4  Geometric Brownian Motion

A random variable \(\tilde{y}\) is lognormally distributed if it can be written as \(\tilde{y}= \mathrm{e}^{\tilde{x}}\) where \(\tilde{x}\) has a normal distribution. Another way of saying this is that \(\log \tilde y\) is normally distributed. We denote the expected value (mean) of any random variable using the symbol \(\mathbb{E}\). The mean of a lognormal random variable is given by the following.

Important Principle

If \(\tilde y = \tilde{x}\) where \(\tilde x\) is normally distributed with mean \(m\) and standard deviation \(s\), then \[\mathbb{E}[\tilde{y}]=e^{m + s^2/2};. \qquad(4.1)\]

An example of the distribution of a lognormal random variable is shown in Figure 4.1.

Code 
import numpy as np
from scipy.stats import lognorm
import plotly.graph_objects as go

# Generate x values
x = np.linspace(0, 5, 1000)

# Lognormal density (s=1, loc=0, scale=1 gives standard lognormal)
pdf = lognorm.pdf(x, s=1, loc=0, scale=1)

# Create plot
fig = go.Figure()
fig.add_trace(
    go.Scatter(
        x=x,
        y=pdf,
        mode='lines',
        name='Lognormal density',
        hovertemplate='x = %{x:.2f}<br>density = %{y:.2f}<extra></extra>'
    )
)

fig.update_layout(
    showlegend=False,
    xaxis_title='',
    yaxis_title='Lognormal density',
    template='plotly_white',
    height=300
)

fig.show()
Figure 4.1: The distribution of the random variable y = exp(x) where x has the standard normal distribution.

An important stochastic process is geometric Brownian motion given by \[ S_t=S_0\mathrm{e}^{\mu t- \sigma^2 t/2 + \sigma B_t} \qquad(4.2)\] for constants \(\mu\) and \(\sigma\), where \(B\) is a Brownian motion. For each time \(t\), the random variable \(S_t\) in Equation 4.2 is a lognormal random variable. Ito’s formula for exponentials of Ito processes implies \[ \frac{\,\mathrm{d}S}{S} = \mu\,\mathrm{d}t+\sigma\,\mathrm{d}B\;. \qquad(4.3)\] When we see an equation of the form Equation 4.3, we should recognize Equation 4.2 as the solution.

The process \(S\) is called a geometric Brownian motion. We interpret Equation 4.3 as stating that \(\mu\,\mathrm{d}t\) is the expected rate of change of \(S\) and \(\sigma^2\,\mathrm{d}t\) is the variance of the rate of change in an instant \(\,\mathrm{d}t\). We call \(\mu\) the drift and \(\sigma\) the volatility. The geometric Brownian motion grows at the average rate of \(\mu\), in the sense that \(\mathbb{E}[S_t] = \mathrm{e}^{\mu t}S_0\). This can be verified with the aid of Equation 4.1.

If \(\mu=0\), then \(\mathbb{E}[S_t] = S_0\). In fact, a geometric Brownian motion \(S\) as in Equation 4.2 and Equation 4.3 is a martingale when \(\mu=0\). It is an example of what is called an exponential martingale. An exponential is always positive, so an exponential martingale is always positive.

Taking the natural logarithm of Equation 4.2 gives an equivalent form of the solution: \[ \log S_t= \log S_0+\left(\mu -\frac{1}{2}\sigma^2\right)t + \sigma B_t\;. \qquad(4.4)\] This shows that \(\log S_t - \log S_0\) is a \((\mu-\sigma^2/2,\sigma)\)–Brownian motion. Given information at time \(t\), the logarithm of \(S(u)\) for \(u>t\) is normally distributed with mean \((u-t)(\mu-\sigma^2/2)\) and variance \((u-t)\sigma^2\). Because \(S\) is the exponential of its logarithm, \(S\) can never be negative. For this reason, a geometric Brownian motion is a better model for stock prices than is a Brownian motion. The differential of Equation 4.4 is \[ \,\mathrm{d}\log S_t = \left(\mu -\frac{1}{2}\sigma^2\right)\,\mathrm{d}t+ \sigma\,\mathrm{d}B_t\;. \qquad(4.5)\]

We can summarize this discussion as follows.

