MA5161. Take home exam

Exercise 1. The Hermite polynomial of order $n$ is defined as
$H_n (x)=(-1)^n e^{\frac{x^2}{2}} \frac{d^n}{dx^n} e^{-\frac{x^2}{2}}.$

Compute $H_0, H_1,H_2,H_3$ .
Show that if $(B_t)_{t \ge 0}$ is a Brownian motion, then the process $\left(t^{n/2}H_n (\frac{B_t}{\sqrt{t}})\right)_{t \ge 0}$ is a martingale.
Show that
$t^{n/2}H_n (\frac{B_t}{\sqrt{t}})=n! \int_0^t \int_0^{t_1} ... \int_0^{t_{n-1}} dB_{s_1}...dB_{s_n}.$

Exercise 2. (Probabilistic proof of Liouville theorem) By using martingale methods, prove that if $f:\mathbb{R}^n \to \mathbb{R}$ is a bounded harmonic function, then $f$ is constant.

Exercise 3. Show that if $(M_t)_{t \ge 0}$ is a local martingale of a Brownian filtration $(\mathcal{F}_t)_{t \ge 0}$ , then there is a unique progressively measurable process $(u_t)_{t \ge 0}$ such that for every $t \ge 0$ , $\mathbb{P} \left(\int_0^t u_s^2 ds < +\infty \right)=1$ and $M_t=\mathbb{E} (M_0)+ \int_0^{t} u_s dB_s.$
Exercise 4 [Skew-product decomposition]
Let $(B_t)_{t \ge 0}$ be a complex Brownian motion started at $z \neq 0$ .

Show that for $t \ge 0$ ,
$B_t=z \exp\left( \int_0^t \frac{dB_s}{B_s} \right).$
Show that there exists a complex Brownian motion $(\beta_t)_{t \ge 0}$ such that
$B_t=z \exp{\left( \beta_{\int_0^t \frac{ds}{\rho_s^2} }\right)},$
where $\rho_t =| B_t |$ .
Show that the process $(\rho_t)_{t \ge 0}$ is independent from the Brownian motion $(\gamma_t)_{t \ge 0}=(\mathbf{Im} ( \beta_t))_{t \ge 0}$ .
We denote $\theta_t=\mathbf{Im}\left( \int_0^t \frac{dB_s}{B_s} \right)$ which can be interpreted as a winding number around 0 of the complex Brownian motion paths. For $r>| z|$ , we consider the stopping time
$T_r =\inf \{ t \ge 0, | B_t | = r \}.$
Compute for every $r>| z|$ , the distribution of the random variable
$\frac{1}{\ln (r/|z|)}\theta_{T_r}.$
Prove Spitzer theorem: In distribution, we have the following convergence
$\frac{ 2 \theta_t}{\ln t} \to_{+\infty} C,$
where $C$ is a Cauchy random variable with parameter 1 that is a random variable with density $\frac{1}{\pi (1+ x^2)}$ .