# Proving Brownian Motion and the Time Integral of Brownian Motion form a 2d Markov Process

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?

• Suggestion: Write $V_{t+s}=V_s+tW_s+\int_s^{t+s}(W_u-W_s)\,ds$. The integral on the right is independent of $\mathcal F_s$. – John Dawkins Apr 27 '17 at 1:31
• @JohnDawkins How do I show the pair together make a markov process? Should I be taking $P(V_{s+t}|F_s)?$ Then by your comment $P(V_{s+t}|F_s)=P(V_s +tW_s + \int_{s}^{t+s}(W_u-W_s)ds|F_s)=P(\int_{s}^{t+s}(W_u-W_s)ds)+P(V_s +tW_s|F_s)$. How do I simplify this further? – user75514 Apr 27 '17 at 1:41

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.