Important Principle

The equation \[\frac{\,\mathrm{d}S}{S} = \mu\,\mathrm{d}t+\sigma\,\mathrm{d}B\] is equivalent to the equation \[\,\mathrm{d}\log S = \left(\mu -\frac{1}{2}\sigma^2\right)\,\mathrm{d}t+ \sigma\,\mathrm{d}B\; .\] The solution of both equations is Equation 4.2 or the equivalent Equation 4.4.

We can simulate a path of a geometric Brownian motion \(S\) by first simulating \(\log S\) and then computing the exponential. We simulate the changes \(\Delta \log S\) as normally distributed random variables with mean equal to \((\mu-\sigma^2/2)\Delta t\) and variance equal to \(\sigma^2\Delta t\). We could also simulate the changes \(\Delta \log S\) using a binomial model as in Section 2.2.

Code 
import numpy as np
import plotly.graph_objects as go

n = 1000     # number of divisions
mu = 0.1    # drift
sigma = 0.3 # volatility
S0 =  100   # initial value
T = 1       # length of simulation
dt = T / n  # Delta t
steps = np.random.normal(
    loc = (mu-0.5*sigma**2)*dt, 
    scale = sigma*np.sqrt(dt),
    size = n
)
logS = np.empty(n+1)
logS[0] = np.log(S0)
logS[1:] = logS[0] + np.cumsum(steps)
S = np.exp(logS)

fig = go.Figure(
    go.Scatter(
        x=np.arange(0, T+dt, dt),
        y=S,
        mode="lines",
        hovertemplate="t = %{x:.2f}<br>S = %{y:.2f}<extra></extra>", #
    )
)

fig.update_layout(
    showlegend=False,
    xaxis_title="Time",
    yaxis_title="Simulated GBM",
    template="plotly_white",
    height=300,
)

fig.show()
Figure 4.2: A path of a simulated geometric Brownian motion with mu=0.1, sigma=0.3, and 1,000 steps.

4.1 Continuously Compounded Returns

Given a rate of return \(r\) over a time period of \(\Delta t\) years, the annualized continuously compounded return is defined to be the number \(x\) such that \(\mathrm{e}^{x\Delta t} = 1+r\); equivalently, \(x = \log(1+r) / \Delta t\). We frequently take “annualized” for granted and just say “continuously compounded return.” As an example, if you earn 1% in a month, then the continuously compounded return is \(\log (1.01) / (1/12) = 0.1194\).

The reason for the name “continuously compounded” is that compounding a monthly rate of interest of \(0.1194/12\) for a total of \(n\) times during a month produces a total return over the month of \[\left(1 + \frac{0.1194/12}{n}\right)^ n - 1\,,\] which converges to \(0.01\) as \(n \rightarrow \infty\). Thus, compounding an infinite number of times during a month at an annual rate of \(0.1194\) (equal to a monthly rate of \(0.1194/12\)) is equivalent to the actual 1% return.1

Given a dividend-reinvested asset price \(S\), the rate of return from date \(t_1\) to \(t_2\) is \(S_{t_2}/S_{t_1} - 1\). Denote this rate of return by \(r\) and defined the corresponding continuously compounded return \[x = \log(1+r) = \log\left(\frac{S_{t_2}}{S_{t_1}}\right) = \log S_{t_2} - \log S_{t_1}\,.\] Using Equation 4.4, we see that the continuously compounded return is \[ x = \log S_{t_2} - \log S_{t_1} = \left(\mu -\frac{1}{2}\sigma^2\right)(t_2-t_1)+ \sigma\,(B_{t_2} - B_{t_1})\;. \qquad(4.6)\] Thus, from the vantage point of date \(t_1\), the continuously compounded return is normally distributed with mean \[\left(\mu -\frac{1}{2}\sigma^2\right)\Delta t\] and variance \(\sigma^2 \Delta t\), where we define \(\Delta t = t_2-t_1\). Given historical data on rates of return, the parameters \(\mu\) and \(\sigma\) can be estimated by standard methods.

