Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 37 / 4

The Brownian reflection principle gives, for $t>0$,For a standard normal random variable $N^{\prime }$, this is also $\mathbb{P}(|N^{\prime }|\ge x/\sqrt{t})=\mathbb{P}(x^2/(N^{\prime }{)}^2\le t)$. Thus the inverse-square Gaussian law of Brownian first passage isIn particular, ${\tau }_x$ is finite almost surely.

The Brownian motions $X,Y$ are independent, so conditional on ${\tau }_x=s$, the ordinate $Y_{{\tau }_x}$ has the same distribution as $\sqrt{s}N$, with $N$ a standard normal independent of the random time. By Brownian scaling and the hitting-time identity,The last equality uses symmetry of $N$ and independence of $N,N^{\prime }$: inserting the independent sign of $N^{\prime }$ does not change the distribution of the numerator. ThereforeFor completeness, the ratio has density ${\int }_{\mathbb{R}}|v|\phi (cv)\phi (v)dv=1/(\pi (1+c^2))$, where $\phi$ is the standard normal density. The scaled exit ordinate has density $x/(\pi (x^2+y^2))$.

Translate the horizontal coordinate by $x$. Then $X_t+x$ is a standard Brownian motion starting at zero, and continuity identifies $\tau$ with its first-passage time to $x$. The vertical coordinate remains an independent standard Brownian motion starting at zero. Part (a) therefore gives the Cauchy exit law from a Brownian half-plane:The TeX transcription's initial value $-z$ and later $Y_T$ are both corrected here to the PDF's $-x$ and $Y_{\tau }$.

Start the planar Brownian motion $X_t+i{Y}_t$ at $-x$ and stop at $\tau =\inf \{t:X_t=0\}$. The complex exponential is holomorphic with nonzero derivative. By conformal invariance of planar Brownian motion, the process $e^{X_t+i{Y}_t}$, run on its conformal Brownian clockis planar Brownian motion started at $\epsilon =e^{-x}$, until its first exit from the unit disk. The clock is strictly increasing before $\tau$, and $A_{\tau }\le \tau <\infty$ almost surely. It is legitimate to use the exponential although it is not globally injective: the Brownian coordinate brackets depend locally on the nonzero derivative, and the clock construction gives the stopped planar Brownian law. Before $\tau$, $|e^{X_t+i{Y}_t}|<1$; at $\tau$ it equals one. Hence the disk exit time in the new clock is exactly $A_{\tau }$.

The continuous lifted argument of this path, starting from angle zero, is $Y_t$. Its total argument at exit is therefore $Y_{\tau }$, which has distribution $xC$ by part (b). If the winding count records only completed integer turns, it differs from $Y_{\tau }/(2\pi )$ by a remainder $R_{\epsilon }$ with $|R_{\epsilon }|\le 1$. If fractional turns are retained, that remainder is zero. In either convention, since $\log \epsilon =-x$,The first term has distribution $-C/(2\pi )$ for every $x$, and the second tends to zero deterministically in absolute value as $x\to \infty$. Symmetry of the Cauchy distribution removes the minus sign, so Slutsky theorem gives the small-radius Brownian winding law: