[Stochastic Process] Geometric Brownian Motion

1. Motivation

Definition: Let $X(t)$ be Brownian motion with drift coefficient $\mu$ and variance parameter $\sigma^2$. Let $Y(t) = e^{-X(t)}$. Then $Y(t)$ is geometric Brownian motion.

Motivation: Let $Y(_n)$ be the price of a stock at time $n$ (where n is distance); Let $X(n) = \frac{Y_n}{Y_+{n-1}}$ be the fractional increase/decrease in the price of the stock from time n-1 to time n.
We suppose that $X_n$ are i.i.d. Then
$Y_n = X_n Y_{n-1} = X_n X_{n-1} X_{n-1} \cdots X_1 Y_0$
$ln Y_n = ln X_n X_{n-1} X_{n-1} \cdots X_1 Y_0 = ln Y_0 + \sum^n_{i=1} ln X_i$

The process $ln(Y_n)$ looks like a random walk.


2. Property of Geometric Brownian Motion

• $E[Y(t) | Y(s) = y_s] = y_s E[e^{{X(t) - X(s)}}]$

Proof: $E[Y(t) | Y(s) = y_s] = E[e^{X(t)} | X(s) = ln y_s]$

                                             = $E[e^{X(t) - X(s) + X(s)} | X(s) = ln y_s]$

                                             = $y_s E[e^{X(t) - X(s)}]$

• E[Y(t) | Y(s) = y_s] = y_s e^{\mu(t-s) + \sigma^2(t-s)/2}

Proof: If W ~ $N(\mu, \sigma^2)$, then $e^W$ is lognormal with mean $E[e^W] = e^{\mu + \sigma^2/2}$
Thus, {X(t) - X(s)} ~ $N[\mu(t-s), \sigma^2(t-s)]$, since $X(t)$ is Brownian motion with drift

• Note: if $\mu = 0$, then $E[Y(t)|Y(0) = y_0 ] = y_0  e^{\sigma^2 t/2}$. Thus $E[Y(t)]$ is increasing even though the jump process with $X(t)$ is symmetric

3. Example
Question: You invest 1000 dollars in the stock market. Suppose that the stock market can be modeled using geometric Brownian motion with an average daily return of 0.03% and a standard deviation of 1.02%. what is the probability that your money increases after 1 year (260 business days?) 10 years? 30 years?

Answer: Let $X(t) = \mu t + \sigma B(t)$, where $B(t)$ is standard Brownian Motion.
Let $Y(t) = Y_0 e^{X(t)}$ where $B(t)$ is standard Brownian motion, $X(t)$ is Brownian motion with drift, and $Y(t)$ is geometric Brownian motion.

$Pr(Y(t) > Y_0) = Pr(Y_0 e^{X(t)} > Y_0)$
                           = $Pr(e^{X(t)} > 1)$
                           = $Pr(X(t) > ln 1 = 0)$
                           = $Pr(N(\mu t, \sigma^2 t) > 0)$
                           = $1 - \Phi(\frac{-\mu t}{\sigma \sqrt{t}})$
                           = $\Phi(\frac{\mu \sqrt{t}}{\sigma})$