Introduction
In this part, we will continue looking at Brownian motions and Gaussian processes. Namely, we will firstly look at zeroes of Brownian motions and introduce the Brownian bridge. Then, we will discuss Brownian motion with a drift term and geometric Brownian motion.
Zeroes of Brownian Motion
Intuition: Zeroes of Brownian Motion
Let $L$ be the last zero in $(0, 1)$ of a Brownian motion. (i.e., $L = \max \ \{t : 0 < t < 1, B_t = 0\}$. Then, $$ L \sim \mathrm{Beta}\left(\frac{1}{2}, \frac{1}{2}\right). $$
Proof: Outline of Proof
$$ \begin{align*} P(L > s) & = \int_{-\infty}^{\infty} P(L > s \mid B_s = t) \ \mathcal{N}(t; 0, s) \ dt \newline & = 2 \int_{-\infty}^{0} P(L > s \mid B_s = t) \ \mathcal{N}(t; 0, s) \ dt \newline & = 2 \int_{-\infty}^{0} P(M_{1 - s} > -t) \ \mathcal{N}(t; 0, s) \ dt \newline & = 2 \int_{0}^{\infty} 2 P(B_{1 - s} > t) \ \mathcal{N}(t; 0, s) \ dt \newline & = 4 \int_{0}^{\infty} \int_{t}^{\infty} \mathcal{N}(r; 0, 1 - s) \ \mathcal{N}(t; 0, s) \ dr \ dt \newline & = \ldots & = \frac{1}{\pi} \int_s^1 \frac{1}{\sqrt{x(1 - x)}} \ dx \newline & = \int_s^1 \mathrm{Beta}\left(x; \frac{1}{2}, \frac{1}{2}\right) \ dx. \newline \end{align*} $$ We can then, let $L_t$ be the last zero in $(0, t)$. Then, $$ \frac{L_t}{t} \sim \mathrm{Beta}\left(\frac{1}{2}, \frac{1}{2}\right). $$
Note
The probability that a Brownian motion has at least one zero in $(r, t)$ for $0 \leq r < t$ is $1 - P(L_t < r)$.
Further, the cumulative distribution for the $\mathrm{Beta}\left(\frac{1}{2}, \frac{1}{2}\right)$ density can be computed with the arcsin function, $$ P(L_t < r) = \int_0^{\frac{r}{t}} \mathrm{Beta}\left(s; \frac{1}{2}, \frac{1}{2}\right) \ ds = \frac{2}{\pi} \arcsin\left(\sqrt{\frac{r}{t}}\right). $$
Brownian Bridge
Definition: Brownian Bridge
Define a Gaussian process $X_t$ by conditioning a Brownian motion $B_t$ on $B_1 = 0$. Then $X_t$ is a Brownian bridge.
If $0 < s < t < 1$, then $(B_s, B_t, B_1)$ is a multivariate normal with, $$ \begin{align*} \mathbb{E}[(B_s, B_t, B_1)] & = (0, 0, 0) \newline \mathrm{Var}((B_s, B_t, B_1)) & = \Sigma = \begin{pmatrix} s & s & s \newline s & t & t \newline s & t & 1 \newline \end{pmatrix}. \newline \end{align*} $$ Conditioning on $B_1 = 0$ and using the properties of the multivariate normal we get that $\mathbb{E}[X_t] = 0$ and, $$ \mathrm{Cov}(X_s, X_t) = s(1 - t) = s - st. $$ It follows that this is identical to the Brownian bridge defined above.
Brownian Motion with Drift and Geometric Brownian Motion
Definition: Brownian Motion with Drift
For any real $\mu > 0$ and $\sigma > 0$ we can define the Gaussian process $X_t$ as, $$ X_t = \mu t + \sigma B_t. $$ This is a Brownian motion with a draft, and is often a more useful model than standard Brownian motion.
Note
Note that $X_t$ is Normal with expectation $\mu t$ and variance $\sigma^2 t$.
This is a Gaussian process with continuous paths and stationary and independent increments.
Definition: Geometric Brownian Motion
The stochastic process, $$ G_t = G_0 e^{\mu t + \sigma B_t}, $$ where $G_0 > 0 $ is called a geometric Brownian motion with drift parameter $\mu$ and variance parameter $\sigma^2$.
$\log(G_t)$ is a Gaussian process with expectation $\log(G_0) + \mu t$ and variance $\sigma^2 t$.
