Sheet 12

Exercise 1 (Brownian Motion Unbounded).

Prove that the Brownian motion is unbounded without using Exercise 3.

Hint.

Borel Cantelli.

Solution.

First note that we have

$\mathbb{P}(\underset{t\to \mathrm{\infty }}{lim\; sup}{X}_t=\mathrm{\infty })=\underset{k\to \mathrm{\infty }}{lim}\mathbb{P}(\underset{t\to \mathrm{\infty }}{lim\; sup}{X}_t\ge k)$ (1)

because of

$\mathbb{P}(\underset{t\to \mathrm{\infty }}{lim\; sup}{X}_t=\mathrm{\infty })$ $=\mathbb{P}(\underset{t\to \mathrm{\infty }}{lim}\left(\underset{s\ge t}{sup}{X}_t\right)=\mathrm{\infty })$

$=\mathbb{P}(\forall k\ge 0,\exists t_0\ge 0,\forall t\ge t_0:(\exists s\ge t:X_t\ge k))$

$=\mathbb{P}(\forall k\ge 0,\forall t\ge 0,\exists s\ge t:X_t\ge k)$

$=\underset{k\to \mathrm{\infty }}{lim}\mathbb{P}(\underset{t\to \mathrm{\infty }}{lim\; sup}{X}_t\ge k),$

where we observed that the claim holds for all if it holds for all $t\ge t_0$ in the third step and then used continuity of the probability distribution to move the first $\forall$ outside in the last step.

We also switch to a discretized version of the Brownian motion ${(B_n)}_{n\in \mathbb{N}}$

$\mathbb{P}(\underset{t\to \mathrm{\infty }}{lim\; sup}|B_t|=\mathrm{\infty })$ $=\underset{k\to \mathrm{\infty }}{lim}\mathbb{P}(\underset{t\to \mathrm{\infty }}{lim\; sup}|B_t|\ge k)$

$\ge \underset{k\to \mathrm{\infty }}{lim}\mathbb{P}(\underset{n\to \mathrm{\infty }}{lim\; sup}|B_n|\ge k)$

$\ge \underset{k\to \mathrm{\infty }}{lim}\mathbb{P}(\underset{n\to \mathrm{\infty }}{lim\; sup}|B_{n+1}-B_n|\ge 2k).$

For the last inequality we used that, if for any $n$

$\underset{m\ge n}{sup}|B_{m+1}-B_m|\ge 2k,$

then for some $m\ge n$ either $|B_m|\ge k$ or $|B_{m+1}|\ge k$ due to the triangle inequality and therefore

$\underset{m\ge n}{sup}|B_m|\ge k.$

But since $\{|B_{n+1}-B_n|\ge 2k\}$ are independent events and

${\displaystyle \sum _{n=0}^{\mathrm{\infty }}}\mathbb{P}(|B_{n+1}-B_n|\ge 2k)={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}\underset{=:c>0}{\underbrace{\mathbb{P}(|B_1-B_0|\ge 2k)}}=\mathrm{\infty },$

we have by Borel-Cantelli

$\mathbb{P}(\underset{n\to \mathrm{\infty }}{lim\; sup}|B_{n+1}-B_n|\ge 2k)$ $=1.\mathit{∎}$

Exercise 2 (Brownian Bridge).

A stochastic process ${(X_t)}_{0\le t\le 1}$ is called a Brownian bridge if it is a continuous centred Gaussian process with covariance function given by

$$\text{Cov}(X_s,X_t)=s(1-t),s\le t.$$

Let $(B_t,t\ge 0)$ be a standard Brownian motion. Show that the process

$$X_t=B_t-t{B}_1,t\in [0,1]$$

is a Brownian bridge. Moreover, prove that $B_1$ is independent of $(X_t,t\in [0,1])$.

Solution.

