I have two $E$-valued processes, $X$ and $Y$, defined on the same probability space $(\Omega, \mathcal{F},(\mathcal{F}_t)_t, P)$. Both $X$ and $Y$ have the Markov property. Under what conditions on $E$ (if any required) is the vector-valued process $(X,Y)$ also Markov?

Since the latter requires bounded and measurable functions in two arguments, I thought maybe some kind of approximation argument can be used (approximating a function with two arguments by two functions with a single argument). That would probably require some kind of regularity on the functions involved. So another argument to drop the regularity assumption would be needed. I don't know how to make this concrete.

The reason why I wanted to know whether this is true is because while showing that $(B_t,M_t)$ (Brownian motion and its running maximum) is Markov, I argued that since $M_t - B_t$ and $B_t$ are Markov, so is $(B_t,M_t-B_t)$. Then I argued since $(B_t,M_t)$ is a linear transformation of $(B_t,M_t-B_t)$, it is Markov as well. My second question is under which conditions, if at all, this argument is valid?


1 Answer 1


1. To show that the process $X=(B,M)$, defined, for every nonnegative $t$, by $$X_t=(B_t,M_t)$$ is a Markov process, the simplest approach is to note that, for every fixed $t$ and every $s>t$, $$X_s=(B_t+\beta_{s-t},\max\{M_t,B_t+\mu_{s-t}\})\tag{$\ast$}$$ where, for every nonnegative $u$, one considers $$\beta_u=B_{t+u}-B_t\qquad\mu_u=\max\{\beta_v;v\leqslant u\}$$ The process $\beta=(\beta_u)_{u\geqslant0}$ is a Brownian motion independent of $\mathcal F^X_t=\sigma(X_r;r\leqslant t)$. Thus, $(\ast)$ shows that the process $(X_s)_{s\geqslant t}$ is a functional of $(X_t,\beta)$, where $\beta$ is independent of $\mathcal F^X_t$, and this implies that $X$ is indeed a Markov process.

2. The second question in your post is whether $A(Z)$ is a Markov process when $Z$ is a Markov process and $A$ is a linear transformation.

If $A$ is invertible, this is direct since then, the filtrations $(\mathcal F^Z_t)_{t\geqslant0}$ and $(\mathcal F^{A(Z)}_t)_{t\geqslant0}$ coincide. If $A$ is not invertible, the answer is no in general, since then $A(Z)$ is the observation process of a hidden Markov chain whose state process is $Z$, and the state processes of hidden Markov chains are well known to be non Markov in general.

In your case, $Z=(B,M-B)$ and $A(b,v)=(b,b+v)$ hence $A$ is invertible. Thus, knowing that $(B,M-B)$ is Markov would indeed imply that $(B,M)$ is.

3. ...But the way to prove that $(B,M-B)$ is a Markov process, that you suggest, is flawed. To wit, it is not true that for every Markov processes $U$ and $V$, the process $(U,V)$ is Markov as well. That is, unless some supplementary condition holds, for example that $U$ and $V$ are independent.

Roughly speaking, the trouble here comes from the fact that the Markovianity of $(U,V)$ requests some properties of conditional distributions conditioned on $\mathcal F^{(U,V)}_t$, while the hypothesis involve conditional distributions conditioned on $\mathcal F^U_t$ and $\mathcal F^V_t$ only.

  • $\begingroup$ Very educational and helpful explanation as usual, especially #3. Thanks. $\endgroup$
    – Olorun
    Commented Mar 27, 2018 at 5:53

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