Distribution of running maximum of a local martingale - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T12:28:47Z http://mathoverflow.net/feeds/question/33923 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/33923/distribution-of-running-maximum-of-a-local-martingale Distribution of running maximum of a local martingale kenneth 2010-07-30T17:08:22Z 2010-09-29T16:29:16Z <p>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 $X^*(T) = \sup_{s\in [0,T]} X(s)$.</p> <p>Question: For a fixed $T$, is this possible to show that $\mathbb{P} ( X^*(T) \ge \beta) = o(\beta^{-1})$ as $\beta \to \infty$?</p> <p>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. </p> <p>Thank you for your time.</p> http://mathoverflow.net/questions/33923/distribution-of-running-maximum-of-a-local-martingale/33938#33938 Answer by George Lowther for Distribution of running maximum of a local martingale George Lowther 2010-07-30T19:55:22Z 2010-07-30T19:55:22Z <p>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.</p> <p>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].</p> <p>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 &amp; Williams book <a href="http://books.google.co.uk/books?id=bDQy-zoHWfcC&amp;lpg=PP1&amp;dq=rogers%2520and%2520williams&amp;pg=PP1#v=onepage&amp;q&amp;f=false" rel="nofollow">Diffusions, Markov Processes and Martingales</a> as an example of a local martingale which is not a proper martingale.</p>