Because $B$ is a centered continuous Gaussian process, we know that $X$ is a centered continuous Gaussian process, because any linear combination of a finite gaussian vector $(X_{t_1},\mathrm{\dots },X_{t_n})$ is a linear combination of $(B_1,B_{t_1},\mathrm{\dots },B_{t_n})$ and therefore gaussian. For $s\le t$, the covariance is

$\text{Cov}(X_s,X_t)=\text{Cov}(B_s-s{B}_1,B_t-t{B}_1)=𝔼\left[(B_s-s{B}_1)(B_t-t{B}_1)\right]=s(1-t).$

To show that $B_1$ is independent from $(X_t,t\in [0,1])$, it is enough to show independence from a finite dimensional marginal $(X_{t_1},\mathrm{\dots },X_{t_n})$. But because $(B_1,X_{t_1},\mathrm{\dots },X_{t_n})$ is a Gaussian vector and

$\text{Cov}(B_1,X_{t_i})$ $=𝔼[B_1(B_{t_i}-t_i{B}_1)]$

$=\text{Cov}(B_1,B_{t_i})-t_i\text{Var}(B_1)$

$=\min (1,t_i)-t_i\min (1,1)$

$=0$

we have the independence between $B_1$ and $(X_{t_1},\mathrm{\dots },X_{t_n})$ by Sheet 8 Exercise 4. ∎

Exercise 3 (Brownian Growth Rate).

Let $(B_t\ge 0)$ be a Brownian motion.

1. Prove that, for any $k>0$, we have

   $$\mathbb{P}(inf\{t>0:B_t\ge k\sqrt{t}\}=0)=1.$$

   Hint.

   Look for a similar statement in the lecture notes.

   Solution.

   This is the same proof as in Theorem 8.2.12, but we drop the absolute value.

   We define the same sets (just without the absolute value)

   $A$ $:=\{inf\{t\ge 0:B_t\ge k\sqrt{t}\}=0\}$

   $A_s$ $:=\{inf\{t\ge 0:B_t\ge k\sqrt{t}\}\le s\}$

   And also have $A={\bigcap }_{s>0}{A}_s\in {ℱ}_{0+}$. Due to Blumenthal’s zero-one law we have $\mathbb{P}(A)\in \{0,1\}$, so we just need to show it is not zero. With the same argument as in the lecture we can finish the proof

   $\mathbb{P}(A)$ $=\underset{s\to 0}{lim}\mathbb{P}(A_s)$

   $=\underset{s\to 0}{lim}\mathbb{P}\left(inf\{t\ge 0:{\displaystyle \frac{B_t}{\sqrt{t}}}\ge k\}\right)$

   $\ge \underset{s\to 0}{lim}\mathbb{P}({\displaystyle \frac{B_s}{\sqrt{s}}}\ge k)$

   $\stackrel{\text{scaling}}{=}\mathbb{P}(B_1\ge k)$

   $>0.\mathit{∎}$

2. Prove that,

   $\underset{t\to \mathrm{\infty }}{lim\; sup}{\displaystyle \frac{B_t}{\sqrt{t}}}=\mathrm{\infty }\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}\underset{t\to \mathrm{\infty }}{lim\; inf}{\displaystyle \frac{B_t}{\sqrt{t}}}=-\mathrm{\infty }\mathit{\hspace{1em}}\text{a.s..}$

   Hint.

   Use time inversion.

   Solution Spirit.

   By applying the previous statement to $t{B}_{1/t}$ we want to essentially argue

   $1$ $=\mathbb{P}(inf\{t>0:{\displaystyle \frac{t{B}_{1/t}}{\sqrt{t}}}\ge k\}=0)$

   For the supremum over times $s$ to be infinite, there can be no largest $s$ for which $\frac{B_s}{\sqrt{s}}\ge k$. There has to be an infinite amount of these $s$ which implies almost surely

   $\underset{s\ge t}{lim\; sup}{\displaystyle \frac{B_s}{\sqrt{s}}}\ge k.$

   Since $k$ is arbitrary we get the claim.

