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\begin{document}

\title{Stat 992: Lecture 22\\
Brownian motion.}
\author{Moo K. Chung \tt{mchung@stat.wisc.edu}}
\maketitle \thispagestyle{empty}

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%\includegraphics[scale=0.8]{homework05-1.eps}
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\begin{enumerate}
\item {\em Spherical harmonics.} By combining the solution from
both ODEs, we have the complete set of solutions to the spherical
Helmholtz equation:
$$H_n^m(\theta,\psi) = \cos (m\psi)P_n^m(\cos \theta),$$
$$\widetilde{H}_n^m(\theta,\psi) = \sin (m\psi) P_n^m(\cos
\theta)$$ where $n=0,1,2,\cdots, m=0,1,\cdots,n.$ The spherical
harmonics satisfy
$$\Delta_{S^2} H_n^m + n(n+1)Y_n^m = \Delta_{S^2} \widetilde{H}_n^m +
n(n+1)Y_n^m=0.$$ Note that $\widetilde{H}_n^0=0$. The
corresponding eigenvalues are $\lambda_n = -n(n+1)$. These
harmonics are orthogonal but not normalized. It can be shown that
$$\|H_n^0\|^2 = \frac{4\pi}{2n+1}, \|H_n^m\|^2 =
\|\widetilde{H}_n^m\|^2 = \frac{2\pi (n+m)!}{(2n+1)(n-m)!}.$$ We
will assume $H_n^m$ and $\widetilde{H}_n^m$ are normalize by
dividing by the norms. For simplicity let us order spherical
harmonics $H_n^m$ and $\widetilde{H}_n^m$ by $H_j$. Then the
spherical harmonics can span any function $f \in L^2(S^2)$ so that
$$f(\theta,\psi) = \sum_{j = 0}^{\infty} c_jH_j$$ where
$c_j = \langle f, H_j \rangle$.

From S. Rosenberg's The Laplacian on a Riemannian Manifold (1997
Cambridge University Press), it can be shown that the spherical
heat kernel can be extended as
$$K_t(p,q) = \sum_{j=0}^{\infty} e^{-\lambda_j t}H_j(p)H_j(q).$$
Hence solution to diffusion smoothing on the unit sphere is
exactly expressed as $$f(t,q) = \sum_{j=0}^{\infty}e^{-\lambda_j
t}H_j(q) \int_0^{2\pi}\int_0^{\pi}  H_j(p)Y(p) \sin\theta \;
d\theta d\psi.$$

{\bf Problem 18.} In MATLAB, simulate 1D and 2D version of
Brownian motion based on the above properties and check ${\bf Var}
B(x) = x$ by simulating 10000 Brownian motions. What is the
corresponding 2D version ?\\
{\em Solution.} (Tulaya Limpiti). Simulation is based on the
property of the independent increment of Brownian motion.

\begin{figure}[t]
\centering
\renewcommand{\baselinestretch}{1}
\includegraphics[scale=0.45]{ima22-1.eps}
\caption{Top: 1D Brownian motion simulation, Bottom: 2D Brownian
motion simulation. {\tt plot3} command in {\tt MATLAB} is used to
plot it in 3D.}
\end{figure}
{\bf Problem 19.} Prove the covariance function of the
$W(x)=\frac{d B(x)}{dx}$ is proportional to the Dirac-delta
function.\\
{\em Solution.} (Tulaya Limpiti) Let $R_W(x,y) = \mathbb{E}
[W(x)W(y)]$. We will show that \bqn &&\int_0^u\int_0^u
f(x)f(y)R_W(x,y) \;dx dy \\
&=& \int_0^u\int_0^u f(x)f(y)\delta(x-y)\;dx dy \\
&=& \int_0^u f^2(y)dy\eqn for all possible smooth test function
$f$. The white noise process $W$ is defined as
$$\int_0^u f(y)W(y)\;dy = -f(u)B(u) + \int_0^u f'(y)B(y)\;dy$$
for all possible smooth test function $f$. Then
\bqn &&\int_0^u\int_0^u f(x)f(y)R_W(x,y) \;dx dy\\
&=& \mathbb{E} \int_0^u\int_0^u f(x)f(y)W(x)W(y) \;dx dy\\
&=& \mathbb{E} \Big[ \int_0^u f(y)W(y)\;dy  \Big]^2\\
&=&\mathbb{E} \Big[-f(u)B(u) +\int_0^u f'(y)W(y)\;dy\Big]^2\\
&=& f^2(u)\mathbb{E}B^2(u) -2f(u)\int_0^u
f'(y)\mathbb{E}[B(u)W(y)]\;dy \\&&+ \int_0^u \int_0^u
f'(x)f'(y)\mathbb{E}[B(x)B(y)] \;dxdy\\
&=&uf^2(u) - 2f(u)\int_0^u yf'(y)\;dy\\
&& + 2\int_0^u f'(y)\;dy\int_0^y xf'(x) \;dx.\eqn In the last
line, we used the fact $\mathbb{E}[B(x)B(y)]=\min(x,y)=x \mbox{ if
} x \leq y$. Now following Tulaya's argument, differentiate the
above expression and get $f^2(u)$. Then taking the integration, we
prove the statement. $\Box$


