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.



Distribution of running maximum of a local martingale
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](http://books.google.co.uk/books?id=bDQy-zoHWfcC&lpg=PP1&dq=rogers%2520and%2520williams&pg=PP1#v=onepage&q&f=false) as an example of a local martingale which is not a proper martingale.


