Let $B$ be a Brownian motion with natural filtration $(\mathcal{F}_t)_{t\geq 0}$ and let $\mathcal{H}_t$ be the $\sigma$-algebra generated by $\mathcal{F}_t$ and $B_1$.


$$A_t = B_t-\int_0^{\min(t,1)} \frac{B_1-B_s}{1-s}ds$$

I'm trying to show that $A_t$ is a Brownian motion with respect to $(\mathcal{H_t})_{t\geq0}$.

As a first step, I'm attempting to show that $A_t$ is a martingale, but haven't made much progress.

Thank you.

  • $\begingroup$ If you have to show that the integral is well-defined for $t=1$? $\endgroup$ – Ilya Sep 24 '11 at 19:40
  • 3
    $\begingroup$ The convergence at $t=1$ is a non-issue since $E(|B_1-B_s|)=\Theta(\sqrt{1-s})$. $\endgroup$ – Did Sep 24 '11 at 21:16
  • 2
    $\begingroup$ Note that $(B_1 - B_s)/(1-s) = B_1 - (B_s - s B_1)/(1-s)$ and the latter term is a Brownian motion under a change of time scale. $\endgroup$ – cardinal Sep 24 '11 at 21:40
  • 1
    $\begingroup$ @cardinal, good point. Let $X_s=B_s-sB_1$, then the process $(X_s)_{0\le s\le 1}$ is a Brownian bridge (from $0$ to $0$ at time $1$) and is independent on $B_1$. $\endgroup$ – Did Sep 24 '11 at 21:48

Look at the quadratic variation of the process.


  • 1
    $\begingroup$ Elaborting on TheBridge's answer you should be aware that one of the 'easy' ways to check a process is a brownian motion is to use Levy's Characterization theorem: almostsure.wordpress.com/2010/04/13/…. Computationally the majority of the proof can be reduced to checking the quadratic variation $[X]_t =t$. $\endgroup$ – user7980 Sep 25 '11 at 1:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.