As a style preference, I’m going to drop all the arguments that are from the probability space.

Fix a probability space (\Omega, \mathcal F, P); we characterize the Brownian motion \{B_t\}_{t \geq 0} via the following properties:

**Independent Increments**: If s < t, the random variable B_t - B_s is independent of \sigma\{B_r: r \leq s\}**Stationary Increments**: If s < t, then B_t - B_s has the same distribution as B_{t-s} - B_0.**Continuity**: The map t \mapsto B_t is almost surely continuous.

**Theorem**: If a process satisfies the above, then
there are \mu, \sigma^2 (respectively
called the drift and the variance parameter, and \sigma is named the volatility) such that
there exist B_t \sim N(\mu t, \sigma^2
t).

*Def*: A stochastic process \{B_t\}_{t \geq 0} is called a (one
dimensional) **Brownian motion** (or Wiener process)
starting from the origin with drift \mu
and variance parameter \sigma^2 if
B_t = 0 and the above three conditions
are satisfied, with the imposition that
B_t - B_s \sim N(\mu (t-s), \sigma^2 (t-s)).

**Prop**: If B_t is a
Brownian motion with \mu = 0, \sigma^2 =
1 (a so-called **standard Brownian motion**), then
Y_t = \sigma B_t + \mu t is a Brownian
motion with parameters \mu,
\sigma^2.

*Proof*: Obvious.

Pick a probability space (\Omega, \mathcal F, P) that is rich enough to support a countable collection of independent standard normal variables. If you are particular, the unit interval with Lesbegue measure is sufficient here.

The strategy is as follows: we define B_t for a countable dense set (in particular the dyadic rationals) of times using our precession of standard normals. then, we find some t \mapsto B_t that agrees on the dense set and is uniformly continuous and then extend by continuity.

Set D_n = \left\{ \frac{k}{2^n}, k = 0, 1, \dots, 2^n \right\} and D = \bigcup_{n=0}^\infty D_n; index our standard normals by \{N_{q}\}_{q \in D}, and set B_0 = 0, B_1 = N_1, and B_{1/2} = \frac{B_1 - B_0}{2} + \frac{1}{2}N_{1/2}. Just continue the same thing for every such dyadic, such that \{B_{1/2^n} - B_0, B_{2/2^n} - B_{1/2^n}, \dots, B_1- B_{(2^n-1) / 2^n} \} are all independent N\left(0, 2^{-n}\right).

**Theorem**: Almost surely, t
\mapsto B_t, t \in D is
uniformly continuous.

*Proof*: Set K_n = \sup \{ |B_s -
B_t| \mid s,t \in D, |s - t| \leq 2^{-n}\}. We just need to show
that K_n \to 0 as n \to \infty. In fact, something even
stronger is true: for \alpha <
\frac{1}{2}, \lim_{n \to \infty}
2^{\alpha n}K_n = 0. Morally speaking, just think that each
Brownian increment is about its standard deviation, which is |t - s|^{1/2}.

Technically, however, we proceed as follows: set Y_n = \max\{B_{1/2^n} - B_0, B_{2/2^n} - B_{1/2^n}, \dots, B_1- B_{(2^n-1) / 2^n} \} and note that the union bound yields \begin{align*} P(Y_n \geq x) &\leq \sum_{j=1}^{2^n} P(|B_{j/2^n} - B_{(j-1)/2^n}| \geq x) \\ &= 2^nP(B_{1/2^n} \geq x) \\ &= 2^{n+1}P(B_1 \geq 2^{n/2} x). \end{align*} If we choose x_n such that \sum_{n=1}^\infty 2^{n+1}P(B_1 \geq 2^{n/2}x_n) < \infty, then by Borel-Cantelli shows that for sufficiently large n, Y_n \leq x_n almost surely. Do any reasonable bound you like on the tail of the normal distribution and take a sufficiently large x and call it a day. In particular, if you choose the easiest bound P(N \geq x) \leq Ce^{-x^2/2}, you can eventually get the bound:

**Prop**:
\limsup_{n \to \infty} \frac{2^{n/2}}{\sqrt{n}}Y_n \leq \sqrt{2 \log
2}.

*Proof*: Look at the sum
\sum_{n=1}^\infty P\left(Y_n > \sqrt{n} \cdot 2 ^{-n/2} \cdot
\sqrt{2 \log 2 (1 + \epsilon)}\right)
and apply Borel-Cantelli. In particular, we have
\begin{align*}
P(Y_n > x_n) &\leq \sum_{j=1^{2^n}} P(|B_{j/2^n} -
B_{(j-1)/2^n}| > x_n) \\
&= 2^{n+1}P(B_{1/2^n} > x_n) \\
&= 2^{n+1}P\left(B_1 > \sqrt{n} \sqrt{2(\log 2 (1 +
\epsilon)}\right) \\
&\leq C 2^n e^{-\frac{\sqrt{2\log 2 n(1+\epsilon)^2}}{2}} \\
&\leq C e^{-n\epsilon}.
\end{align*}

**Prop**: Set K_n = \sup \{
|B_s - B_t| \mid s,t \in D, |s - t| \leq 2^{-n}\}; then, there is
C such that with almost surely,
\limsup_{n \to \infty} \frac{2^{n/2}}{\sqrt{n}}K_n \leq C.

*Proof*: It’s easy to see that K_n
\leq 2 \sum_{j=n+1}^\infty Y_j (this is just the triangle
inequality). Then for sufficiently large n, we get
K_n \leq 2 \cdot 2 \sum_{j=n+1}^\infty 2^{-j/2}\sqrt{j}
with full probability, and so
\sup_{\substack{s, t \in D \\ s < t}} \frac{|B_t -
B_s|}{\sqrt{(t-s)|\log((t-s)^{-1})|}} < \infty.

Now we may set B_t for t \in [0, 1] by B_t = \lim_{\substack{s \to t \\ s \in D}}B_s, and check that this is in fact a genuine Brownian motion, which is not bad; and of course this construction can extend to [0, \infty) easily as well.

*Def*: A function f: [0, 1] ->
\mathbb R is called **Hölder continuous** of order
\beta \geq 0 if there is some C < \infty such that for all s, t, |f(t) - f(s)|
\leq C|t-s|^\beta. Futher, f is
**weakly Hölder continuous** of order \beta if it is Hölder continuous of order
\alpha for all \alpha < \beta. In both cases, we will say
Hölder-\beta continuous for short.

**Prop**: Brownian motion paths are weakly Hölder-\frac{1}{2} continuous.

*Proof*: Omitted.

**Theorem**: The function t
\mapsto B_t is nowhere differentiable almost surely.

*Proof*: Assume |f'(t)| <
K; there exists \delta > 0
such that if |s - t| \leq \delta, |f(t) - f(s)| \leq 2K|s-t|; in particular
there is an N such that for all n > N, |s - t|
\leq n^{-1}, |r - t| \leq n^{-1}, |f(s)
- f(r)| \leq 4Kn^{-1}. Then, set
Z_{k, n} = \max \left\{ |B_{k/n} - B_{(k-1)/n}|, |B_{(k+1)/n} -
B_{k/n}|, |B_{(k+2)/n} - B_{(k+1)/n}|\right\}
and
Z_n = \min \{ Z_{k, n} \mid k = 1, \dots, n \}.

If B is differentiable, then there is some M such that Z_n \leq Mn^{-1} for all n. Now set E_M to be the event that Z_n \leq Mn^{-1} for all sufficiently large; our theorem reduces to showing that P(E_M) = 0 for all M. In fact, we will show that \lim_{n \to \infty} P(Z_n \leq Mn^{-1}) = 0.

Consider the union bound \begin{align*} P(Z_n \leq Mn^{-1}) &\leq \sum_{j=1}^{n}P(Z(n,k) \leq Mn^{-1}) \\ &\leq nP\left( \max \left\{ |B_{1/n}|, |B_{2/n} - B_{1/n}|, |B_{3/n} - B_{2/n}|\right\} \right) \\ &\leq nP(|B_{1/n}| \leq Mn^{-1})^3 \\ &\leq nP(|B_1| \leq Mn^{-1/2})^3 \end{align*} and just do literally the stupidest estimate you can, e.g. just look at the density and say that the probability is bounded by 2CMn^{-1/2}, so that the above is sent to zero as n \to \infty.