   But the transformation $t=\frac{1}{s}$ is of course not quite formally correct. Which is why there is a formal solution below. ∎

   Formal Solution.

   Using (1) and continuity of probability measures, we have

   $\mathbb{P}(\underset{t\to \mathrm{\infty }}{lim\; sup}{\displaystyle \frac{B_t}{\sqrt{t}}}=\mathrm{\infty })$ $=\underset{k\to \mathrm{\infty }}{lim}\underset{t\to \mathrm{\infty }}{lim}\mathbb{P}(\underset{s\ge t}{sup}{\displaystyle \frac{B_s}{\sqrt{s}}}\ge k).$

   It is therefore sufficient to show for every $k,t$ that

   $\mathbb{P}(\underset{s\ge t}{sup}{\displaystyle \frac{B_s}{\sqrt{s}}}\ge k)=1.$ (2)

   So assume some given $k,t$. Because $\tilde{s}{B}_{1/\tilde{s}}$ is a Brownian motion, we have almost surely

   $inf\{\tilde{s}>0:\sqrt{\tilde{s}}{B}_{1/\tilde{s}}\ge k\}=0.$

   So there exists for all $\omega \in N^c$ where $N$ is a null set, such that

   $\sqrt{\tilde{s}}{B}_{1/\tilde{s}}(\omega )\ge k.$

   Defining $s:=1/\tilde{s}$ implies $s\ge t$ and results in

   ${\displaystyle \frac{B_s}{\sqrt{s}}}(\omega )\ge k.$

   But this implies that forall $\omega \in N^c$ we have

   $\underset{s\ge t}{sup}{\displaystyle \frac{B_s(\omega )}{\sqrt{s}}}\ge k.$

   This proves (2). And because $-B_t$ is also a brownian motion we also have

   $\underset{t\to \mathrm{\infty }}{lim\; inf}{\displaystyle \frac{B_t}{\sqrt{t}}}$ $=-\left(\underset{t\to \mathrm{\infty }}{lim\; sup}{\displaystyle \frac{-B_t}{\sqrt{t}}}\right)=-\mathrm{\infty }.\mathit{∎}$

Exercise 4 (Compound Poisson is Lévy).

Show that the compound Poisson process $X={(X_t)}_{t\ge 0}$ is a Lévy Process. You can assume that the Poisson process $N={(N_t)}_{t\ge 0}$ is a Lévy Process.

Hint.

Show $𝔼\mathit{}\mathrm{[}f\mathit{}\mathrm{(}{X}_{\overline{t}}\mathrm{-}{X}_{\underset{\mathrm{¯}}{t}}\mathrm{)}\mathit{}g\mathit{}\mathrm{(}{X}_{\overline{s}}\mathrm{-}{X}_{\underset{\mathrm{¯}}{s}}\mathrm{)}\mathrm{]}$ factorizes.

Solution.

1. Start in 0 (because $N_0=0$).

2. Right continuity with left limits

   For any $t$ we have ${lim}_{s\downarrow t}{N}_s=N_t$ and $N_s\in \mathbb{N}$. Therefore there exists $ϵ>0$ such that $N_s=N_t$ for all $s\in [t,t+ϵ]$. But then by definition for all $s\in [t,t+ϵ]$ we have

   $X_s={\displaystyle \sum _{i=1}^{N_s}}{Y}_i={\displaystyle \sum _{i=1}^{N_t}}{Y}_i=X_t$

   therefore $X$ is right continuous in $t$. Let $N_{-t}:={lim}_{s\uparrow t}{N}_s$ be the left limit of the process $N$ in $t$. Again because $N_s\in \mathbb{N}$ there exists $ϵ>0$ such that $N_s=N_{-t}$ for all $s\in [t-ϵ,t)$. For all $s_1,s_2\in [t-ϵ,t)$ we have $X_{s_1}=X_{s_2}$ so we can define the left limit of $X$ in $t$ as $X_{-t}:=X_{t-ϵ}$ and get convergence

   $\underset{s\uparrow t}{lim}{X}_s=X_{-t}.$

   Because $t$ was arbitrary we are done.

3. Independent increments

   We first observe that for $s\le t$, we have $N_s\le N_t$ since for independent waiting times ${\xi }_i\sim \text{Exp}(\lambda )$ we have

   $N_s=\max \underset{A(s)}{\underbrace{\{k\in {\mathbb{N}}_0:{\displaystyle \sum _{i=1}^k}{\xi }_i\le s\}}}\le \max \underset{A(t)}{\underbrace{\{k\in {\mathbb{N}}_0:{\displaystyle \sum _{i=1}^k}{\xi }_i\le t\}}}=N_t$

   due to $A(s)\subseteq A(t)$.

   So to show that $X_{\overline{t}}-X_{\underset{¯}{t}}$ is independent of $X_{\overline{s}}-X_{\underset{¯}{s}}$ for $\underset{¯}{s}\le \overline{s}\le \underset{¯}{t}\le \overline{t}$ we use

   $𝔼\left[f(X_{\overline{t}}-X_{\underset{¯}{t}})g(X_{\overline{s}}-X_{\underset{¯}{s}})\right]$

   $=𝔼\left[f\left({\displaystyle \sum _{i=N_{\underset{¯}{t}}+1}^{N_{\overline{t}}}}{Y}_i\right)g\left({\displaystyle \sum _{i=N_{\underset{¯}{s}}+1}^{N_{\overline{s}}}}{Y}_i\right)\right]$

   $\stackrel{N⟂⟂Y}{=}{\displaystyle \sum _{k_{\underset{¯}{s}},k_{\overline{s}},k_{\underset{¯}{t}},k_{\overline{t}}}}𝔼\left[f\left({\displaystyle \sum _{i=k_{\underset{¯}{t}}+1}^{k_{\overline{t}}}}{Y}_i\right)g\left({\displaystyle \sum _{i=k_{\underset{¯}{s}}+1}^{k_{\overline{s}}}}{Y}_i\right)\right]\mathbb{P}(N_{\underset{¯}{s}}=k_{\underset{¯}{s}},N_{\overline{s}}=k_{\overline{s}},N_{\underset{¯}{t}}=k_{\underset{¯}{t}},N_{\overline{t}}=k_{\overline{t}}).$

   Since $\overline{s}\le \underset{¯}{t}$ implies $N_{\overline{s}}\le N_{\underset{¯}{t}}$ we can assume $k_{\overline{s}}\le k_{\underset{¯}{t}}$, which means that the sums contain different $Y_i$ which are all independent and identically distributed so the expectation factorizes like so

   $𝔼\left[f\left({\displaystyle \sum _{i=k_{\underset{¯}{t}}+1}^{k_{\overline{t}}}}{Y}_i\right)g\left({\displaystyle \sum _{i=k_{\underset{¯}{s}}+1}^{k_{\overline{s}}}}{Y}_i\right)\right]$ $=𝔼\left[f\left({\displaystyle \sum _{i=k_{\underset{¯}{t}}+1}^{k_{\overline{t}}}}{Y}_i\right)\right]𝔼\left[g\left({\displaystyle \sum _{i=k_{\underset{¯}{s}}+1}^{k_{\overline{s}}}}{Y}_i\right)\right]$

   $=𝔼\left[f\left({\displaystyle \sum _{i=1}^{k_{\overline{t}}-k_{\underset{¯}{t}}}}{Y}_i\right)\right]𝔼\left[g\left({\displaystyle \sum _{i=1}^{k_{\overline{s}}-k_{\underset{¯}{s}}}}{Y}_i\right)\right].$

   Defining ${\mathrm{\Delta }}_s:=k_{\overline{s}}-k_{\underset{¯}{s}}$ and ${\mathrm{\Delta }}_t:=k_{\overline{t}}-k_{\underset{¯}{t}}$ we can also write

   $\mathbb{P}(N_{\underset{¯}{s}}=k_{\underset{¯}{s}},N_{\overline{s}}=k_{\overline{s}},N_{\underset{¯}{t}}=k_{\underset{¯}{t}},N_{\overline{t}}=k_{\overline{t}})$

   $=\mathbb{P}(N_{\underset{¯}{s}}=k_{\underset{¯}{s}},N_{\underset{¯}{t}}=k_{\underset{¯}{t}},N_{\overline{s}}-N_{\underset{¯}{s}}={\mathrm{\Delta }}_s,N_{\overline{t}}-N_{\underset{¯}{t}}={\mathrm{\Delta }}_t)$

   so if we sum over $k_{\underset{¯}{s}},{\mathrm{\Delta }}_s$ instead of $k_{\underset{¯}{s}},k_{\overline{s}}$ and $k_{\underset{¯}{t}},{\mathrm{\Delta }}_t$ instead of $k_{\underset{¯}{t}},k_{\overline{t}}$ and use independence of the increments of $N$, we get

   $𝔼\left[f(X_{\overline{t}}-X_{\underset{¯}{t}})g(X_{\overline{s}}-X_{\underset{¯}{s}})\right]$

   $={\displaystyle \sum _{{\mathrm{\Delta }}_s,{\mathrm{\Delta }}_t}}𝔼\left[f\left({\displaystyle \sum _{i=1}^{{\mathrm{\Delta }}_t}}{Y}_i\right)\right]𝔼\left[g\left({\displaystyle \sum _{i=1}^{{\mathrm{\Delta }}_s}}{Y}_i\right)\right]\mathbb{P}(N_{\overline{s}}-N_{\underset{¯}{s}}={\mathrm{\Delta }}_s,N_{\overline{t}}-N_{\underset{¯}{t}}={\mathrm{\Delta }}_t)$

   $=\underset{=𝔼[f(X_{\overline{t}}-X_{\underset{¯}{t}})]}{\underbrace{{\displaystyle \sum _{{\mathrm{\Delta }}_t}}𝔼\left[f\left({\displaystyle \sum _{i=1}^{{\mathrm{\Delta }}_t}}{Y}_i\right)\right]\mathbb{P}(N_{\overline{t}}-N_{\underset{¯}{t}}={\mathrm{\Delta }}_t)}}\underset{=𝔼[g(X_{\overline{s}}-X_{\underset{¯}{s}})]}{\underbrace{{\displaystyle \sum _{{\mathrm{\Delta }}_s}}𝔼\left[g\left({\displaystyle \sum _{i=1}^{{\mathrm{\Delta }}_s}}{Y}_i\right)\right]\mathbb{P}(N_{\overline{s}}-N_{\underset{¯}{s}}={\mathrm{\Delta }}_s)}}.$

   Therefore the increments of $X$ are independent.

4. Stationary Increments

   With the stationarity of the increments of $N$ and the previous section we get

   $𝔼[f(X_{\overline{t}}-X_{\underset{¯}{t}})]$ $={\displaystyle \sum _{{\mathrm{\Delta }}_t}}𝔼\left[f\left({\displaystyle \sum _{i=1}^{{\mathrm{\Delta }}_t}}{Y}_i\right)\right]\mathbb{P}(N_{\overline{t}}-N_{\underset{¯}{t}}={\mathrm{\Delta }}_t)$

   $={\displaystyle \sum _{{\mathrm{\Delta }}_t}}𝔼\left[f\left({\displaystyle \sum _{i=1}^{{\mathrm{\Delta }}_t}}{Y}_i\right)\right]\mathbb{P}(N_{{\mathrm{\Delta }}_t}-N_0={\mathrm{\Delta }}_t)$

   $=𝔼[f(X_{{\mathrm{\Delta }}_t}-X_0)].\mathit{∎}$