I want to show the following:
Let $W_t$ be a 1 dimensions brownian motion and $V_t= \int_{0}^{t} W_sds.$ Prove that the pair $(W_t,V_t)$ is a two-dimensional Markov process.
I know that the Brownian Motion alone is a Markov Process. The time integral of Brownian Motion is not a Markov Process, but I know it is normally distributed from these two questions: previous question 1 and previous question 2.
In order to show that $(W_t,V_t)$ is a Markov Process I know I need to verify that is $s,t \ge 0$ and then $P(X_{s+t}|F_s)=P(X_{s+t}|X_s)$ almost surely. How can I apply the information I know about $W_t$ alone being a Markov Process and $V_t$ being normally distributed to verify this fact?
The suggestion lets one see that conditional on $\mathcal F_s$, the random pair $(W_{t+s},V_{t+s})$ has a bivariate normal distributed with mean vector $\left[\matrix{W_s\cr V_s + tW_s\cr}\right]$ and covariance matrix $\left[\matrix{ t&t^2/2\cr t^2/2&t^3/3}\right]$. Because this conditional distribution depends on $\mathcal F_s$ only through the pair $(W_s,V_s)$, we see that the pair process$\{(W_t,V_t): t\ge 0\}$ is Markovian.