Brownian Motion and Stochastic Calculus Chapter 1 Section 1

PDF of Solutions

Problem 1.1.5

Assume for each $t$ that $X_{t}(\omega)-Y_{t}(\omega)=0$ for $\omega\in A_t$ such that $P(A_t^C)=0$ (that is, Y is a modification of X). Let $B= \mathbb Q \cap [0,\infty)$ . For $t_k \in B$ let $A_k = \{ \omega | X_{t_k}(\omega)-Y_{t_k}(\omega)= 0\}$ . Note that $\cup A_k^C$ is a countable union of null sets and so is also a null set. Now let $\omega \in (\cup A_k^C)^C = \cap A_k$ . Then $X_{t_k}(\omega) - Y_{t_k}(\omega)=0$ for every rational number $t_k$ .

Fix $\omega \in \cap A_k$ . Each $t\in [0, \infty)$ is the limit of a sequence of rational numbers $\{t_k\}_{k=1}^\infty$ in $(t, \infty)$ and since $\omega \in \cap A_k$ we have $X_{t_k}(\omega) - Y_{t_k} (\omega) =0$ for each $t_k$ . By right continuity we see that it must be that:

$X_t(\omega)- Y_t(\omega) = \lim_{k\to \infty} X_{t_k}(\omega) - Y_{t_k} (\omega) =0$

Then for this $\omega$ we see that $X_t(\omega) = Y_t(\omega)$ for all $t$ . Since $\omega \in \cap A_k$ was chosen arbitrarily, this is true for all such $\omega\in \cap A_k$ . Recall $(\cap A_k)^C$ is a null set, so we have $X_t(\omega) = Y_t(\omega)$ for all $t$ , except on a null set of $\Omega$ . In other words:

$P(X_t =Y_t\ \forall\ 0\leq t< \infty) =1$

Exercise 1.1.7

It is enough to show continuity at the rational numbers in $(0, t_0)$ is in $\mathcal F_{t_0}^X$ . Continuity everywhere in $(0,t_0)$ will follow from right continuity and fact that left hand limits exist.

Let $s$ be any real number in $(0,t_0)$ . $X_t(\omega)$ is right continuous at $s$ , it will be continuous if the left hand limit $\lim_{t\to s^+} X_t(\omega) = X_s(\omega)$ . We can represent this in set notation using the following:

$A= \bigcap_{j=1}^\infty \bigcup_{k=1}^\infty \bigcap_{n=k}^\infty \{\omega : |X_{s-{1\over n}}(\omega) - X_s(\omega)| \leq {1\over j} \}$

Explanation: Fix $j$ . Then $\cup \cap \{ \omega : |X_{s-{ 1\over n }}(\omega) - X_s(\omega)| \leq {1\over j} \}$ gives all $\omega$ such that $|X_{s-{1\over n}}(\omega) - X_s(\omega)|$ is eventually less that ${1\over j}$ . Taking the intersection over all $j$ we have all $\omega$ such that $|X_{s-{1\over n}}(\omega) - X_s(\omega)|$ is eventually less than all ${1\over j}$ . Meaning, the $\omega$ such that $\lim_{t\to s^+} X_{t}(\omega) = X_s(\omega)$ . So $A$ is the event that $X_t$ is continuous at $s$ .

Clearly, $X_s \in \mathcal F_s^X$ . And for each $n$ , $X_{s-{1\over n}}\in \mathcal F_s^X$ since $\mathcal F_{s-{1\over n}}^X \subset \mathcal F_s^X$ . Thus $|X_{s-{1\over n}} -X_s|\in \mathcal F_s^X$ . Then $\{ \omega : |X_{s-{1\over n}}(\omega) -X_s(\omega)|\leq {1\over j} \}\in\mathcal F_s^X$ for each $j$ . Since $\sigma$ -algebras are closed under countable unions and intersections we have that $A\in \mathcal F_s^X\subset \mathcal F_{t_0}^X$ .

We have shown the result for general $s\in (0, t_0)$ , so it holds for rational numbers in particular. The advantage of using rational numbers is they are countable. So, the event $B$ that $X_t$ is continuous at all rational numbers is a countable intersection of sets in $\mathcal F_{t_0}^X$ . So $B\in \mathcal F_{t_0}^X$ .

Exercise 1.1.10

Similar to previous exercise

Problem 1.1.16 (not mine)

Let $\tau: \Omega\to [0, \omega) \times \Omega$ be defined as $\tau(\omega) = (T(\omega), \omega)$ . Then $\tau$ is measurable since $T$ is measurable (each component of $\tau$ is measurable). Now let $A \in \mathcal B ( \mathbb R^d)$ . Note that $X(T) = X\circ \tau$ . The composition of measurable functions is measurable, thus $X(T)$ is measurable.

Problem 1.1.17

Let $B= \{X_T \in A\}$ . Note $B^C = \{X_T \in A^C\}$ . Similarly if $B = \{X_T \in A \}\cup \{T=\infty\}$ , then $B^C = \{X_T \in A^C\} \cap \{T<\infty\} = \{X_T \in A^C\}$ (I think).

Now let $B_n = \{X_T \in A_n\}$ and $C_n = \{X_T \in A_n\} \cup \{ T = \infty\}$ . Then the union of a countable number of such sets is

$\{\bigcup \{X_T \in A_n\} \} \cup \{ T= \infty\}$

$= \{X_T \in \cup A_n \} \cup \{T= \infty\}$

So $\mathcal F_T$ is closed under complements and countable unions. Thus it is a $\sigma$ -algebra. And clearly $\mathcal F_T$ is a subset of $\mathcal F$ .

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