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 $t=1$? $\endgroup$
    – SBF
    Sep 24, 2011 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, 2011 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, 2011 at 21:40
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    $\begingroup$ @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, 2011 at 21:48

1 Answer 1


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, 2011 at 1:56

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