Integral of white noise is a continuous space Markov process?

Question

I just read the chapter 4.1 in the Gardiner. He defines that the so called "white noise" $\xi$ has the properties: $\left<\xi\right> = 0$, $\xi(t)$ is independent of $\xi(t')$ for $t\neq t'$. The last property also implies that $\left<\xi(t)\xi(t')\right> = \delta(t-t')$. Now he defines $u(t')$ as

$$u(t') = \int_0^{t_0}dt' \xi(t').$$

He claims that assuming that $u(t)$ is continuous, $u(t)$ is a Markov Process. The argument goes as follows:

$$u(t') = \lim_{\epsilon\rightarrow0}\left[ \int_0^{t-\epsilon} \xi(s) ds\right] + \int_t^{t'}ds \xi(s).$$

1. For any $\epsilon>0$ the $\xi(s)$'s in the first integral are independent of the $\xi(s)$'s in the second.

(OK, so far I agree.)

2. Hence by continuity $u(t)$ and $u(t')-u(t)$ are statistically independent.

(I also agree on that. It follows from the preceding argument by noting that as $u(t) = \lim_{\epsilon\rightarrow0}\left[ \int_0^{t-\epsilon} \xi(s) ds\right]$ and $u(t')-u(t) = \int_t^{t'}ds \xi(s)$)

3. Further $u(t') - u(t)$ is independent of $u(t'')$ $\forall t''<t$.

(nothing new, it follows from point 1)

4. This means that $u(t')$ is fully determined (probabilistically) from the knowledge of the values $u(t)$ and not by any past value. Hence, $u(t)$ is a Markov process.

I don't get the last point. Why does this follow?

Answer 1

It follows from the definition of a Markov process (Gardiner)

The conditional probability is determined entirely by the knowledge of the most recent condition, i.e.

$p(x_1,t_1;...|y_1,\tau_1;y_2,\tau_2;...)=p(x_1,t_1;...|y_1,\tau_1)$

(ordered times). In our case we have to demonstrate that for $t''<t<t'$

$p(u(t'),t'|u(t),t;u(t''),t'')=p(u(t'),t'|u(t),t)$

From Gardiner

$u(t')=u(t)+\int_t^{t'}ds\xi(s)$

As you stated the integral is independent from u(t) and u(t'') for every $t''<t$, this means that knowing that at time t'' the value of u was u(t'') it doesn't matter since u(t') is only given by u(t) added to something that doesn't depend on u(t'') (remember this is true for every $t''<t$). So in the conditional probability you can just remove the last term.