Note
One can show that, $$ \begin{align*} \mathbb{E}[G_t] & = G_0 e^{t(\mu + \frac{1}{2} \sigma^2)} \newline \mathrm{Var}(G_t) & = G_0^2 e^{2 t(\mu + \sigma^2)} (e^{\sigma^2 t} - 1). \newline \end{align*} $$
Proof
Using that $\log(G_t) \sim \mathcal{N}(\log(G_0) + \mu t, \sigma^2 t)$ we have, $$ \begin{align*} \mathbb{E}[G_t] & = \mathbb{E}[e^{\log(G_t)}] \newline & = \int_{-\infty}^{\infty} e^x \ \mathcal{N}(x; \log(G_0) + \mu t, \sigma^2 t) \ dx \newline & = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi \sigma^2 t}} \exp\left(-\frac{(x - (\log(G_0) + \mu t))^2}{2 \sigma^2 t} + x\right) \ dx \newline & = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi \sigma^2 t}} \exp\left(-\frac{(x - (\log(G_0) + \mu t - \sigma^2 t))^2}{2 \sigma^2 t} + \log(G_0) + \mu t + \frac{1}{2} \sigma^2 t\right) \ dx \newline & = e^{\log(G_0) + \mu t + \frac{1}{2} \sigma^2 t} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi \sigma^2 t}} \exp\left(-\frac{(x - (\log(G_0) + \mu t - \sigma^2 t))^2}{2 \sigma^2 t}\right) \ dx \newline & = e^{\log(G_0) + \mu t + \frac{1}{2} \sigma^2 t} \newline & = G_0 e^{t(\mu + \frac{1}{2} \sigma^2)}. \newline \end{align*} $$ Similarly, we can compute $\mathbb{E}[G_t^2]$ and use $\mathrm{Var}(G_t) = \mathbb{E}[G_t^2] - (\mathbb{E}[G_t])^2$ to get the variance, $$ \begin{align*} \mathbb{E}[G_t^2] & = \mathbb{E}[e^{2 \log(G_t)}] \newline & = \int_{-\infty}^{\infty} e^{2x} \ \mathcal{N}(x; \log(G_0) + \mu t, \sigma^2 t) \ dx \newline & = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi \sigma^2 t}} \exp\left(-\frac{(x - (\log(G_0) + \mu t))^2}{2 \sigma^2 t} + 2x\right) \ dx \newline & = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi \sigma^2 t}} \exp\left(-\frac{(x - (\log(G_0) + \mu t - 2 \sigma^2 t))^2}{2 \sigma^2 t} + \log(G_0) + \mu t + 2 \sigma^2 t\right) \ dx \newline & = e^{\log(G_0) + \mu t + 2 \sigma^2 t} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi \sigma^2 t}} \exp\left(-\frac{(x - (\log(G_0) + \mu t - 2 \sigma^2 t))^2}{2 \sigma^2 t}\right) \ dx \newline & = e^{\log(G_0) + \mu t + 2 \sigma^2 t} \newline & = G_0^2 e^{t(2 \mu + 2 \sigma^2)}. \newline \end{align*} $$ Thus, $$ \begin{align*} \mathrm{Var}(G_t) & = G_0^2 e^{t(2 \mu + 2 \sigma^2)} - \left(G_0 e^{t(\mu + \frac{1}{2} \sigma^2)}\right)^2 \newline & = G_0^2 e^{2 t(\mu + \sigma^2)} (e^{\sigma^2 t} - 1). \newline \end{align*} $$ $_\blacksquare$
Modeling with Brownian Motion and Geometric Brownian Motion
Example: Stock Prices
From what we have seen so far, one can model the price of a stock with:
- Use a continuous-time stochastic model.
- Consider the factor with which it changes, not the differences in prices.
- Consider the log-normal distribution for such factors.
- Use a parameter for the trend of the price (drift), and one for the volatility (variance) of the price.
- Make a Markov assumption (although, we have to reflect on this choice).
This leads to using a geometric Brownian motion as a model, $$ G_t = G_0 e^{\mu t + \sigma B_t}. $$ In this context $\sigma$ is called the volatility of the stock.
Example
A stock price is modeled with $G_0 = 67.3$, $\mu = 0.08$, and $\sigma = 0.3$. What is the probability that the price is above 100 after 3 years?
Solution
We want to compute,
$$
\begin{align*}
P(G_3 > 100) & \coloneqq P\left(G_0 e^{\mu \cdot 3 + \sigma B_3} > 100\right) \newline
& = P\left(e^{\mu \cdot 3 + \sigma B_3} > \frac{100}{G_0}\right) \newline
& = P\left(\mu \cdot 3 + \sigma B_3 > \log\left(\frac{100}{G_0}\right)\right) \newline
& = P\left(B_3 > \frac{\log\left(\frac{100}{G_0}\right) - \mu \cdot 3}{\sigma}\right) \newline
\end{align*}
$$
Since $B_3 \sim \mathcal{N}(0, 3)$ we can compute this as 1 - pnorm((log(100 / 67.3) - 0.08 * 3) / 0.3, mean = 0, sd = sqrt(3)) in R.