1
$\begingroup$

Let $X(t)$ be a martingale w.r.t. filtration generated by Brownian motion $B(t)$. There is a well-known theorem that states that there is a unique adapted process $H(t)$ such that $$ X(t) = \int_0^t H(s)dB(s).$$ Let's define step process $$M(t)=2 \sum_{j=1}^{\lfloor t \rfloor} 1_{\{ B(j)-B(j-1) > 0 \}} - \lfloor t \rfloor.$$ It's just a simple martingale, generated by fair coin tosses, just written in another way. One can see that $B(t)$ is the only source of uncertainty. What is $H(t)$ for $M(t)$ then? Or am I missing something and M.R.T. is not applicable here?

$\endgroup$
  • $\begingroup$ $M(t)$ is not continuous. But for distrete time, we can use the stoping time $T=\inf{t:B_t =\pm 1}$ and $H(s)=1_{s\leq T}$ $\endgroup$ – RaphaelB4 Dec 4 '18 at 10:06
0
$\begingroup$

I've caught myself into a trap here: the reason why I couldn't find a martingal representation is that $M(t)$ is not a martingale. For example: $$ \mathbb{E}[M(1) | M(1/2)] = 2 \mathbb{P}(B(1) > 0 |B(1/2)) - 1. $$ And if $B(1/2) \neq 0$, which is true a.s., $\mathbb{E}[M(1) | M(1/2)] \neq M(1/2).$

$\endgroup$

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.