questions/790106/does-finite-expectation-imply-bounded-random-variable – Thomas Kojar Nov 08 '23 at 20:43
  • @ThomasKojar Thank you very much, I think this is the condition I need. – David Nov 08 '23 at 20:52

1 Answers1

1

So as mentioned in the comments, if you can prove that

$E[f(\omega,t)]<\infty$,

then as mentioned here Does finite expectation imply bounded random variable?, we get $|f(\omega,t)|<\infty$ a.s.. Therefore, by Fubini and the tower property for

$$Y=\int E[f(\omega,t+c)\mid\mathcal{F}_{t}]dt$$

we have the condition

$$E[Y]=\int Ef(\omega,t+c)]dt<\infty$$

implies $|Y|<\infty$.

Thomas Kojar
  • 7,349
  • I don't understand how to show the $\int_{0}^{T-c} \mathbb{E}[f(t + c)\vert \mathcal{F}_t] dt < \infty$ according to $X_t < \infty$ a.s. – David Nov 09 '23 at 11:23
  • @David I added more steps. – Thomas Kojar Nov 09 '23 at 16:37
  • 1
    Yes, I agree with you. Since the Fubini theorem is used here, I think there remains one point to be clarified: the conditional expectation $\mathbb{E}[f(t + c) \vert \mathcal{F}_t]$ is still measurable. This might be guaranteed by the projection theorem. Thank you for your answer! – David Nov 09 '23 at 16:59