*Def*: A filtration \{ \mathcal F
\}_{t \geq 0} is an incerasing collection of sub \sigma-algebras. Further, we put
\mathcal F_{\infty} = \bigcup_{t \geq 0} \mathcal F_t.

*Def*: A stochastic process \{X_t
\}_{t \geq 0} is adapted to \{ \mathcal
F_t \}_{t \geq 0} if for each t,
X_t is \mathcal F_t-measurable.

*Def*: A process \{ B_t \}_{t \geq
0} is a standard Brownian motion start at 0 w.r.t. \{ \mathcal F_t \}_{t \geq 0} if

- B_0 = 0,
- \{ B_t \}_{t \geq 0} is adapted to \{ \mathcal F_t \}_{t \geq 0},
- if s < t then B_t - B_s is independent of \mathcal F_s,
- B_t - B_s \sim N(0, t-s),
- and with probability 1 t \mapsto B_t is continuous.

*Def*: A random variable \tau
taking values in [0, \infty] is called
a stopping time with respect to \{ \mathcal
F_t \}_{t \geq 0} if for every t, the event \{\tau
\leq t\} \in \mathcal F_t.

*Examples*: The following are all stopping times:

- constants;
- \tau = \inf\{ t \mid B_t \in V \} where V is Borel;
- \tau_1 \land \tau_2, \tau_1 \lor \tau_2, where \tau_1, \tau_2 are both stopping times.

*Def*: If \tau is a stopping
time, then \mathcal F_\tau is the \sigma-algebra corresponding to the
collection of events A such that for
each t, A
\cap \{ \tau \leq t \} \in \mathcal F_t.

*Def*: For some stochastic process \{X_t\}_{t \geq 0} (or any other indexed set)
with filtration \{F_t\}_{t \geq 0},
X_t has the **Markov
property** if it saitisfies that
E[f(X_t) \mid \mathcal F_s] = E[f(X_t) \mid \sigma(X_s)].

*Def*: In general, if X_t is
a stochastic process and \tau is a
stopping time, both adapted to \{ \mathcal F_t
\}_{t \geq 0} with P(\tau < \infty)
= 1, then X_t has the
**strong Markov property** if X_{\tau + t} is independent of \mathcal F_{\tau}.

**Prop**: Suppose B_t
is a Brownian motion and \tau is a
stopping time, both with respect to \{
\mathcal F_t \}, and assume P(\tau <
\infty) = 1. Set
Y_t = B_{\tau + t} - B_{\tau}.
Then Y_t is a Brownian motion
independent of \mathcal F_t,
e.g. Brownian motion has the strong Markov property and the new process
is also a Brownian motion.

*Proof*: You proceed by doing successive approximations.

- First, let \tau take a finite amount of values, and use the normal Markov property.
- Then, approximate any \tau by stopping times taking a finite amount of values, such as by \tau_n = \begin{cases} \frac{k}{2^n} & \frac{k - 1}{2^n} \leq \tau \leq \frac{k}{2^n} \leq n \\ n & \tau > n \end{cases} for example.
- Take a limit by continuity.

In particular, the following is clear for Brownian motion:

**Prop**: If \{B_t\}_{t \geq
0} is a Brownian motion, t a
fixed time, and Y_s = B_{t+s} - B_t,
then \{Y_s\}_{s \geq 0}, is a BM and
independent of \mathcal F_t = \sigma\{B_s \mid
s \leq t\}.

**Theorem**: Set B_t to
a Brownian motion with drift zero; **the reflection
principle** is that
P\left(\max_{0 \leq s \leq t} B_s \geq a\right) = 2P(B_t \geq a).
Set \tau_a = \min \{ s \mid B_s = a
\}; we also have
P(\tau_a \leq t) = 2P(B_t \geq a)
or equivalently
P(B_t \geq a \mid \tau_a \leq t) = \frac{1}{2}.

**Prop**: If 0 < r < s
< \infty,
q(r, s) = P(B_t = 0 \text{ for some } r \leq t \leq s) = 1 -
\frac{2}{\pi} \arctan\left(\sqrt{\frac{r}{s - s}}\right).

*Proof*: First, I claim that q(r, s)
= q(1, s/r) just by a change of variables, so we only need to
compute q(t) = q(1, 1 + t). Set A = \{ B_s = 0 \text{ for some } 1 \leq s \leq 1 + t
\}, so that
\begin{align*}
q(t) &= \frac{2}{\sqrt{2 \pi}}\int_0^\infty P(A \mid B_1 = x)
e^{-x^2 / 2}dx.
\end{align*}
However, by the reflection principle, we have that
\begin{align*}
P(A \mid B_1 = x) &= P\left(\max_{0 \leq s \leq t} B_s \geq x
\right) \\
&= P \left( \min_{0 \leq s \leq t} B_s \leq -x \right) \\
&= 2P(B_t \geq x) = 2P\left(B_1 \geq \frac{x}{\sqrt{t}}\right)
\end{align*}
upon which we can just compute the integral.

**Corollary**: One dimensional standard Brownian motion
is (pointwise) recurrent (that is, the zero set of Brownian motion is
unbounded).

**Corollary**: Since Y_t =
t^{-1}B_{1/t} is a standard Brownian motion, this shows that for
any \epsilon > 0, Z_\epsilon = \{ t \mid B_t = 0, 0 \leq t \leq
\epsilon \} has more elements that just 0.

*Def*: A process \{ M_t \}_{t \geq
0} is a **supermartingale** (resp.
**submartingale**) w.r.t. \{
\mathcal F_t \} if - E[|M_t|] <
\infty, - M_t is \{ \mathcal F_t \} adapted, - and if E[M_t \mid \mathcal F_s] \leq M_s (resp.
\geq M_s) almost surely for s \leq t. A process which is both a
submartingale and supermartingale is just a martingale, it is continuous
if M_t is a continuous function of
t almost surely, and is square
integrable (or simply L^2) if it has
finite second moment for all t.

As an aside, I’m going to stop saying with probability 1 or almost surely because it’s annoying!

**Prop**: Brownian motion (without drift) is an L^2 continuous martingale.

**Theorem (Kolmogorov Zero-One)**: Tail events of
independent \sigma-algebras happen with
probability 0 or 1.

*Proof*: Set \mathcal F_n =
\sigma(X_1, \dots, X_n), and \mathcal
T_n = \sigma(X_{n+1}, \dots), \mathcal
F_\infty = \bigcup_{n} \mathcal F_n, and \mathcal T_\infty = \bigcap_n \mathcal
T_n.

Then, one can see that if A \in \mathcal F_\infty and \epsilon > 0, there is n and A_n \in \mathcal F_n such that P(A_n \Delta A) < \epsilon; even better, there exists A_n independent of A \in \mathcal T_\infty such that P(A \Delta A_n) < \epsilon, and so we conclude that P(A) = P(A)P(A).

**Theorem (Blumenthal Zero-One)**: Let B_t be a Brownian motion with the standard
filtration, and set \mathcal F_{0+} =
\bigcap_{\epsilon > 0} \mathcal F_{\epsilon}; then, if A \in \mathcal F_{0+}, either P(A) = 0 or 1.

Let B_t be a standard Brownian motion; a partition \Pi of [0, 1] is a sequence 0 = t_0 < t_1 < \dots < t_k = 1, and the mesh of the partition is just \| \Pi \| = \max_{i = 1, \dots, k} t_i - t_{i - 1}. Now take a sequence of partitions \Pi_n, e.g. 0 = t_{0, n} < \dots < t_{k_n, n}, and define the quantity Q(t, \Pi) = \sum_{t_{j} <= t} (B_{t_j} - B_{t_{j-1}})^2 and Q_n(t) = Q(t, \Pi_n) and Q_n = Q_n(1).

**Theorem**: If \| \Pi_n \|
\to 0, then Q_n \to 1 in
probability. Furthermore, if \sum_{n=1}^\infty
\| \Pi_n \| < \infty, then almost surely \lim_{n \to \infty} Q_n = 1.

