2
$\begingroup$

Let $(B_t)_{t\geq 0}$ be a standard Brownian motion in $\mathbb{R}^d$. It is intuitive that, for fixed $s<t<u$

$$\mathbb{E}[B_t\mid \sigma(B_s,B_u)]=B_s+\frac{t-s}{u-s}(B_u-B_s).$$

However, I cannot think of a way to show this rigorously.

If first attempted to take $A\in\sigma(B_s,B_u)$ and show that $\mathbb{E}[1_A B_t]=\mathbb{E}[1_A(B_s+\frac{t-s}{u-s}(B_u-B_s))]$. But I cannot manage to show this equality.

I'd be very thankful for any ideas and suggestions on how to tackle this problem.

$\endgroup$
  • 2
    $\begingroup$ "Intuitive"? How so? (The late Paul Lévy would have declared that, we mere mortals would have been forced to believe him, but...) $\endgroup$ – Did May 22 '14 at 21:02
  • $\begingroup$ So $B_s$, $B_t-B_s$, and $B_u-B_t$ are independent normally distributed random variables with respective variances $s$, $t-s$, and $u-t$. It's just about three random variables, now about a whole Brownian process. $\endgroup$ – Michael Hardy Jun 2 '14 at 22:39
  • 2
    $\begingroup$ I agree that it's intuitive. Given that one hour after the clock starts, a particle is one mile west of where it started, what is its expected location 20 minutes after it started. Given that the distribution is suitably symmetric, it's $1/3$ mile west of the starting point. $\endgroup$ – Michael Hardy Jun 2 '14 at 22:48
4
$\begingroup$

Actually, it just might be that OP is the next Paul Lévy, as the result holds!

Unfortunately, I am far from being as bright as Lévy, so I'll apply the same kind of tricks that I used in this answer.

Set $$ Z_t=B_t-\frac{u-t}{u-s}B_s-\frac{t-s}{u-s}B_u, $$ which you can check is a Gaussian random variable independent of $B_s$ and $B_u$.

It follows that \begin{align*} E[B_t\,|\,\sigma(B_s,B_u)] &=E[Z_t\,|\,\sigma(B_s,B_u)]+\frac{u-t}{u-s}B_s+\frac{t-s}{u-s}B_u\\ &=E[Z_t]+\frac{u-t}{u-s}B_s+\frac{t-s}{u-s}B_u\\ &=\frac{u-t}{u-s}B_s+\frac{t-s}{u-s}B_u\\ &=B_s+\frac{t-s}{u-s}\left(B_u-B_s\right). \end{align*}

$\endgroup$
  • 1
    $\begingroup$ That, or their TA? $\endgroup$ – Did May 23 '14 at 23:12
3
$\begingroup$

As a side note for further reading, this property of the Brownian motion is called the harness property. Processes satisfying this property have been studied in the literature, for instance in the paper

Harnesses, Levy bridges and Monsieur Jourdain

$\endgroup$
  • $\begingroup$ This really is a lovely paper, thank you for posting it. $\endgroup$ – Chris Janjigian Jun 3 '14 at 17:23

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.