3

I am trying to compute the following question:

Let $(X_t,F_t)_{t \in \mathbb{R}}$ be a martingale with continuous realizations. For $0 \le s \le t$ find $E(\int_{0}^{t} X_u du | F_s).$

I am confused how to compute the conditional expectation with the integral inside.

user75514
  • 1,425
  • I think it should be computed based on: $\mathbb{E}\left(\int_0^tX_u\mathrm{d}u|\mathcal{F}_s\right)=\mathbb{E}\left(\int_s^t(X_u-X_s)\mathrm{d}u|\mathcal{F}_s\right)+(t-s)X_s+\int_0^sX_s\mathrm{d}s$ – meowmeow Mar 28 '17 at 01:57
  • Duplicate of https://math.stackexchange.com/q/1720385 – xFioraMstr18 Mar 20 '20 at 22:41

1 Answers1

1

First show that $E\int_0^t |X_u|du < \infty$, (see Exchange integral and conditional expectation for why). This is possible by Fubini and using that $E[|X_u|] \leq E[|X_t|]$ since $u \leq t$.

Then we have $$E[\int_0^t X_u\,du| F_s] = \int_0^t E[X_u|F_s]\,du = \int_0^s X_u\,du + X_s(t-s).$$

nullUser
  • 28,703