*Proof*: A simple computation gives us that
E(Q_n) = \sum_{j=1}^{k_n}E[(B_{t_j} - B_{t_{j-1}})^2] =
\sum_{j=1}^{k_n}(t_j - t_{j-1}) = 1
and
\mathop{\mathrm{Var}}(Q_n) = \sum_{i=1}^{k_n}
\mathop{\mathrm{Var}}((B_{t_j} - B_{t_{j-1}})^2) = \sum_{j=1}^{k_n}
(t_j - t_{j-1})^2 \mathop{\mathrm{Var}}(B_1^2) = 2\sum_{j=1}^{k_n}(t_j -
t_{j-1})^2.
Then,
\mathop{\mathrm{Var}}(Q_n) \leq \| \Pi_n \| 2 \sum_{j=1}^{k_n}(t_j -
t_{j-1}) = 2 \| \Pi_n \|
and
P(|Q_n - 1| \geq \epsilon) \leq
\frac{\mathop{\mathrm{Var}}(Q_n)}{\epsilon^2} \leq \frac{2 \| \Pi_n
\|}{\epsilon^2}.

The latter half of the theorem follows from Borel-Cantelli.

**Theorem**: In general, if \sum_{n=1}^\infty \| \Pi_n \| < \infty,
then almost surely we have Q_n(t) \to
t.

*Proof*: With probability 1 this holds for rational t, but by construction Q_n(t) is monotone, so it holds
everywhere.

*Def*: In general, if X_t is
a process, then its quadratic variation is
\left\langle X\right\rangle_t = \lim_{n \to \infty} \sum_{t_{j,n} \leq
t} (X_{t_{j, n}} - X_{t_{j-1, n}})^2
(sort of, since sometimes this depends on the partition).

*Def*: Alternatively, if M_t
is a continuous L^2-martingale, its
quadratic variation is the unique predictable process \left\langle M \right\rangle_t that makes
M_t^2 - \left\langle M \right\rangle_t
a martingale (in particular, this exists by Doob decomposition).

We showed above that \left\langle B_t \right\rangle_t = t.

**Prop**: If B_t is a
standard Brownian motion and Y_t = \mu t +
\sigma B_t, then
\left\langle Y \right\rangle_t = \sigma^2 t.

*Proof*: Just check directly.

**Lemma (Relaxed Borel-Cantelli)**: Let A_1, A_2, \dots be a sequence of events, and
set \mathcal F_n = \sigma(A_1, \dots,
A_n); if there is q_n with
\sum_{n=1}^\infty q_n = \infty
such that P(A_n \mid \mathcal F_{n-1})
\geq q_n, then A_n happens
infinitely often almost surely.

*Proof*: Same as usual, but just with a little more
caution.

Recall that the second Borel-Cantelli lemma tells us that if they are independent and \sum_{n=1}^\infty P(A_n) = \infty, then A_n occurs infinitely often with probability 1; this lemma is stronger than that.

**Theorem**: If B_t is
a standard Brownian motion, then
\limsup_{t \to \infty} \frac{B_t}{\sqrt{2 t \log \log t}} = 1.

**Corollary**: By symmetry,
\liminf_{t \to \infty} \frac{B_t}{\sqrt{2 t \log \log t}} = -1.

*Proof*: Let \mathcal T_t = \sigma\{
B_{s + t} - B_t \mid s \geq 0\}, and \mathcal T_\infty = \bigcap_t \mathcal T_t;
one can adapt the arguments from the Kolmogorov 0-1 law to show that
everything in \mathcal T_\infty happens
with probability 0 or 1. Then,
A_\epsilon = \left\{ \omega \mid \limsup_{t \to \infty}
\frac{B_t}{\sqrt{2 t (1 + \epsilon) \log \log t}} \leq 1\right\}
is a tail event (e.g. in \mathcal
T_\infty) and thus P(A_\epsilon) =
0 or 1. In fact, if \epsilon < 0 then it’s 0, and if \epsilon > 0 then it’s 1. In fact, by this
0-1 law and symmetry one may see
P \left( \limsup_{t \to \infty} \frac{|B_t|}{\sqrt{2 t (1 + \epsilon)
\log \log t}} \leq 1 \right) = P \left(\limsup_{t \to \infty}
\frac{B_t}{\sqrt{2 t (1 + \epsilon) \log \log t}} \leq 1\right)
as well.

First take \epsilon > 0 and take some \rho > 1 to be specified later. Then, let V_n = \left\{ |B_{\rho^n}| \geq \sqrt{2 \rho^n(1-\epsilon) \log \log \rho^n} \right\}; we want to to show that V_n occurs infinitely often. I claim that P(V_{n+1} \mid V_1, \dots, V_n) \geq P \left( B_{\rho^{n+1}} - B_{\rho^n} \geq \sqrt{2 \rho^{n+1}(1-\epsilon) \log \log \rho^n} \right). To see this, compute \begin{align*} &P \left( \frac{B_{\rho^{n+1}} - B_{\rho^n}}{\sqrt{\rho^{n+1} - \rho^n}} \geq \frac{\sqrt{2\rho^{n+1}(1-\epsilon)\log \log \rho^n}}{\sqrt{\rho^n (\rho - 1)}} \right) \\ &= P \left( \frac{B_{\rho^{n+1}} - B_{\rho^n}}{\sqrt{\rho^{n+1} - \rho^n}} \geq \sqrt{\frac{2 \rho}{\rho - 1}(1-\epsilon)(\log n + \log \log \rho)} \right) \\ \end{align*} and choose \rho large enough so that \frac{2 \rho}{\rho - 1}(1-\epsilon) < 1, and use the estimate P(B_1 > x) \sim \exp(-x^2 / 2) and conclude by the earlier lemma. The other direction for \epsilon < 0 is similar (in fact, easier since we may conclude from the first Borel-Cantelli lemma).

*Def*: Set B_t a standard
Brownian motion, Z = \{ t \mid B_t = 0
\}, and Z_t = Z \cap [0, t].
Then, t \in Z is right-isolated if
t \in Z, and \exists \epsilon > 0 such that (t, t + \epsilon) \cap Z = \emptyset; similar
for left-isolated. A point which is both left and right isolated is just
isolated.

With probability 1, 0 is not right-isolated; further, Z_1 is homeomorphic to the Cantor set.

With probability 1, the sets of left and right isolated points are countable, and there are no isolated points.

We can show that if q \in \mathbb{Q}_{\geq 0}, P(B_q = 0) = 0, and \tau_q = \min \{ t \geq q \mid B_t = 0\} is a stopping time; further, the left-isolated points are just \{ \tau_q \mid q \in \mathbb{Q}_{\geq 0}\}. And by strong Markov property, no \tau_q is a right-isolated point, so there are no isolated points.

Set \sigma_q = \max_t \{ t < q \mid B_t = 0 \} (this is well-defined, but not a stopping time). Then associate every q to an interval (\sigma_q, \tau_q); then Z = [0, \infty) \setminus \bigcup_q (\sigma_q, \tau_q) and has Lebesgue measure 0. To see this, we just interchange integrals: E[ \lambda(Z_1) ] = E\left[ \int_0^1 1_{\{B_s = 0\}} ds \right] = \int_0^1 P\{ B_s = 0 \}ds = 0.

Look at Z \cap [1, 2]; now cover [1, 2] by intervals of length n^{-1}; put the number of these intervals that intersect Z as X_n. Then, E[X_n] = \sum_{j=1}^n P\left(Z \cap \left[1 + \frac{j-1}{n}, 1 + \frac{j}{n} \right]\neq \emptyset \right) \sim Cn^{1/2}.

The box or Minkowski dimension of a set is given by the exponent above (sort of).

*Def*: The **Hausdorff dimension** comes from the
**Hausdorff measure**: for \epsilon > 0 and \alpha, set
\mathcal H^\alpha_\epsilon = \left\{ \inf \sum_{j=1}^\infty
(\operatorname{diam} U_j)^\alpha \right\}
where the infimum is over all coverings \bigcup_j=1^\infty U_j of V with each \operatorname{diam} (U_j) < \epsilon. Then
the Hausdorff measure is given by
\mathcal H^\alpha(V) = \lim_{\epsilon \to 0} \mathcal
H_{\epsilon}^\alpha (V).
Then, there is some number D
such that
\mathcal H^{\alpha}(V) = \begin{cases}
\infty & \alpha < D \\
0 & \alpha > D
\end{cases}
and we call this D the
Hausdorff dimension. In general, the Hausdorff dimension is at most the
Minkowski dimension, but in this case we actually do have equality.

