Is $(\int_0^t W_s ds, W_t)$ Markov?

Question

Approximating $I_t = \int_0^t W_s ds$ by Riemann sums I have convinced myself that it is not Markov, but I have been met by the claim that $(I,W)$ is and I cannot figure out why. Do you guys have any ideas?

Answer 1

Let $f: \mathbb{R}^2 \to \mathbb{R}$ be a bounded continuous function. In order to show that $$M_t := (I_t,W_t) := \left( \int_0^t W_s \, ds, W_t \right)$$ is Markov, we have to show that there exists $g: \mathbb{R}^2 \to \mathbb{R}$ such that

$$\mathbb{E}^x(f(M_t) \mid \mathcal{F}_s) = g(M_s)$$

for any $s \leq t$. To this end, write

$$\begin{align*} f(M_t) &= f \left( \int_s^t W_r \, dr + \int_0^s W_r \, dr, (W_t-W_s)+W_s \right) \\ &= f \left( \int_s^t (W_r-W_s) \, dr + (t-s) W_s + I_s, (W_t-W_s)+W_s \right). \end{align*}$$

Now use that $(W_r-W_s)_{r \geq s}$ and $(W_t-W_s)$ are independent from $\mathcal{F}_s$ in order to calculate the conditional expectation.