Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider the stochastic process defined by

$$ Z_t = \frac{1}{\sqrt{1-t}} \exp \left( \frac{-B^2_t}{2\left( 1-t\right)} \right ) , t \geq 0$$ where $ \left(B_t\right)_{t\geq 0}$ is a real standard brownian motion starting from zero.

I'd like to show that $\left(Z_t\right)_{t\geq 0} $ is a exponential martingale for $t \in \left [ 0, 1\right [$.

For it, I'd like to use the following theorem:

Theorem: If $ M = \left(M_t\right)_{t\geq 0}$ is an adapted, continuous process, null at zero and $ V = \left(V_t\right)_{t\geq 0}$ is a increasing, continuous and adapted process, null at zero, then are equivalent:

(i) $ M = \left(M_t\right)_{t\geq 0}$is a local continuous martingale and $ \langle M \rangle_t= V_t,\ \forall{t\geq 0}$

(ii) For all $\lambda \in R$, $ X^\lambda_t = \exp \left(\lambda M_t -\frac{\lambda}{2}V_t \right)$ is a positive local continuous martingale.

Note that $$ Z_t = \exp \left( \frac{-B^2_t}{2\left( 1-t\right)} - \frac{1}{2} \log \left( 1-t\right) \right ) $$

and by Itô's lemma

$$ M_t :=\frac{-B^2_t}{2\left( 1-t\right)} = - \int_0 ^t \frac{B_s}{\left( 1-s\right)}dB_s -\int_0 ^t \frac{-B^2_s}{2\left( 1-s\right)} ds - \int_0 ^t \frac{1}{\left( 1-s\right)}ds .$$

Also, we have

$$ \langle M \rangle_t = \int_0 ^t \frac{B^2_s}{\left( 1-s\right)^2}d\langle B \rangle_s =\int_0 ^t \frac{B^2_s}{\left( 1-s\right)^2}ds$$

and $$ \mathbb E \langle M \rangle_t = \int_0 ^t \frac{ \mathbb E \left[ B^2_s\right ] }{\left( 1-s\right)^2}ds = \int_0 ^t \frac{ s }{\left( 1-s\right)^2}ds = - \log(1-t) < \infty , \ \forall t \in \left[0,1 \right[ $$

Nevertheless, I'm still not capable to see how to manipulate $Z_t$ in order to apply the theorem. Perhaps, there is an alternative to usage of this theorem.

Someone could help me please?

share|cite|improve this question
Be careful, $\dfrac{B_t^2}{(1-t)}$ is not a local martingale. – Siméon Jan 17 '13 at 21:16
@Ju'x: I've not said it was I've just cald it $M_t$. Anyway, thank you very much for your answer that undid my confusion. – Paul Jan 17 '13 at 22:54
I know, it was just a small warning. Anyway, glad to help. – Siméon Jan 17 '13 at 23:07
up vote 2 down vote accepted

According to Kiyoshi Itō's lemma: $$ \frac{B_t^2}{2(1-t)} = \int_0^t\frac{B_s}{1-s}dB_s + \frac{1}{2}\int_0^t\frac{ds}{1-s}+ \frac{1}{2}\int_0^t \frac{B_s^2}{(1-s)^2}ds. $$

The term $M_t = \displaystyle\int_0^t \frac{B_s}{1-s}dB_s$ is a continuous local martingale with quadratic variation given by: $$ \langle M \rangle_t = \int_0^t\frac{B_s^2}{(1-s)^2}ds. $$

In other words, $$ \exp\left\{-\frac{B_t^2}{2(1-t)} - \frac{1}{2}\ln(1-t)\right\} = \exp\left\{-M_t - \frac{1}{2}\langle M\rangle_t\right\} $$ is an exponential local martingale.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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