The local time is the amount of time the Brownian motion spends at 0 by a certain time. It is sort of like the Cantor measure, insofar as if s < t, then L_t - L_s > 0 \iff (s, t) \cap Z \neq \emptyset.

*Def*: We define the **local time** of Brownian
motion at x \in \mathbb{R}, L_t, by first setting
L_{t, \epsilon}(x) = \frac{1}{2\epsilon} \int_0^t 1_{\{|B_s - x| \leq
\epsilon\}} ds
and then letting L_t(x) =
\lim_{\epsilon \to 0} L_{t, \epsilon}. We abbreviate L_t = L_t(0).

We can compute the expectation \begin{align*} E[L_{t, \epsilon}] &= \frac{1}{2\epsilon} \int_0^t P(|B_s| \leq \epsilon)ds \\ &\sim \frac{1}{2\epsilon} \int_0^t 2e \frac{1}{2 \pi s}ds \\ &= \sqrt{\frac{2}{\pi}}t^{1/2} \end{align*}.

**Theorem**: With probability 1, L_t exists for all t, and this holds in L^2 as well.

There are more facts: L_t is continuous in t and non-decreasing, L_t - L_s > 0 \iff (s, t) \cap Z \neq \emptyset, and L_t is weakly \frac{1}{2}-Hölder continuous.

**Theorem (Scaling Rule)**: L_t has the same distribution as t^{1/2} L_1. Further, M_t = \max_{ 0 \leq s \leq t}B_s has the same
distribution as L_t.

*Def*: If B_t^1, \dots, B_t^d
are independent standard Brownian motions, then
B_t = (B_t^1, \dots, B_t^d)
is a standard Brownian motion in \mathbb{R}^d.

**Prop**: B_0 = \mathbf
0, has independent increments, if s
< t, B_t - B_s \sim N(0, (t-s)I), and B_t has continuous paths.

In particular, if we have the above with B_t - B_s \sim N(\mu, \Gamma), then B_t is a Brownian motion with dift \mu \in \mathbb{R}^d and covariance matrix \Gamma. Further, if AA^T = \Gamma, then we may write B_t = AY_t + t \mu for a standard Brownian motion Y_t.

Consider the open annulus with inner radius r and outer radius R, e.g. D(r, R) = \{ y \in \mathbb{R}^d \mid r < |y| < R \}. Start a Brownian motion at x, and let \tau = \tau(r, R) be the first time with |B_t| = r or |B_t| = R. What is the probability that |B_\tau| = R?

In one dimension, this is easy: stop the Brownian motion at \tau and look at what happens at infinity.

In higher dimensions, we need a quick detour.

For this section, a domain is an open connected subset of \mathbb{R}^d.

*Def*: For a domain D, we say
f: D \to \mathbb{R} is harmonic if it
is continuous (or merely locally integrable) and satisfies the mean
value property
f(x) = MV(f, x, \epsilon) = \int_{|x - y| = \epsilon} f(y) ds
where s is the surface measure,
normalized so that \int_{|x - y| = \epsilon}ds
= 1.

In a probabilistic vein, if B_t is a standard d-dimensional Brownian motion starting at x, and \tau = \min \{ t \mid |B_t - x| = \epsilon \}, since B_t is rotation invariant, the above is just MV(f, x, \epsilon) = E[f(B_\tau)].

*Def*: The Laplacian of f is
\nabla f = \sum_{j=1}^d \frac{\partial ^2 f }{\partial x_j^2}.

**Prop**: If f is C^2 in D,
then
\frac{1}{2d}\nabla f = \lim_{\epsilon \to 0} \frac{MV(f, x, \epsilon)
- f(x)}{\epsilon^2}.

*Proof*: Look at the Taylor expansion.

**Theorem**: f is
harmonic in D if and only if it is in
C^2 and \nabla f = 0 everywhere.

Let \tau = \tau_{r, R} = \min \{ |B_t| = r \text{ or } R \} = \min \{ t \mid B_t \in \partial D \}. We will let a superscript x denote that B_0 = x, and let \varphi(x) = P^x(|B_\tau| = R). By rotational invariance, \varphi(x) = \varphi(|x|), and it is continuous at the boundary; we also clearly have \varphi(x) = 0 if |x| = r and \varphi(x) = 1 if |x| = R. The strong Markov property furnishes that \varphi is harmonic. Set \tau_\epsilon = \min \{ t \mid |B_t - x| = \epsilon\}; then P^X(|B_\tau| = R \mid \mathcal F_{\tau_\epsilon}) = \varphi(B_{\tau_\epsilon}) and P^X(|B_\tau| = R) = E^x[P(|B_\tau| = R \mid \mathcal F_{\tau_\epsilon})] = E^x[\varphi(B_{\tau_\epsilon})] = MV(\varphi, x, \epsilon). This list of properties gives a unique function (solve an ODE in polar coordinates), and is given by \varphi(x) = \frac{|x|^{2-d} - r^{2-d}}{R^{2-d} - |r|^{2-d}} for d \neq 2 and \varphi(x) = \frac{\log|x| - \log r}{\log R - \log r} for d = 2.

Let B_t be a standard Brownian motion in \mathbb{R}^d.

**Theorem**: With probability 1, Brownian motion is
transient in dimensions at least 3; that is,
\lim_{t \to \infty} |B_t| = \infty.

*Proof*: If d \geq 3, and we
start at x with |x| > r, if we set T_r = \min \{ t \mid |B_t| = r \},
P^x(T_r < \infty) = \lim_{R \to \infty} P^X(|B_{\tau_{r, R}}| = r)
= \left( \frac{r}{|x|} \right)^{d-2} < 1

**Theorem**: With probability 1, Brownian motion is
neighborhood recurrent; that is, \forall z \in
\mathbb{R}^2, \epsilon > 0, Brownian motion visits the disk of
radius \epsilon about z infinitely often. However, it is not point
recurrent, e.g. it hits z \neq 0 with
probability 0.

*Proof*: If d = 2, then
P^x(T_r < \infty) = \lim_{R \to \infty} P(|B_{\tau_{r, R}}| = r) =
1
But if T = \min\{ t \mid B_t = 0
\},
P^x(T < \infty) \leq \lim_{R \to \infty} \lim_{r \to 0}
P^x(|B_{\tau_{r, R}}| = r) = 0

A fun fact is that if d \geq 2, then \{ B_t, t \geq 0 \} has Hausdorff dimension 2 but also zero Hausdorff-2 measure.

Take a bounded domain D \subset \mathbb{R}^d, and a continuous function F: \partial D \to \mathbb{R}; the Dirichlet problem is to find the unique continuous f: \overline D \to \mathbb{R} that agrees with F on \partial D and is harmonic on D.

In fact, uniqueness follows from the maximum principle, which says that the maximum of f is attained on the boundary (think about the mean value principle). Then subtract two solutions and see that it is 0.

Let T = \min\{t \mid B_t \in \partial D \}, and f(x) = E^x[F(B_T)]. This satisfies the mean value principle, and in fact continuous (which is kinda hard) but it is obviously locally integrable, so f here is harmonic.

In general, such a harmonic function is not necessarily continuous on the boundary: take the example of the punctured unit disk, with F(x) = 1 for |x| = 1 and F(0) = 0. And so everything is fine except at the origin, since f(x) = P^x(|B_T| = 1) = 1.

*Def*: If x \in \partial D,
let \sigma = \inf \{ t > 0 \mid B_t \in
\partial D\}; x is regular if
P^x(\sigma = 0) = 1.

**Prop**: f as defined
above is continuous at every regular boundary point. If we relax the
boundary to be only bounded and measurable, then the above is continuous
at every regular point at which F is
continuous.

Therefore the Dirichlet problem has a solution for every continuous F if and only if every point on \partial D is regular.

*Def*: If D is a domain, then
**harmonic measure** on D
(or on \partial D) at z \in D, is the hitting measure of Brownian
motion, starting at z and stopped at
\partial D. That is, if V \subset \partial D,
\mathop{\mathrm{hm}}_D(V, z) = P^z(B_T \in V).

