Past exam of the stochastic differential equations and Brownian motion in complex analysis course of the University of Oxford / 2009 / B / 5

A continuous nondecreasing function $\Phi$ with $\Phi (0)=0$ has moderate growth if for some fixed $a>1$ and finite $K$,Iteration and monotonicity then give such an inequality, with a different constant, for every fixed dilation factor. Powers $\Phi (x)=x^p$, $p>0$, are standard examples.

The Burkholder-Davis-Gundy inequalities say that for a real continuous local martingale $M$ with $M_0=0$ and any stopping time $\tau$, there are positive finite constants $c_{\Phi },C_{\Phi }$, depending only on $\Phi$, such thatHere $M_{\tau }^{\ast }={\sup }_{s\le \tau }|M_s|$. Expectations may be infinite, with the inequalities understood by localization. For $\Phi (x)=x^p$, these compare the $p$th moment of the running maximum with the $p/2$th moment of quadratic variation.

Set $a=\delta \lambda$, $t_0={\lambda }^2$, and $h=({\beta }^2-1){\lambda }^2$. On the event $\{{\tau }^{1/2}>\beta \lambda ,X_{\tau }^{\ast }\le a\}$, one has $\tau >t_0$ and the Brownian path remains in $[-a,a]$ throughout the next $h$ time units after $t_0$.

For Brownian motion started at $x\in (-a,a)$, let $H$ be its first exit time from $(-a,a)$. The martingale $B_t^2-t$, stopped at $H\wedge n$, gives ${\mathbb{E}}_x(H\wedge n)\le a^2-x^2$. Monotone convergence first proves finite mean exit time, and dominated convergence for the bounded stopped position then givesThus Markov's inequality yields ${\mathbb{P}}_x(H>h)\le a^2/h$. Remaining in the closed interval for a fixed positive duration has the same probability as avoiding boundary exit: Brownian motion crosses immediately after first hitting an endpoint.

Condition on ${\mathcal{F}}_{t_0}$. The event $\{\tau >t_0\}$ is measurable there because $\tau$ is a stopping time, and future Brownian increments are independent of this sigma-algebra. The preceding bound is uniform in the current point inside the interval. ThereforeThis is a good-lambda bound for Brownian stopping time and maximum.

Put $Y=\sqrt{\tau }$ and $V=X_{\tau }^{\ast }$. Splitting according to whether $V>\delta \lambda$, the good-lambda bound for Brownian stopping time and maximum givesIntegrate against the positive measure $d\Phi (\lambda )$. The tail integral formula for expectation in Stieltjes form yieldsBy moderate growth, choose $K_{\beta }$ with $\Phi (\beta x)\le K_{\beta }\Phi (x)$. ThenFix $\beta >1$ and choose $\delta$ small enough that $K_{\beta }\eta <1/2$. Absorb the last term and apply moderate growth once more to the fixed dilation $1/\delta$. This provesTo avoid subtracting infinities, perform the argument first for $\tau \wedge n$, where Brownian maxima have all finite polynomial moments and moderate growth is at most polynomial, and then use monotone convergence. Applying the Dambis-Dubins-Schwarz theorem transfers this Brownian estimate to continuous local martingales, giving the lower-maximum side of the Burkholder-Davis-Gundy inequalities.