Sheet 13

Prof. Leif Döring, Felix Benning

^†^†course: Wahrscheinlichkeitstheorie 1^†^†semester: FSS 2022^†^†tutorialDate: 30.05.2022^†^†dueDate: 10:15 in the exercise on Monday 30.05.2022

Exercise 1 (Brownian Bridge Time Change).

Let $(B_{t},t\geq 0)$ be a standard Brownian motion. Set $X_{1}=0$ and

X_{t}=(1-t)B_{\frac{t}{1-t}},\quad t\in[0,1).

Show that $(X_{t},t\in[0,1])$ is a Brownian bridge (with continuous paths).

Solution.

$\tilde{B}:t\mapsto tB_{1/t}$ is a Brownian motion by Proposition 8.2.8 (time inversion) and therefore almost surely continuous on $[0,\infty)$ . So due to $\lim_{t\to 1}\tfrac{1-t}{t}=0$ , we have

\displaystyle\lim_{t\to 1}\tilde{B}_{\tfrac{1-t}{t}}=\tilde{B}_{0}=0.

But this implies continuity of $X_{t}$ in $t=1$ , because

\displaystyle X_{t}=t\tfrac{1-t}{t}B_{\tfrac{t}{1-t}}=t\tilde{B}_{\tfrac{1-t}{% t}}\to 0=X_{1}.

Further, we know that $X$ is a centred continuous Gaussian process because $B$ is a centred continuous Gaussian process. For $0\leq s\leq t\leq 1$ , we have

\displaystyle s-st\leq t-st

\displaystyle\implies s(1-t)\leq t(1-s)\implies\frac{s}{1-s}\leq\frac{t}{1-t}

which implies

\displaystyle\text{Cov}(X_{s},X_{t})

\displaystyle=\mathbb{E}\left[(1-s)B_{\frac{s}{1-s}}(1-t)B_{\frac{t}{1-t}}% \right]=(1-s)(1-t)\min\{\tfrac{s}{1-s},\tfrac{t}{1-t}\}=s(1-t).

So $X_{t}$ is a centred continuous Gaussian process with the right covariance and endpoints and is therefore a Brownian Bridge. ∎

Exercise 2 (Converging Covariance).

If $X_{1},X_{2},\dots$ are centred Gaussian processes with covariance functions $K_{1},K_{2},\dots$ , and $K$ is another covariance function, then there exists a Gaussian process $X$ with covariance function $K$ with $X_{n}\stackrel{{\scriptstyle\text{fdd}}}{{\implies}}X$ if and only if $\lim_{n\to\infty}K_{n}(t,s)=K(t,s)$ for all $t,s\in I$ .

Solution.

If there exists such an $X$ , we have for any $t, s$ , $(X_{n}(t),X_{n}(s))\to(X(t),X(s))$ in distribution. Therefore

	$\displaystyle\varphi_{(X_{n}(t),X_{n}(s))}(x)$	$\displaystyle=\exp\left(-\frac{1}{2}x^{T}\begin{pmatrix}K_{n}(t,t)&K_{n}(t,s)% \\ K_{n}(s,t)&K_{n}(s,s)\end{pmatrix}x\right)$
		$\displaystyle\to\exp\left(-\frac{1}{2}x^{T}\begin{pmatrix}K(t,t)&K(t,s)\\ K(s,t)&K(s,s)\end{pmatrix}x\right)=\varphi_{(X(t),X(s))}(x).$

Taking the logarithm we get for all $x$

\displaystyle x^{T}\begin{pmatrix}K_{n}(t,t)&K_{n}(t,s)\\ K_{n}(s,t)&K_{n}(s,s)\end{pmatrix}x\to x^{T}\begin{pmatrix}K(t,t)&K(t,s)\\ K(s,t)&K(s,s)\end{pmatrix}x

Taking $x=(1,0)^{T}$ we get $K_{n}(t,t)\to K(t,t)$ , while $x=(0,1)^{T}$ results in $K_{n}(s,s)\to K(s,s)$ . And, due to symmetry of the covariance matrix, we get for $x=(1,1)^{T}$

	$\displaystyle 2K_{n}(s,t)$	$\displaystyle=(K_{n}(t,t)+2K_{n}(s,t)+K_{n}(s,s))-K_{n}(t,t)-K_{n}(s,s)$
		$\displaystyle=x^{T}\begin{pmatrix}K_{n}(t,t)&K_{n}(t,s)\\ K_{n}(s,t)&K_{n}(s,s)\end{pmatrix}x-K_{n}(t,t)-K_{n}(s,s)$
		$\displaystyle\to x^{T}\begin{pmatrix}K(t,t)&K(t,s)\\ K(s,t)&K(s,s)\end{pmatrix}x-K(t,t)-K(s,s)$
		$\displaystyle=2K(s,t).$

Since $s, t$ were arbitrary we have $\lim_{n\to\infty}K_{n}(s,t)=K(s,t)$ .

A Gaussian process $X$ with covariance function $K$ exists, because we can construct consistent finite dimensional distributions of $(X(t_{1}),\dots,X(t_{n})$ using

\displaystyle\Sigma=\begin{pmatrix}K(t_{1},t_{1})&\cdots&K(t_{1},t_{n})\\ \vdots&&\vdots\\ K(t_{n},t_{1})&\cdots&K(t_{n},t_{n})\end{pmatrix}

i.e. find $B$ such that $BB^{T}=\Sigma$ and apply $B$ to iid standard normal random variables, resulting in a gaussian vector with the desired distribution. For similarly defined $\Sigma_{n}$ , we have that all the entries of $\Sigma_{n}$ converging to $\Sigma$ by assumption. And therefore by continuity of sums and products $x^{T}\Sigma_{n}x\to x^{T}\Sigma x$ for all $x\in\mathbb{R}^{2}$ . But this implies that the characteristic functions of all finite dimensional marginals converge and therefore $X_{n}\stackrel{{\scriptstyle\text{fdd}}}{{\implies}}X$ . ∎