**Prop**: Note that \mathop{\mathrm{hm}}_D(\partial D, z) = P^z(T <
\infty) so if D is bounded, this
is a probability measure. Then,
E^z[F(B_T)] = \int_{\partial D} F(w) \mathop{\mathrm{hm}}_D(dw, z)
and if \partial D is smooth,
then \mathop{\mathrm{hm}}_D(\cdot, z)
is absolutely continuous with respect to surface measure; if H_D(z, w) is the Poisson kernel,
\mathop{\mathrm{hm}}_D(v, z) = \int_V H_D(z, w)S(dw)
where S is the surface
measure.

A nice example of all of the above is the unit ball. Then, H_D(z, w) = c_d^{-1}\frac{1 - |z|^2}{|z - w|^d} where c_d is the surface measure of the unit sphere.

**Prop**: The following are true.

- If V \subset \partial D and h(z) = \mathop{\mathrm{hm}}_D(V, z) = P^z(B_T \in V), then h is the unique harmonic function in D with boundary condition F(w) = \begin{cases} 1 & w \in V \\ 0 & w \notin V \end{cases} In particular, one could define the harmonic measure this way and show existence, but this is unreasonably hard compared to just using the Brownian motion.
- If D contains the closed unit ball B, and f is harmonic in D, then for every |z| < 1, f(z) = \int_{|w| = 1} f(w) H_B(z, w)S(dw).

**Corollary (Harnack Inequality)**: For every r \in (0, 1), there is C = C(r, d) such that if f: B \to (0, \infty) is harmonic, then for
all |z|, |z'| \leq r,
f(z) \leq Cf(z')
and
C = \max_{\substack{|z|, |z'| \leq r \\ |w| = 1}} \frac{H_B(z,
w)}{H_B(z', w)}

**Corollary**: For every k there exists c =
c(k, d) such that if f: B \to
\mathbb{R} is harmonic, then for any k-th order partial derivative,
|\partial^k f(0)| \leq c\| f \|_\infty.
Moreover, there is C = C(k, d)
such that if f: D \to \mathbb{R} is
harmonic and z \in D, then
|\partial^k f(z)| \leq \frac{C_k}{(\operatorname{dist}(z, \partial
D)^k} \left( \max_{|z - w| \leq \operatorname{dist}(z, \partial D) / 2}
|f(w)| \right).

**Theorem (Harnack principle)**: if D is a domain and K
\subset D is compact, then there exists some C = C(K, D) such that if f: D \to (0, \infty) is harmonic and z, w \in K, then
f(z) \leq Cf(w).

Now let D and \mathbb{R}^{d} \setminus D = K; if F: K \to \mathbb{R} is continuous, we can ask about the existence of f: \mathbb{R}^d \to \mathbb{R} which is harmonic on D, coincides with F on K, and is continuous on \mathbb{R}^d. Such a thing is obviously not unique (think!).

Let both F and f bounded but otherwise as above. Set T = \min\{ t \geq 0 \mid B_t \in K \}, and assume for every z \in D, P^z(T < \infty) > 0. If d = 1 or 2, there is a unique f, given by f(z) = E^z[f(B_T)].

Now if d \geq 3 and K is bounded, then g(z) = P^z(T = \infty) > 0; g is harmonic and bounded, and if g = 0 on K then g is continuous.

**Theorem**: All solutions to the above problem are
given by
f(z) = E^z[(F(B_t) 1_{\{T < \infty\}}] + cP^z(T = \infty).

Intuitively, we just add a point at infinity which has value c and get this formula; but this only works on \mathbb{R}^d and \mathbb{Z}^d.

Consider a random walk on an infinite binary tree; such a walk is clearly recurrent, since it moves away from the starting point with probability 2/3. But now there are lots of infinities.

The heat equation is described by some initial function u_0: D \to \mathbb{R}, some boundary condition u(t,x) = F(x) for x \in \partial D, and \dot i(t,x) = \frac{1}{2} \Delta u(t,x). Then, u_t(x) = E^x[1_{\{T \leq t\}}F(B_T) + u_0(B_t)1_{\{T > t\}}].

We can also handle Green’s functions. Let B_t be a Brownian motion in \mathbb{R}^d, d \geq 3; G(x,y) is the “expected amount of time spent at y starting at x; that is, G(x, y) = \int_0^\infty P_t(x,y)dt = G(y,x) = G(0, y-x). Further, G(x) = G(0, x) = \int_0^\infty \frac{1}{(2\pi t)^{d/2}}\exp(-|x|^2/2t)dt; and when one computes this integral, we get C_d |x|^{2-d} away from the origin. This is no coincidence: in d=2, you get a(x) = C_2\log(|x|), which is also the unique radially symmetric harmonic function.

In a domain D (in any dimension), let G_D(x,y) = \int_0^\infty P_t^D(x,y)dt so in d \geq 3, G(x,y) - E^x[G(B_T, y)] and in d = 2, E^x[a(B_T, y)] - a(x,y) where a(x,y) = a(y - x).

For fixed y, the function h(x) = G_D(x,y) is harmonic in D \setminus \{ y \} as x \to y.

Let \frac{d}{dt} F(t) = C(t, F(t)); we then write dF(t) = C(t, F(t))dt, so F(t) = F(0) + \int_0^t C(s, F(s))ds. Here we are interested in the case where we allow things to be random, e.g. dX_t = R_tdt + A_tdB_t. Intuitively, we require that X_t looks like a BM with some drift R_t and variance A_t^2. Then, X_t = X_0 + \int_0^tR_sds + \int_0^tA_sdB_s. Of course, we still need to define the (Itô) stochastic integral.

Let B_t be a Brownian motion with a filtration \mathcal F_t.

*Def*: A process A_t is a
**simple process** (with respect to \mathcal F_t) if there exists a finite number
of times 0 = t_0 < t_1 < \dots < t_n
< t_{n+1} = \infty and random variables Y_0, Y_1, \dots, Y_n such that Y_j is \mathcal
F_{t_j}-measurable, and A_t =
Y_j on t_j \leq t < t_{j+1}.
We can have L^2 or bounded simple
processes, and those are just requirements on the Y_i.

*Def*: If A_t is a simple
process, we define the stochastic integral of A_t to be
Z_t = \int_0^t A_sdB_s = \sum_{k=0}^{j-1}Y_k[B_{t_{k+1}} - B_{t_k}]
+ Y_j[B_t - B_{t_j}].

We have some properties immediately:

- If A_t and C_t are simple, and a, c \in \mathbb{R}, then aA_t + cC_t is simple and \int_0^t (aA_t + cC_t)dB_s = a\int_0^tA_tdB_s + c\int_0^tC_tdB_s.
- If A_t is L^1, then Z_t is a martingale (just compute).
- The Itô isometry is, if A_t is in L^2, \mathop{\mathrm{Var}}(Z_t) = E[Z_t^2] = \int_0^t E[A_s^2]ds = E \left[ \int_0^t A_s^2ds \right]. Again, just compute and use the orthogonality of martingale increments.
- With probability 1, t \mapsto Z_t is a continuous function.

Now let A_t be a bounded, continuous
process, adapted to \mathcal F_t.
*Def*: We define
\int_0^t A_sdB_s = \lim_{n \to \infty}\int_0^t A_s^{(n)} dB_s
where A_s^{(n)} is a sequence
of simple processes approaching A_s.

**Lemma**: For every t,
we can find a sequence of simple processes A_t^{(n)} with |A_t^{(n)} \leq K and such that
\lim_{n \to \infty} \int_0^t E[(A_s^{(n)} - A_s)^2]ds = 0.

*Proof*: For ease, take t =
1. Then,
A_t^{(n)} = n \int_{\frac{k-1}{n}}^{\frac{k}{n}}A_sds
does the job. Scale appropriately.

Now if Z_t^{(n)} = \int_0^t A_s^{(n)}dB_s, then E[(Z_t^{(n)} - Z_t^{(m)})^2] = \int_0^tE[(A_s^{(n)} - A_s^{(m)})^2]ds, so Z_t^{(n)} is a Cauchy sequence in L^2(\Omega), and there is therefore an L^2 limit that we call Z_t. Further, there is a continuous modification of Z_t, which we can show exists by defining it on dyadics and using continuity.

