# Conditional expectation of integral of Ornstein-Uhlenbeck process

Given that $$X(t)$$ is an Ornstein-Uhlenbeck process with $$X(0) = x_0$$, which is a Markov process, but not a Martingale, how could I go forward if I would like to calculate

$$E[\int_0^T X(s)ds | \mathcal{F}_t]$$?

I have been twisting my brain for hours, but can't seem to find any reasonable approach.

• What is $\mathcal{F}_t$? – d.k.o. Mar 12 at 21:02

Let $$(\Omega,\mathcal{F},\lbrace{\mathcal{F_t}\rbrace}_{t\in\mathbb{R}_+},P)$$ be a filtered probability space where we defined a $$\lbrace{\mathcal{F_t}\rbrace}-$$brownian motion $$\lbrace{W_t\rbrace}_{t\in\mathbb{R}_+}$$ starting from 0. We suppose that the process $$X$$ satisfies the following SDE: $$$$dX_t = \theta(\mu-X_t)dt + \sigma dW_t$$$$ where $$\sigma >0, \theta >0, \mu \in \mathbb{R}$$ and $$X_0 \in L^2$$ and independant to $$W$$. We can show that it exists a unique strong solution to this SDE and is given by: $$\begin{equation*} \forall t\in \mathbb{R}_+, \quad X_t = e^{-\theta t}X_0 + \mu(1+e^{-\theta t}) +\sigma\int_0^te^{-\theta(t-s)}dW_t \end{equation*}$$ The problem was to compute: for $$T>t$$, $$E\left[\int_0^TX_sds|\mathcal{F_t}\right]$$ \begin{align*} E\left[\int_0^TX_sds|\mathcal{F_t}\right] &= E\left[\int_0^T\left(e^{-\theta s}X_0 + \mu(1+e^{-\theta s}) +\sigma\int_0^se^{-\theta(s-u)}dW_u\right)ds|\mathcal{F}_t\right] \\ &=\int_0^T\left(e^{-\theta s}X_0 + \mu(1+e^{-\theta s})\right)ds + \sigma E\left[\int_0^T\int_0^se^{-\theta(s-u)}dW_uds|\mathcal{F}_t\right] \quad \\ &= \int_0^T\left(e^{-\theta s}X_0 + \mu(1+e^{-\theta s})\right)ds + \sigma E\left[\int_0^T\int_u^Te^{-\theta(s-u)}dsdW_u|\mathcal{F}_t\right] \text{Fubini stochastic} \\ &= \int_0^T\left(e^{-\theta s}X_0 + \mu(1+e^{-\theta s})\right)ds + \sigma E\left[\int_0^T\phi(u)dW_u|\mathcal{F}_t\right] \text{where} \ \phi(u) := \int_u^Te^{-\theta(s-u)}ds \\ &= \int_0^T\left(e^{-\theta s}X_0 + \mu(1+e^{-\theta s})\right)ds + \sigma \int_0^t\phi(u)dW_u \quad \text{a.s.} \end{align*} The last equality holds because the function $$\phi$$ is $$L^2(\mathbb{R}_+, dt)$$. Thus the stochastic integral is a Wiener integral (which is a martingale).