How would you show that Brownian motion is Markov only using the fact that the Brownian bridge is Markov?

Question

Try to forget everything you know about Brownian motion.

You are told there is a process called Brownian bridge on $[0,1]$. It is Gaussian with mean $0$ and covariance $s(1-t)$ for $s<t$ and can be written as $B_t=W_t - tW_1$ where $(W_t)$ is a continuous process with $W_0=0$, called Brownian motion. Of course, from what you were told about the Brownian bridge, you know that $(W_t)$ is Gaussian with mean $0$ (this is not true, see remark of John Dawkins and EDIT below).

You are told $(B_t)$ is Markov with regard to its own filtration. How would you show that $(W_t)$ is Markov with regard to its own filtration ?

EDIT: In light of the remark of John Dawkins, we should indeed say that $(W_t)$ is itself Gaussian with mean $0$, otherwise, other processes than the real Brownian motion could be solution of $B_t = W_t - t W_1$.

Answer 1

1. You don't need to be told that $(B_t)$ is Markov; this follows from the form of the covariance function.

2. Why does it follow from $B_t=W_t-tW_1$, $0\le t\le 1$, that $(W_t)$ is Gaussian, or even that $E[W_t]=0$?

3. If $(W_t)$ is one solution of $B_t=W_t-tW_1$, $0\le t\le 1$, then so is $\tilde W_t:=W_t+tZ$, where $Z$ is any random variable on the probability space where $B$ and $W$ are defined. If $Z$ were to depend on the "future" of $W$ in some way, this might disturb the Markov property of $\tilde W$.