**Prop**: Let Z_t = \int_0^t
A_sdB_s, with A_s continuous in
L^2. Then the above properties all
hold.

We do not necessarily need continuity. We do need progressive measurability, e.g. \{A_s(\omega)\} is measurable on \Omega \times [0, t].

If A_s is continuous, however, without any other boundedness assumptions, let \tau_n = \min \{ |A_s| \geq n \}. Then let A_s^{(n)} = A_{s \land \tau_n} so that Z_t^{(n)} = \int_0^t A_s^{(n)}dB_s is well-defined, and moreover if n > \sup_{0 \leq s \leq t} |A_s|, then Z_t^{(n)} = \int_0^t A_sdB_s. Thus, we define Z_t = \int_0^t A_sdB_s = \lim_{n \to \infty}Z_s^{(n)} which is a pointwise limit in \Omega (so we have to be careful). With this definition, linearity, continuity, and the Itô isometry holds (allowing for \infty as a possible value), but Z_t is merely a local martingale.

**Prop**: If M_t is a
continuous square integrable martingale starting from 0, the for all
\epsilon > 0,
P \left(\max_{0 \leq s \leq t}|M_s| \geq \epsilon\right) \leq
\frac{E(M_t^2)}{\epsilon^2}.

Now, set M_t = Z_t - Z_t^{(n)} is a continuous square-integrable martingale; thus by the above proposition, P(\max_{0 \leq t \leq 1} |M_t| \geq \epsilon ) \leq \frac{E(M_t^2)}{\epsilon^2}.

**Theorem**: Suppose A_s is a continuous process with \int_0^1 E[A_s^2)ds < \infty, and \Pi^{(n)} is a sequence of partitions of
[0, 1] such that
\sum_{n=1}^\infty \int_0^1 E[|A_t - A_t^{(n)}|^2]dt < \infty.
If Y^{(n)} = \max_{0 \leq t \leq
1}|Z_t - Z_t^{(n)}|, then Y^{(n)} \to
0 with probability 1.

*Proof*: Apply Borel-Cantelli to the above.

If Z_t = \int_0^t A_sdB_s, then the quadratic variation is \left\langle Z\right\rangle_t = \int_0^t A_s^2ds and Z_t - \left\langle Z \right\rangle_t is a martingale. In fact, the quadratic variation is the unique increasing process such that it is a martingale.

**Theorem**: Suppose that f:
\mathbb{R}\to \mathbb{R} is C^2,
and B_t is a standard Brownian motion.
Then
f(B_t) - f(B_0) = \int_0^t f'(B_s)dB_s + \frac{1}{2}\int_0^t
f''(B_s)ds.

*Proof*: Taylor approximate; prove for a countable dense set
of t, and conclude for f with compact support. Otherwise, take
something that agrees with f on [-n. n] send n \to
\infty.

People often write this in differential form: dX_t = R_tdt + A_tdB_t.

Suppose that f(t, x) is a function from [0, \infty) \times \to \mathbb{R} that is C^1 in t and C^2 in x. Then, f(t, B_t) - f(0, B_0) = \int_0^t \partial_s f(s,B_s)ds + \int_0^t \partial_x f(s, B_s) + \frac{1}{2} \int_0^t\partial^2_{xx}f(s, B_s)ds

Suppose f is C^2, but not all of \mathbb{R}, merely on an open interval (a, b), and B_t is a Brownian motion starting in the interval. Then, if T = \inf \{ t \mid B_t = a, b\}, then if t < T, then Ito’s formula holds.

*Def*: Suppose that dX_t = R_tdt +
A_tdB_t; that is,
X_t = X_0 + \int_0^t R_sds + \int_0^t A_sdB_s.
Then, we define
\int_0^t Y_sdX_s = \int_0^t Y_sR_sds + \int_0^t Y_sA_sdB_s
and
\left\langle X \right\rangle_t = \lim_{n \to \infty} \sum_{j <
t/n} (X_{j/n} - X_{(j-1)/n})^2.

If you look closely at X, you can see that the quadratic variation becomes \left\langle X \right\rangle_t = \lim_{n \to \infty} (A_{j/n} - A_{(j-1)/n})^2 = \int_0^t A_s^2ds.

**Prop**: If f(t,x) is
C^1 in t and C^2 in
x, and X_t is as above,
f(t, X_t) - f(0, X_0) = \int_0^t \partial_t f(s,X_s)ds + \int_0^t
\partial_x f(s, X_s)dX_s + \frac{1}{2} \int_0^t \partial_{xx}f(s,
X_s)A^2_sds.

*Def*: **Diffusions** are SDEs of the form
dX_t = m(t, X_t)dt + \sigma(t, X_t)dB_t
where m, \sigma are
deterministic. Something satisfies the above if
X_t = y_0 + \int_0^t m(s, X_s)ds + \int_0^t \sigma(s, X_s)dB_s.
for some initial condition y_0.

**Prop**: When m,
\sigma are uniformly Lipschitz in the latter arguments, a
solution to the above exists.

*Proof*: Let m, \sigma be
uniformly Lipschitz with constant \beta. For ease, assume that there is no
s-dependence (you don’t need this, but
writing it is annoying). We proceed by Picard iteration.

Let X_t^{(0)} = y_0. Define X_t^{(n)} by X_t^{(n)} = y_0 + \int_{0}^t m(X_t^{(n-1)})ds + \int_0^t \sigma(X_s^{(n-1)})dB_s.

We wish to show X_t = \lim_{n \to \infty} X_t^{(n)} exists (in L^2). Look at \begin{align*} E[|X_t^{(k+1)} - X_t^{(k)}|^2] &= E \left[ \left| \int_0^t (m(X_s^{(k)}) - m(X_s^{(k-1)}))ds + \int_0^t (\sigma(X_s^{(k)}) - \sigma(X_s^{(k-1)}))dB_s \right|^2 \right] \\ &\leq 2E \left[ \left| \int_0^t (m(X_s^{(k)}) - m(X_s^{(k-1)}))ds \right|^2 \right] + E\left[ \left| \int_0^t (\sigma(X_s^{(k)}) - \sigma(X_s^{(k-1)}))dB_s \right|^2 \right] \\ &\leq E \left[ \left(\int_0^t \beta^2 |X_s^{(k)} - X_s^{(k-1)}|\right)^2 \right] + \int_0^t E[\sigma(X_s^{(k)}) - \sigma(X_s^{(k-1)})]^2dB_s \\ &\leq \beta^2 E[t \int_0^t |X_s^{(k)} - X_s^{(k-1)}|^2ds] + \beta^2 \int_0^t E[|X_s^{(k)} - X_s^{(k-1)}|^2]ds \\ &\leq Ct \int_0^t E[|X_s^{(k)} - X_s^{(k-1)}|^2] \end{align*} and specifically, one can now show E[|X_t^{k+1} - X_t^{(k)}|] \leq \frac{\lambda^{k+1}t^{k+1}}{(k+1)!} which vanishes as k \to \infty.

*Def* Let X_t be a diffusion
as above. The **generator** of X_t is
Lf(x) = \lim_{t \to 0} \frac{E^x[f(X_t)] - E^x[f(X_0)]}{t}.
Let f be C^2; then
\begin{align*}
f(X_t) - f(X_0) &= \int_0^t f'(X_s)dX_s + \frac{1}{2}
\int_0^t f''(X_s)d \left\langle X \right\rangle_s \\
&= \int_0^t f'(X_s)\sigma(X_s)dB_s + \int_0^t
(f'(X_s)m(X_s) + \frac{1}{2}f''(X_s)\sigma^2(X_s))ds.
\end{align*}
Taking the expectation, under suitable regularity conditions (we
will take \sigma, m bounded), we get
that
E[f(X_t)] - E[f(X_0)] = 0 + E[t(f'(X_0)m(X_0) +
\frac{1}{2}f''(X_0)\sigma^2(X_0) + o(t^2)]
and
Lf(x) = m(x)f'(x) + \frac{\sigma^2(x)}{2} f''(x).

