4
$\begingroup$

I am trying to solve problem 2.16 from the book "Continuous Martingales and Brownian Motion" by Revuz and Yor. There are two things that confuse me from the exercise so hopefully someone can shed some light into the subject.

For the Standard Brownian Motion $B$, set $\mathcal G = \sigma\,(B_u, u \geq t)$. I have to prove that for every real $\lambda$, the process $$M_t = \exp\left(\frac{\lambda B_t}{t}-\frac{\lambda^2}{2t}\right), \,\,\,\, t > 0$$ is a martingale with respect to the decreasing family $\mathcal G_t$.

This confuses me, as normally these proofs involve proving that a stochastic process is a martingale with respect to a natural (increasing) filtration adapted to the process in question, so I wonder what could be the applications of working with such family.

The exercise also includes the following hint: observe that $\left(B_s - \frac{s}{t}B_t\right)$ is independent of $\mathcal G_t$ for $s<t$ OR use time-inversion.

How can one use time-inversion to prove the above argument? Thanks everyone in advance.

$\endgroup$
2
  • 1
    $\begingroup$ @ Iliana : Using time inversion $s=1/T$ you get $B_s$ is a brownian motion in the filtration $\mathcal{G}_s=\mathcal{F}_{1/t}$ which puts you back in the usual situation. Best regards $\endgroup$
    – TheBridge
    Jan 29, 2015 at 18:48
  • 1
    $\begingroup$ @ Iliana : It's $X_s=1/s.B_s=t.B_{1/t}$ to be precise which is a Brownian Motion in the filtration $\mathcal{G}_s$. Best regards. $\endgroup$
    – TheBridge
    Jan 29, 2015 at 22:05

1 Answer 1

3
$\begingroup$

It is not difficult to see that the process

$$W_t := \begin{cases} t B_{1/t}, & t > 0, \\ 0, & t=0 \end{cases}$$

defines a Brownian motion and $$\mathcal{G}_t := \sigma(B_u; u \geq t) = \sigma\left(W_u; u \leq \frac{1}{t} \right) =: \mathcal{H}_{\tilde{t}}$$ for ${\tilde{t}} := 1/t$ is the canonical filtration of $(W_\tilde{t})_{\tilde{t} \geq 0}$. Moreover, we can write

$$\begin{align*} M_t &= \exp \left( \lambda \frac{B_t}{t} - \lambda^2 \frac{1}{2t} \right) \\ &= \exp \left( \lambda W_{1/t} - \lambda^2 \frac{1}{2t} \right) \\ &= \exp \left( \lambda W_{\tilde{t}} - \frac{\lambda^2}{2} {\tilde{t}} \right). \end{align*}$$

Now let $s \leq t$. Then $\tilde{s} \geq \tilde{t}$ and since

$$\left( \exp \left[ \lambda W_{\tilde{t}}- \frac{\lambda^2}{2} \tilde{t} \right], \mathcal{H}_{\tilde{t}} \right)_{\tilde{t} \geq 0}$$

is a martingale, we get

$$\begin{align*} \mathbb{E}(M_s \mid \mathcal{G}_t) &= \mathbb{E} \left( \exp \left[ \lambda W_{\tilde{s}} - \frac{\lambda^2}{2} \tilde{s} \right] \mid \mathcal{H}_{\tilde{t}} \right) \\ &= \exp \left[ \lambda W_{\tilde{t}} - \frac{\lambda^2}{2} \tilde{t} \right] \\ &= M_t. \end{align*}$$

This shows that $(M_t,\mathcal{G}_t)$ is a (backwards) martingale.

Backward martingales are often useful because they are closable and therefore nice martingale convergence theorem may be applied. For example one can prove the strong law of large numbers using backward martingales.

$\endgroup$
1
  • $\begingroup$ Dear @saz, this also implies that the process is a martingale under the natural filtration, right? $\endgroup$
    – Nobody
    Jun 8, 2022 at 17:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .