Brownian Motion and Stochastic Calculus Chapter 1 Section 2

PDF of Solutions

Problem 1.2.2 (not mine)

Let $\mathcal C$ be the collection of subsets $A$ of $\Omega$ such that $1_A(\omega_1) = 1_A(\omega_2)$ . Note that $\mathcal C$ is a $\sigma$ -algebra (simple to show). Assume $X_s (\omega_1) = X_s (\omega_2)$ for some $s$ . Then $1_A (\omega_1)= 1_A(\omega_2)$ for all $A\in \sigma(X_s)$ so $\sigma(X_s) \subset \mathcal C$ . Since $X_t(\omega_1) = X_t(\omega_2)$ for all $t\in [0, T(\omega_1)]$ have $\sigma(X_{T(\omega_1)}) \subset \mathcal C$ . So $1_A(\omega_1) = 1_A(\omega_2)$ for all $A$ in $\sigma(X_{T(\omega_1)})$ . Note that $B= \{ \omega: T(\omega) = T(\omega_1)\}$ in $\mathcal F_{T(\omega)}^X$ and $1_B(\omega_1) = 1$ . So $1_B (\omega_2) = 1$ which implies $\omega_2 \in B$ . Thus $T(\omega_1) = T(\omega_2)$ as well.

Problem 1.2.6

We wish to show that $\{ H_\Gamma < t \} \in \mathcal F_t$ for all $t$ . For $t=0$ , $\{H_\Gamma < 0\} = \emptyset$ so is in $\mathcal F_0$ . Let $t>0$ . Consider the set $\{ H_\Gamma <t\}$ .

$\{ H_\Gamma < t \} = \{ \omega \in \Omega : H_\Gamma ( \omega) <t \}$

$= \{ \omega \in \Omega : inf ( s \geq 0 : X_s ( \omega) \in \Gamma) < t \}$

Let $\omega \in \{H_\Gamma < t\}$ . By the definition of infimum, there is some $s<t$ such that $X_s(\omega) \in \Gamma$ . Since $\Gamma$ is open there is some open ball $B_\epsilon$ such that $X_s(\omega) \in B_\epsilon \subset \Gamma$ . Since $X_t(\omega)$ is right continuous there is some rational number $r$ such that $s \leq r < t$ and $X_r(\omega) \in B_\epsilon \subset \Gamma$ .

In fact, we can see that $\omega \in \{H_\Gamma < t\}$ if and only if there is some rational number $r<t$ such that $X_r(\omega) \in \Gamma$ . (We have shown one direction, other direction is trivial). Thus:

$\{H_\Gamma < t\} = \bigcup_{r\in \mathbb Q \cap [0, t)} \{ X_r \in \Gamma\}$

If $\{ X_r \in \Gamma\} \in \mathcal F_t$ for each r, we are done (because $\mathcal F_t$ is closed under countable unions). Note $X_t$ is adapted so $\{ X_r \in \Gamma\} \in \mathcal F_r \subset \mathcal F_t$ .

Problem 1.2.7

We wish to show that $\{H_\Gamma \leq t\} \in \mathcal F_t$ for all $t$ . Begin with $t=0$ .

$\{ H_\Gamma \leq 0\} = \{ \omega \in \Omega: X_0(\omega) \in \Gamma\}$

$= \{X_0 \in \Gamma\}$

Since $X$ is adapted, $\{X_0 \in \Gamma\} \in \mathcal F_0$ . (Does not rely on fact that $\Gamma$ is closed).

Now assume $t>0$ . Let us instead show that $\{H_\Gamma \leq t\}^C = \{ H_\Gamma > t\} \in \mathcal F_t$ . We argue that for closed set $\Gamma$ that

$\{ H_\Gamma > t\} = \bigcap_{s\leq t} \{X_s \in \Gamma^C\}$

If $\omega \in \{H_\Gamma > t\}$ then it is certainly true that $X_s(\omega) \in \Gamma^C$ for all $s\leq t$ , so $\{ H_\Gamma > t\} \subset \cap\{X_s \in \Gamma^C\}$

Now assume $\omega \in \cap \{X_s \in \Gamma^C\}$ . It is clear that $H_\Gamma(\omega) \not < t$ . But is it possible that $X_t(\omega) \in \Gamma^C$ and $t= H_\Gamma(\omega)$ ? No. If $t= H_\Gamma(\omega)$ there are two possibilities. Either $X_t(\omega) \in \Gamma$ (but have already seen assumed that $X_t(\omega) \in \Gamma^C$ so contradiction). OR, there is a sequence $\{t_n\}$ such that $t\leq t_n$ , $t_n \to t$ and $X_{t_n}(\omega) \in \Gamma$ for each $t_n$ . Then by continuity of sample paths $X_{t_n}(\omega) \to X_t(\omega)$ . Thus $X_t(\omega)$ is a limit point of $\Gamma$ . Since $\Gamma$ is closed, $X_t(\omega) \in \Gamma$ . Thus, $X_t(\omega) \in \Gamma \cap \Gamma^C$ , again a contradiction. Hence $\cap_{s\leq t} \{X_s \in \Gamma^C\}\subset \{ H_\Gamma > t\}$ .

Now we just need to show $\cap_{s\leq t} \{X_s \in \Gamma^C\}$ is in fact a countable union of measurable sets. This follows directly from continuity of sample paths.

$\bigcap_{s\leq t} \{X_s \in \Gamma^C\}= \bigcap_{r\in \mathbb Q \cap [0,t]} \{X_r \in \Gamma^C\}$

And $\{X_r \in \Gamma^C\} \in \mathcal F_r \subset \mathcal F_t$ for each $r$ by the fact that $X_t$ is adapted.

Problem 1.2.10

i). We wish to show that $\{ S+T \leq t \} \in \mathcal F_t$ . This is equivalent to showing that $\{S + T > t \} \in \mathcal F_t$ . Note that :

$\{S + T > t \} = \{ S \geq t , T > 0\} \cup \{ S > 0, T \geq t \} \cup \{ S<t, T<t, S+T > t \}$

The first two sets on the right hand side are in $\mathcal F_t$ by our preliminary assumptions. Consider the last set.

Assume that $S,T <t$ and let $\{ q_n \}_{n=1}^\infty$ be the set of all rational numbers in $[0,t)$ . Note that

$\{ T> q_n \} = \cup_{k=m}^\infty \{T \geq q_n + {1\over k}\}$

where $m$ can be chosen to be sufficiently large so that $q_n + {1\over m} <t$ . Then $\{ T> q_n\} \in \mathcal F_t$ for each $q_n$ and so $\{S\geq t- q_n , T> q_n\}\in \mathcal F_t$ .

Finally note that

$\{ S,T < t, S+T > t \} = \cup_{n=1}^\infty \{ S \geq t- q_n , T > q_n \} \in \mathcal F_t$

ii).

$\{ S+ T > t\} = \{ T>t\} \cup \{T< t, S\geq t\} \cup \{ S, T <t, S+T > t\}$

Recall if $T$ is a stopping time it is also an optional time, so we can use the same work in previous part to show last set on right hand side is in $\mathcal F_t$ .

Problem 1.2.13

Assume $A\in \mathcal F_T$ , i.e $A\cap \{T \leq t\} \in \mathcal F_t$ for all $t$ . Note

$A^C \cap \{T \leq t\} = \{ T\leq t\} \setminus \{ A\cap \{T \leq t\} \}$

The right hand side is in $\mathcal F_t$ be definition of $\mathcal F_T$ . So $A^C \in \mathcal F_T$ . Now let $A_n \in \mathcal F_T$ for each $n$ . Note that

$\bigg{(} \bigcup A_n \bigg{)} \cap \{T \leq t\} = \bigcup (A_n \cap \{ T\leq t\})$

Since $\mathcal F_t$ is closed under countable unions, $\cup A_n \in \mathcal F_T$ . So $\mathcal F_T$ is closed under complements and countable unions, so it is a $\sigma$ -algebra.