**Prop**: Let
dX_t = R_tdt + A_tdB_s
and
dY_t = S_tdt + C_tdB_s.
Then if we define the covariation as
\left\langle X, Y \right\rangle_t = \lim_{n \to \infty} \sum_{j <
tn}(X_{j/n} - X_{(j-1)/n})(Y_{j/n} - Y_{(j-1)/n}) = \int_0^t A_sC_sds
we have
dX_tY_t = X_tdY_t + Y_tdX_t + d\left\langle X, Y\right\rangle_t.

Let B_t = (B_t^1, \dots, B_t^d) be a Brownian motion. Then, \left\langle B^j, B^k \right\rangle_t = 0 for k \neq j. You can then write multidimensional stochastic integrals as stochastic integrals in each of the different dimensions; that is, dX_t^{i} = R_t^{i}dt + \sum_{j=1}^d A_t^{(i,j)} dB_t^j. The covariation is then \left\langle X^{(j)}, X^{(k)} \right\rangle_t = \sum_{i=1}^d A^{(j, i)}_tA^{(k, i)}_t.

**Theorem (Multivariate Itö)**: Suppose f(t, x) is a map from \mathbb{R}^{n + 1} \to \mathbb{R}, and is
C^1 in t and C^2 in
x. Then,
X_t = (X_1^{(1)}, \dots, X_t^{(n)})
satisfies
df(t, X_t) = \partial_t f(t, X_t)dt + \sum_{j=1}^{n}
\partial_{x_j}f(t, X_t)dX_t^{(j)} + \frac{1}{2}\sum_{j=1}^n\sum_{k=1}^n
\partial_{x_j x_k}f(t, X_t)d \left\langle X^{(j)}, X^{(k)}
\right\rangle_t.

If Z_t = \int_0^t A_sdB_s, then Z_t is not necessarily a martingale, but it is a local martingale.

*Def*: A process Z_t is a
**continuous local martingale** on [0, \tau) (where \tau could be \infty) if there is a sequence of stopping
times \tau_1 \leq \tau_2 \leq \cdots
such that a.s. \lim_{n \to \infty} \tau_n =
\tau and for each n, Z_{t \land \tau_n} is a continuous
martingale.

**Prop**: Stochastic integrals are local
martingales.

*Proof*: Take hitting times as stopping times.

Let X_t be a geometric Brownian motion dX_t = mX_tdt + \sigma X_t dB_t.

Suppose we have a European option, such that at some fixed time T > 0 and fixed price S, we can exercise the option to buy some asset at T for S. Then the value of an option at time T is F(X_t), where F(x) = (x - s)_+ = \max \{x - s, 0\}.

Let \varphi(t, x) be the value of this option at time t < T; that is, \varphi(t, x) = E[e^{-r(T - t)}F(X_T) \mid X_t = x]. What PDE does \varphi(t, x) satisfy?

More generally, let \begin{gather*} dX_t = m(t, X_t)dt + \sigma(t, X_t)dB_t \\ dR_t = r(t, X_t)R_tdt \\ R_t = R_0 \exp\left( \int_0^t r(s, X_s)ds \right) \\ \varphi(t, x) = E \left[ \exp\left(-\int_{t}^Tr(s, X_s)ds\right) F(X_T) \mid X_t = x \right] \end{gather*} Now define M_t = E[R_T^{-1}F(X_T) \mid \mathcal F_t] = R_t^{-1}E \left[ \exp \left(-\int_t^T r(s,X_s)ds\right) F(X_T) \mid \mathcal F_t\right] so M_t = R_t^{-1}\varphi(t, X_t). is a martingale.

Assume for now that \varphi is sufficiently regular to apply Ito; then \begin{align*} d\varphi(t, X_t) &= \partial_t \varphi dt + \partial_x \varphi dX_t + \frac{1}{2}\partial_{xx} \varphi d \left\langle X \right\rangle_t \\ &= \partial_t \varphi dt + \partial_x \varphi(mdt + \sigma dB_t) + \frac{1}{2}\partial_{xx} \varphi\sigma^2 dt \end{align*} and if you compute you get the following.

**Theorem**: Let X_t be
a geometric Brownian motion as above. Then, if \varphi is as above and is C^1 in t and
C^2 in x, then \varphi satisfies the PDE
\partial_t \varphi(t, x) = -m(t,x) \partial_x \varphi(t,x) -
\frac{1}{2}\sigma(t,x)^2 \partial_{xx} \varphi(t,x) + r(t,x)
\varphi(t,x)
with terminal condition \varphi(T,x) =
F(x).

Suppose that we have some SDE dX_t = m(X_t)dt + \sigma(X_t)dB_t with m, \sigma Lipschitz. Then the generator is Lf(x) = m(x)f'(x) + \frac{\sigma^2(x)}{2}f''(x) and, if we have initial condition F(x), then u(t, x) = E^x[F(X_t)] and \partial_t u(t,x) = L_x u(t,x), \ \ u(0, x) = F(x). On the other hand, if we have a terminal condition at t = T, we have \partial_t\varphi(t, x) = L_x(x, t), \ \ \varphi(T, x) = F(X) and \varphi(t, x) = u(T - t, x), \ \ \partial_t \varphi(t, x) = -L_x \varphi(t, x).

*Example*: Suppose we have a geometric Brownian motion with
m(x) = mx, \sigma(x) = \sigma x,
alongside some interest rate r. Then,
\varphi(t, x) = E[e^{-r(T - t)}F(X_T)]
and
\partial_t \varphi(t, x) = r\varphi(t, x) - mx\varphi'(t,x) -
\frac{\sigma^2}{2}x^2 \varphi''(t, x).

*Def*: If we have f:[0, 1] \to
\mathbb{R}, its **variation** is
V_f = \sup_{0 = t_0 < \dots < t_n = 1} \sum_{j=1}^n |f(t_j) -
f(t_{j-1})|.
We say that f has
**bounded variation** if V_f <
\infty.

**Prop**: Let M_t be a
continuous martingale with respect to the filtration \{ \mathcal F_t \} and M_0 = 0. If M_t has paths of bounded variation, then
M_t = 0 for all t with probability 1.

*Proof*: We will show that E[M_1^2]
= 0. Then, in the case V_M(1) \leq K
< \infty,
E[M_1^2] = E \left[ \sum_{j=1}^n (M_{j/n} - M_{(j-1)/n})^2 \right].
Bound the inner sum by \delta_n
\sum_{j=1}^n |M_{j/n} - M_{(j-1)/n}| \leq \delta_n K where \delta_n is the maximal increment. Since
M_t is continuous, \delta_n K \to 0. If we do not have that
uniform bound, set \tau_K = \min \{ t: V_M(t)
= K \}. Then we have E[M_{1 \land
\tau_K}^2] = 0 and we conclude by monotone convergence.

*Def*: The quadratic variation \left\langle M \right\rangle_t is the unique
increasing process such that M_t^2 -
\left\langle M \right\rangle_t is a martingale.

**Theorem**: If M_t is
a continuous martingale with respect to \mathcal F_t with quadratic variation \left\langle M \right\rangle_t, then M_t is a standard Brownian motion ith respect
to \mathcal F_t.

*Proof*: Certainly M_0 = 0
and we have continuous paths. We only need to show that E[e^{i\lambda (M_t - M_s)} \mid \mathcal F_s] =
e^{-\frac{\lambda^2(t-s)}{2}}. Since the definitions of the Ito
integral and Ito’s formula go through with only the quadratic variation
assumption, we apply the Ito formula to f(x) =
e^{i\lambda x}, so
f(M_t) - f(M_0) = \int_0^t f'(M_s)dB_s + \frac{1}{2} \int_0^t
f''(M_s)ds
and in expectation
E[f(M_t) - f(M_0)] = -\frac{\lambda^2}{2} \int_0^t E[f(M_s)]ds
since f'' =
-\lambda^2f. Then just solve the differential equation.

Take a standard Brownian motion B_s, 0 \leq s \leq 1; the Wiener measure \mathcal W is a measure on the Borel \sigma-algebra of C[0,1], the space of continuous functions with the \sup norm. Let B(f, \epsilon) = \{ g \mid |f - g| < \epsilon\}. Then, \mathcal W[B(f, \epsilon)] = P(|f(t) - B_t| < \epsilon). Put \mathcal W_{m, \sigma^2} if we have a drift/variance term.

