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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 $B_t = W_t - t W_1$.

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  • $\begingroup$ The law of Brownian bridge is the regular conditional distribution of Brownian motion. Maybe u can use this to calculate the probability and then show the Markov property. $\endgroup$ May 31, 2021 at 3:06
  • $\begingroup$ You cannot use that as you know nothing about Brownian motion besides the information above. $\endgroup$
    – W. Volante
    May 31, 2021 at 13:27
  • $\begingroup$ N.B. The covariance of the Brownian bridge is $s(1-t)$ for $0\le s\le t\le 1$. $\endgroup$ May 31, 2021 at 15:42
  • $\begingroup$ Thank you, I corrected the typo $\endgroup$
    – W. Volante
    May 31, 2021 at 16:03

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  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 $\tilde W$.

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  • $\begingroup$ You are right, I should say that $(W_t)$ is Gaussian, otherwise we can indeed create counter-examples, thank you! $\endgroup$
    – W. Volante
    May 31, 2021 at 17:58

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