Now assume that $T(\omega) =t$ for some fixed $t$ and all $\omega\in \Omega$ . Let $A \in \mathcal F_t$ . Then $A\in \mathcal F$ as well. Note that for $\{ T \leq s\} = \emptyset$ for $s< t$ and $\{ T \leq s\} = \Omega$ for $s\geq t$ . Then, for $s <t$ we have:

$A\cap \{T\leq s\} = A\cap \emptyset = \emptyset \in \mathcal F_s$

On the other hand, for $s\geq t$ , (noting that $\mathcal F_t \subset \mathcal F_s$ ) we have

$A \cap \{T\leq s\} = A \cap \Omega = A \in \mathcal F_s$

Thus, if $A\in \mathcal F_t$ we see that $A\in \mathcal F_T$ . Trivial to show that $\mathcal F_T \subset \mathcal F_t$ .

Problem 1.2.14

Wish to show that $\{ S\leq t\} \in \mathcal F_t$ for all $t$ . One equivalent definition of $S\in \mathcal F_T$ is that $\{ S\leq s \} \in \mathcal F_T$ for all $s\in \mathbb R$ . Note $\{ S\leq s \} \in \mathcal F_T$ by definition means $\{S\leq s\} \cap \{T \leq t\} \in \mathcal F_t$ for all $t$ . So $\{S\leq s\} \cap \{T \leq t\} \in \mathcal F_t$ for all $s, t$ . In particular, let $s=t$ . Then for all $t$ we have:

$\{S \leq t\} \cap \{T \leq t\} \in \mathcal F_t$

But since $S\geq T$ , $\{S \leq t\} \cap \{T \leq t\} = \{ S \leq t\}$ . Thus, $\{S \leq t\} \in \mathcal F_t$ for all $t$ . Thus, $S$ is a stopping time.

Problem 1.2.17 (not mine)

i). Recall definition of restriction to $\{ T\leq S\}$ . For all $A\in \mathcal F$ , we take $A\cap \{T \leq S\}$ . Note that since $\{T \leq S\} \in \mathcal F_T$ we have $A\cap \{T\leq S\} \in \mathcal F_T$ for all $A\in \mathcal F_T$ . Similarly, $A\cap \{T\leq S\} \in \mathcal F_S$ . Thus $A\cap \{T\leq S\} \in \mathcal F_{S\wedge T}$ .

Want to show that $E[ Z| \mathcal F_T] = E[ Z| \mathcal F_{T\wedge S}]$ so pick a set $A$ in larger $\sigma$ -algebra, $\mathcal F_T$ . Then for such an $A$ (and on $\{T\leq S\}$ ) we have:

$E[ 1_A 1_{T\leq S} E[ Z| \mathcal F_T] ] =E[ 1_{A \cap \{T\leq S\}} E[Z|\mathcal F_T] ]$

$=E[ 1_{A \cap \{T\leq S\}} Z ]$

$= E[ 1_{A \cap \{T \leq S \} } E[Z| \mathcal F_{T \wedge S } ] ]$

$E[ 1_A 1_{T\leq S} E[ Z| \mathcal F_{T\wedge S}] ]$

ii). Similar to previous, we can show that $E[Z| \mathcal F_S ] = E[ Z| \mathcal F_{T\wedge S}]$ on $\{S<T\}$ . Why ? In previous, didn’t use specific form of inequality ( $T\leq S$ versus $T<S$ ), just used fact that these sets are in $\mathcal F_{T\wedge S}$ .

Now to show (ii), will show it on the two disjoint sets $\{ T\leq S\}$ and $\{ S<T\}$. By (i) we see that

$1_{\{ T\leq S\}}E[ E[ Z | \mathcal F_T ] | \mathcal F_{T\wedge S}] = E[ 1_{\{ T\leq S\}} E[ Z | \mathcal F_T ] | \mathcal F_{T\wedge S}]\ \ since\ \ \{ T\leq S\} \in \mathcal F_{T\wedge S}$