**Prop**: \mathcal W_{0, 1}
\perp \mathcal W_{0, \sigma^2} for \sigma^2 \neq 1. On the other hand, \mathcal W_{0, 1} is mutually absolutely
continuous with \mathcal W_{m, 1}.

*Proof*: The first part is easy. Look at A, the functions with quadratic variation 1;
then \mathcal W_{0, 1}(A) = 1, \mathcal W_{0, \sigma^2}(A) = 0. In the
second case, we go to a weak version of the Girsanov theorem.

**Theorem**: \mathcal W
and \mathcal W_{m, 1} are mutually
absolutely continuous and
\frac{d\mathcal W_{m,1}}{d \mathcal W} = \exp\left(mB_1(\omega) -
\frac{1}{2}m^2\right).

*Proof*: Let B_t be a
standard Brownian motion, and define M_t =
\exp(mB_t - \frac{m^2t}{2}) so that dM_t = mM_t dB_t. On \mathcal F_t, define a probability measure
Q_t such that
Q_t[V] = E[1_V M_t]
for all V \in \mathcal F_t; by
conditioning on \mathcal F_s, we can
see that if s < t and V \in \mathcal F_s, then Q_s(V) = Q_t(V). It is clear that the
Radon-Nikodym derivatives are just given by M_t and 1/M_t.

Now define Q a probability measure on \mathcal F_\infty such that if \mathcal F_{t}, Q(V) = E[1_V M_t], which is well defined because of the above. Moreover, if X is \mathcal F_t measurable, this gives that E_Q[X] = E[X M_t]. Now we claim that B_t is a Brownian motion under Q with variance parameter 1 and drift m. Certainly it holds that B_0 = 0 and t \mapsto B_t is continuous with probability 1 under Q and P (since they are mutually absolutely continuous), and if s, t > 0 then the conditional distribution of B_{t + s} - B_s given \mathcal F_s is N(mt, t). In fact, all we need to show then is that E_Q[\exp(\lambda(B_{t+s} - B_s)) \mid \mathcal F_s] = \exp\left(\lambda mt + \frac{\lambda^2 m^2 t }{2}\right). This essentially boils down to the definition of the conditional expectation: write it down and unwrap in terms of P-expectations and win.

In the above omitted computation, we essentially do a simple version of the following theorem.

**Theorem**: Let B_t be
a Brownian motion on a probability space (\Omega, \mathcal F, P). Suppose we have a
nonnegative martingale M_t, M_0 = 1
satisfying
dM_t = A_tM_t dB_t.
Then, M_t = M_0e^{Y_t} where
Y_t = \int_0^t A_sdB_s - \frac{1}{2} \int_0^t A_s^2 ds.
This follows by a simple application of Ito. Define a measure
Q_t on \mathcal F_t, given by
Q_t[V] = E[M_t 1_V]
so that \frac{dQ_t}{dP} = M_t
and if s < t and V \in \mathcal F_s, Q_s[V] = Q_t[V]. Then, we define Q as before. Let X_t = B_t - \int_0^t A_sds; then X_t is a standard Brownian motion on (\Omega, \mathcal F, Q).

*Proof*: We just do a heuristic argument. As an approximation,
B_{t + \Delta t} - B_t = \begin{cases}
\sqrt{\Delta t} & \text{probability } 1/2 \\
-\sqrt{\Delta t} & \text{probability } 1/2 \\
\end{cases}
and similarly M_{t + \Delta t} = M_t(1
\pm A_t \sqrt{\Delta t}) equiprobably as well. In the new
measure, this is essentially tilting the probabilities by 1 \pm A_t \sqrt{\Delta t} and so E_Q[B_{t + \Delta t} - B_t] = A_t \Delta
t.

Check that X_t is a Q-martingale and conclude by noting that it has quadratic variation t.

Let B_t = (B_t^1, B_t^2) be a two dimensional Brownian motion; identify it to B_t^1 + iB_t^2.

Let f(B_t) = u(B_t) + iv(B_t). Then du(B_t) = \nabla u \cdot dB_t + \frac{1}{2} (\Delta u) dt and dv(B_t) = \nabla v \cdot dB_t + \frac{1}{2} (\Delta v) dt but if f is holomorphic, then the Laplacians vanish and we use the Cauchy-Riemann equations to simplify to du(B_t) = u_x(B_t)dB_t^{1} + u_y(B_t)dB_t^2 and dv(B_t) = v_x(B_t)dB_t^{1} + v_y(B_t)dB_t^2 which have quadratic variations d\langle u(B_t) \rangle_t = d\langle v(B_t) \rangle = (u_x^2 + u_y^2)dt = |f'(B_t)|^2 dt.

Define T(t) = \int_0^T |f'(B_s)|^2ds = T. Then Y_t = f(B_{T(t)}) is a standard complex Brownian motion.

*Def*: A *Levy process* is a stochastic process that
satisfies

- X_0 = 0 almost surely;
- stationary increments: X_t - X_s \sim X_{t - s};
- independent increments;
- continuity in probability: X_t \to 0 in probability as t \to 0.

*Def*: A *compound Poisson process* is composed of a
Poisson process N_t with some rate
\lambda > 0, alongside Y_1, Y_2, \dots mutually independent of
themselves and N_t and identically
distributed random variables with distribution \mu^\sharp. Then,
X_t = \sum_{i=1}^{N_t} Y_i
is the compound Poisson process.

*Def*: A random variable X
has an *infinitely divisible* distribution if for each n we may find Y_1,
\dots Y_n iid such that X \sim
\sum_{i=1}^n Y_i.

**Prop**: Levy processes are infinitely divisible at any
time.

Now look at the characteristic functions of a Levy process. In particular define the characteristic exponent \Psi(s) where \varphi_{X_1}(s) = \exp(\Psi(s)) and note that \varphi_{X_t}(s) = (\varphi_{X_1}(s))^t.

Let X_t, Y_1, \dots be as in a compound Poisson process. Then, if we set \varphi(s) = \varphi_{Y_j}(s) = \int_{-\infty}^\infty \exp(isx)\mu^\sharp (dx) we have \varphi_{X_1}(s) = \sum_{n=0}^\infty P(N_1 = n) E[e^{sX_1} \mid N_1 = n] = \sum_{n=0}^\infty \frac{\lambda^n}{n!} e^{-\lambda} (\varphi(s))^n = \exp(\lambda(\varphi(s) - 1)) or equivalently, \Psi(s) = \int_{-\infty}^\infty (e^{isx} - 1)\mu(dx) where \mu = \lambda \mu^\sharp. We call \mu the Levy measure for X_t.

**Prop**: There are a few properties that we can say: as
t \to 0,

- P(N_t = 0) = 1 - \lambda t + o(t);
- P(N_t \geq 2) = o(t);
- P(X_t \in (a, b)) = \mu(a,b)t + o(t) for 0 \notin (a, b).

*Def*: We define the generator of a Levy process as
Lf(x) = \lim_{t \to 0}\frac{E^x[f(X_t) - f(x)]}{t}
= \int_{-\infty}^\infty (f(x + y) - f(x)) \mu(dy).

**Prop**: If X_t, Y_t
are independent Levy processes, then X_t +
Y_t is a Levy process with characteristic exponent \Psi_X + \Psi_Y and generator L_x + L_y.

*Def*: Let X_t be a Levy
process with measure \mu, and put
m = \int_{-\infty}^\infty x\mu(dx), \ \sigma^2 =
\int_{-\infty}^\infty x^2 \mu(dx).
Then E[X_t] = mt and \mathop{\mathrm{Var}}(X_t) = \sigma^2 t.

*Def*: A function f:[0, \infty) \to
\mathbb{R} is called cadlad (continue a droite, limite a gauche)
if the paths are right-continuous and for all t, f(t-)
exists.

*Def*: If X_t is compound
Poisson, then
\int_0^t A_s dX_s = \sum A_s (X_s - X_{s-})
where the sum is over jump times.

The issue is that if A_t is adapted, then A_t = X_t shows that \int A_s dX_s = 1 at the first jump, and is not a martingale; in fact we need A_t adapted to \mathcal F_{t-} instead.