MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

Let $(\Omega, \mathcal{F}, \mathbb{P}, \mathcal{F}_t)$ be a given probability space with usual conditions, on which $W$ is a standard Brownian motion. For $x \ge 0$, consider $$X(t) = x + \int_0^t \sigma (X(s)) dW(s)$$ Assume $\sigma \in C^{0,1/2}_{loc}$, $\sigma(0) = 0$, $\sigma>0$ on $(0,\infty)$. By [Karatzas and Shereve 98], there exists a unique strong solution with absorbing state at zero. Denote the running maximum by $X^*(T) = \sup_{s\in [0,T]} X(s)$.

Question: For a fixed $T$, is this possible to show that $\mathbb{P} ( X^*(T) \ge \beta) = o(\beta^{-1})$ as $\beta \to \infty$?

I am trying to use time-changed Brownian motion, i.e. $X(t) = x + B([X]_t)$, where $B$ is BM, and $[X]$ is quadratic variation. There is also density function available for running maximum $B^* (T)$, i.e. $\mathbb{P}(B^*(T) \ge \beta) = 2 - 2 \Phi(\beta/\sqrt{T}) = o(\beta^{-1})$, where $\Phi(\cdot)$ is c.d.f of standard normal distribution. But, I could not succeed using those facts to prove it.

Thank you for your time.

share|cite|improve this question
It is pretty clear that the estimate you are after will depend strongly on the behavior of $\sigma$ near 0. Can you be more specific about what $C^{0,1/2}_{loc}$ is? – Jeff Schenker Jul 30 '10 at 18:50
Actually, it is not going to depend on what $\sigma$ is like near zero much at all. It is a local martingale and, once it gets very close to zero, it is unlikely to escape. It depends more on how fast $\sigma$ grows as x goes to infinity. $C^{0,1/2}$ is the class of Holder continuous functions of exponent 1/2. This does guarantee a unique strong solution, but I don't think that has much bearing on the question. – George Lowther Jul 30 '10 at 19:12
The bound $\mathbb{P}(X^*_T>K)\le x/K$ follows from the local martingale property. Stopped at the first time it hits K, its expectation is x, but is equal to K with probability at least $\mathbb{P}(X^*_\infty>K)$, giving the inequality (actually, it is Doob's maximal inequality). It is not possible to achieve this bound, so the question can be understood as asking if we can get very close to it in some sense. – George Lowther Jul 30 '10 at 19:21
Doh! Of course this has nothing to do with the behavior near zero. I read it all too quickly and imagined the question was about the typical time to reach zero. Thanks for explaining the notation. – Jeff Schenker Jul 30 '10 at 20:08
up vote 4 down vote accepted

No. It is true that $\mathbb{P}(X^*_T>\beta)=O(\beta^{-1})$, but you don't have a`little-o' bound. In fact it fails, and $\beta\,\mathbb{P}(X^*_T>\beta)$ converges to a strictly positive value, precisely when $X$ fails to be a martingale.

If $S$ is the first time at which $X$ hits $\beta>x$ then continuity gives $$ X_{S\wedge T} = \beta 1_{\{X^*_T>\beta\}}+1_{\{X^*_T\le\beta\}}X_T $$ Take expectations, and use $\mathbb{E}[X_{S\wedge T}]=x$, which follows from the fact that the first term is a local martingale stopped at time $S$, so is bounded (and hence a proper martingale). $$ x=\beta\,\mathbb{P}(X^*_T>\beta)+\mathbb{E}[1_{\{X^*_T\le\beta\}}X_T]. $$ The final expectation converges to $\mathbb{E}[X_T]$ as $\beta$ goes to infinity, by monotone convergence. This gives $$ \lim_{\beta\to\infty}\beta\,\mathbb{P}(X^*_T>\beta)=x-\mathbb{E}[X_T]. $$ Now, it is a well known result that if $X$ is a nonnegative local martingale and $X_0$ is integrable then it is a supermartingale, so $\mathbb{E}[X_T]\le\mathbb{E}[X_0]$, and equality holds precisely when it is a martingale over the range $[0,T]$. So, in our case, $\mathbb{P}(X^*_T>\beta)=o(\beta^{-1})$ exactly when $\mathbb{E}[X_T]=x$ and $X$ is a martingale over the range $[0,T]$.

An example when solutions to your SDE fails to be a martingale is $\sigma(x)=x^2$, $dX=X^2\,dW$. The solution to this SDE can be written as $X=1/\Vert B\Vert$ for a 3-dimensional Brownian motion $B$ started from the point $(x^{-1},0,0)$. You can calculate $\mathbb{E}[X_t]$ and determine that it is decreasing in $t$, so $X$ is not a martingale - just a local martingale. This example appears in Roger's & Williams book Diffusions, Markov Processes and Martingales as an example of a local martingale which is not a proper martingale.

share|cite|improve this answer
George, Thank you very much. – kenneth Jul 30 '10 at 23:41

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.