# More Cauchy distribution

I’ve mentioned earlier the first cases I had found of (almost) mod-Cauchy convergence (see this post for the definition).

Yesterday, following from a citation in Sarnak’s note to one in a paper of Guivarc’h and Le Jan, I ended up looking at this paper of F. Spitzer, which implicitly proves a true mod-Cauchy convergence result. Morever, this result is very interesting from the probabilistic point of view, since it concerns that most beautiful of objects, the (complex) Brownian motion.

Here is the result: consider such a complex Brownian motion

$W(t)=W_1(t)+iW_2(t),$

for non-negative t, where the real and imaginary parts are themselves independent standard Brownian motions on the real line, except that the real part is started at

$W_1(0)=R,$

for some fixed R> 0 (for instance R=1).

Because almost surely the Brownian motion is not zero, one can define a continuous process

$\theta_R(t)=\mathrm{Im} \log W(t),\quad\quad \theta_R(0)=0,$

which measures the argument of the Brownian motion and its winding around the origin. Spitzer’s result is then the fact that

$\frac{\theta_R(t)}{\log \sqrt{t}}$

converges in law, as t goes to infinity, to a Cauchy distribution with parameter 1 (it had already been noticed by P. Lévy that the argument of the Brownian motion is not square-integrable).

However, his argument is more precise: using relations with suitable PDE’s, he manages to compute exactly the Fourier transform of the law of the argument in terms of Bessel functions: for u≥0 (this characteristic function is even), we have

$\mathbf{E}(e^{iu\theta_R(t)})=\sqrt{\frac{\pi R^2}{8t}}e^{-R^2/(4t)}\Bigl\{I_{(u-1)/2}\Bigl(\frac{R^2}{4t}\Bigr)+I_{(u+1)/2}\Bigl(\frac{R^2}{4t}\Bigr)\Bigr\}$

where Iν(x) is the I-Bessel function of index ν.

Since the first term in the power series expansion (at 0) of the Bessel function is given by

$I_{\nu}(x)=\frac{(x/2)^{\nu}}{\Gamma(\nu+1)}+O(x^{\nu+1}),$

it follows easily that we have a mod-Cauchy convergence:

$\exp(A(t)|u|) \mathbf{E}(e^{iu \theta_R(t)})\longrightarrow \Phi(u),$

as t goes to infinity, with parameters and limiting function given by

$A(t)=\log\frac{\sqrt{8t}}{R},\quad\quad \Phi(u)=\frac{\sqrt{\pi}}{\Gamma((|u|+1)/2)}=\frac{\Gamma(1/2)}{\Gamma((|u|+1)/2)}.$