{\bf Problem 21.} Let $X$ be a Brownian motion with covariance
function given by $R(x,y) = \min(x,y)$. Find the eigenvalues and
eigenfunctions. Based on this eigensystem, expand the Brownian
motion via the Karhunen-Lo$\grave{e}$ve expansion. Based on the
Karhunen-Lo$\grave{e}$ve expansion, simulate Brownian motion in
$\tt{MATLAB}$.\\
{\em Solution.} (Tulaya Limpiti). This is a standard example in
the Karhunen-Lo$\grave{e}$ve expansion. For simplicity assume that
the Brownian motion is defined on $[0,1]$. We need to solve \bqn
\lambda \phi(x) &=& \int_0^u R(x,y)\phi(y)\;dy\\
&=& \int_0^1 \min(x,y)\phi(y)\;dy\\
&=& \int_0^x y\phi(y)\; dy + \int_x^1 x\phi(y)\; dy\eqn
Differentiating both sides with respect to $x$, we get
$$\lambda \phi'(x)= \int_x^1 \phi(y)\;dy,$$
$$\lambda \phi''(x) = -\phi(x).$$
Two independent solutions are $\sin \frac{x}{\sqrt \lambda}, \cos
\frac{x}{\sqrt \lambda}$. From the boundary condition $\phi(0)=0$,
$\phi(x) = \sin \frac{x}{\sqrt \lambda}$. From $\phi'(1)=0$, we
must have eigenvalues $$\lambda_j = \Big[\frac{2}{(2j
+1)\pi}\Big]^2.$$ Note that $\int_0^1 \phi^2(x) \;dx =
\frac{1}{2}$. So normalizing the eigenfunctions, we have
$\phi_j(x) = \sqrt 2 \sin \frac{x}{\sqrt \lambda_j}$. From the
Karhunen-Lo$\grave{e}$ve expansion, Brownian motion is written as
$$B(x) = \sum_{j=0}^{\infty} Z_j \psi_j(x)$$
where i.i.d. $Z_j \sim N(0,\lambda_j)$.


\begin{figure}[t]
\centering
\renewcommand{\baselinestretch}{1}
\includegraphics[scale=0.45]{ima22-2.eps}
\caption{Top: 1D Brownian motion simulation based on the
Karhunen-Lo$\grave{e}$ve expansion.}
\end{figure}


{\bf Problem 34.} When you smooth Brownian motion with the
covariance function $R(x,y)=\min(x,y)$ via kernel smoothing with
$K_{\sigma}$, what do you get?



\end{enumerate}
\end{document}


 \item {\em Excursion probability.}


Consider 1-dimensional smooth Gaussian random process $Y(x), x \in
\Omega=[0,1] \subset \mathbb{R}$. Let $N_h$ to be the number of
upcrossings over $h$ in $[0,1]$. Then \bqn P\Big(\sup_{x \in
[0,1]} \geq h\Big) &=& P(N_h \geq 1 \mbox{ or } X(0) > h)\\
&\leq& P(N_h \geq 1) + P(X(0) >h)\\
&\leq& \mathbb{E} N_h + P(X(0) >h).\eqn It can be shown that
$$\mathbb{E} N_h = \frac{}{2\pi\mathbb{E}}$$