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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?

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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

1 Answer 1

up vote 1 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.

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