Quantitative Finance Stack Exchange is a question and answer site for finance professionals and academics. 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 a probability filtred space $(\Omega, \mathcal F, \mathbb F, \mathbb P)$, where $\mathbb F = (\mathcal F_t)_{0\leq t\leq T}$ satisfing the habitual conditions and isgenerated by $1 d $- Brownian Motion (with $\mathcal F_T = \mathcal F$).

Also, consider a finantial market where the interest rate is nul, $r=0$, and the dynamics of the risky asset $S$ is given by $$S_t= S_0 + \int_0^t \mu_s ~ds +\int_0^t \sigma_s ~dW_s \quad , t \geq 0$$

where $t \in [0,T] \mapsto \mu_t$ and $t \in [0,T] \mapsto \sigma_t \geq 0$ are deterministic and continuous functions.

Show that:

  1. If the absence of arbitrage opportunity hypothesis is verified, then $B:=\{t \in [0,T] : \sigma_t=0 \ \text{and} \ \mu_t \neq 0\}$ is a Lebesgue nul-measure set (ie, $\int_0^T\mathbf1_{t \in B} dt=0$).
  2. $\nu_\sigma(O):= \int_0^T\mathbf1_{t \in O}\sigma_t ~dt$ dominates $\nu_\mu (O):= \int_0^T\mathbf1_{t \in O} \mu_t ~dt$, where $O$ is a borelian of $[0,T]$ ( ie, $\nu_\mu \ll \nu_\sigma$) and deduce from it that there is a measurable function $\lambda$ such that $\mu = \sigma \lambda$.
share|improve this question
Can you please tell from which book is this question? – Alexey Kalmykov Apr 8 '13 at 21:59
@AlexeyKalmykov: I really don't know, but I'd like too if it makes part of a book. It was part of an exam question that I couldn't answer. If you find it somewher, please let me know. – Paul Apr 8 '13 at 22:08
Since it's an exam question, what have you tried so far? – chrisaycock Apr 8 '13 at 23:43
In your second question, should it be $1_{t \in O}$? – quasi Apr 9 '13 at 0:24
For your second question, do you need to assume that $\sigma_t \geq 0$? – quasi Apr 9 '13 at 2:19
up vote 2 down vote accepted

For the first one absurd reasoning allows you to construct an arbitrage (as r=0) by investing (or short selling according to the sign of $\mu$) at the time where $\sigma$ is null, or if you prefer as soon as $t$ is in $B$ (which is not a Lebesgue negligible set by hypothesis) which is absurd as no-arbitrage holds. The details that remain to be proved is that such a strategy is an admissible one.

For the second question, as soon as the first part is done, it is only the application of Lebesgue's decomposition theorem.

For the first part as any borelian set such that $\nu_\sigma(O):= \int_0^T\mathbf1_{t \in O}\sigma_t dt $ doesn't dominate $\nu_\mu (O):= \int_0^T\mathbf1_{t \in O} \mu_t dt$ is included in $B$ and as $B$ is of null probability the conclusion holds true.

Best regards

share|improve this answer
@Paul : Better late than never;-) – TheBridge May 10 '13 at 13:59

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.