4.2 Tail Probabilities of Geometric Brownian Motions

For each of the numeraires discussed in the previous section, we have \[ \,\mathrm{d}\log S = \alpha\,\mathrm{d}t + \sigma\,\mathrm{d}B\;, \qquad(4.7)\]

for some \(\alpha\) and \(\sigma\), where \(B\) is a Brownian motion under the probability measure associated with the numeraire. Specifically, \(\sigma=\sigma\), \(B=B^*\), and

  1. for the risk-neutral probability, \(\alpha = r-q-\sigma^2/2\),
  2. when \(\mathrm{e}^{qt}S_t\) is the numeraire, \(\alpha = r-q +\sigma^2/2\),
  3. when another risky asset price \(Y\) is the numeraire, \(\alpha = r-q+\rho\sigma\phi-\sigma^2/2\).

We will assume in this section that \(\alpha\) and \(\sigma\) are constants. The essential calculation in pricing options is to compute \(\text{prob}(S_t>K)\) and \(\text{prob}(S_t<K)\) for a constant \(K\) (the strike price of an option), where \(\text{prob}\) denotes the probabilities at date \(0\) (the date we are pricing an option) associated with a particular numeraire.

Equation 4.7 gives us \[\log S_t = \log S_0 + \alpha T + \sigma B_t\; .\] Given this, we deduce

\[ S_t > K \quad\Longleftrightarrow\quad \log S_t > \log K \] \[ \quad\Longleftrightarrow\quad \sigma B_t > \log K - \log S_0-\alpha T \] \[ \quad\Longleftrightarrow\quad \frac{B_t}{\sqrt{T}} > \frac{\log K - \log S_0-\alpha T}{\sigma\sqrt{T}} \] \[ \quad\Longleftrightarrow\quad -\frac{B_t}{\sqrt{T}} < \frac{\log S_0-\log K + \alpha T}{\sigma\sqrt{T}} \] \[ \quad\Longleftrightarrow\quad -\frac{B_t}{\sqrt{T}} < \frac{\log \left(\frac{S_0}{K}\right) + \alpha T}{\sigma\sqrt{T}}\;. \qquad(4.8)\]

The random variable on the left-hand side of Equation 4.8 has the standard normal distribution—it is normally distributed with mean equal to zero and variance equal to one. As is customary, we will denote the probability that a standard normal is less than some number \(d\) as \(\mathrm{N}(d)\). We conclude:

Key Result

Assume \(\,\mathrm{d}\log S = \alpha\,\mathrm{d}t + \sigma\,\mathrm{d}B\), where \(B\) is a Brownian motion. Then, for any number \(K\), \[ \text{prob}(S_t>K) = \mathrm{N}(d)\;, \qquad(4.9)\] where \[ d = \frac{\log \left(\frac{S_0}{K}\right) + \alpha T}{\sigma\sqrt{T}}\;. \qquad(4.10)\]

The probability \(\text{prob}(S_t<K)\) can be calculated similarly, but the simplest way to derive it is to note that the events \(S_t>K\) and \(S_t<K\) are complementary—their probabilities sum to one (the event \(S_t=K\) having zero probability). Therefore \(\text{prob}(S_t<K) = 1-\mathrm{N}(d)\). This is the probability that a standard normal is greater than \(d\), and by virtue of the symmetry of the standard normal distribution, it equals the probability that a standard normal is less than \(-d\). Therefore, we have:

Key Result

Assume \(\,\mathrm{d}\log S = \alpha\,\mathrm{d}t + \sigma\,\mathrm{d}B\), where \(B\) is a Brownian motion. Then, for any number \(K\), \[ \text{prob}(S_t<K) = \mathrm{N}(-d)\;, \qquad(4.11)\] where \(d\) is defined in Equation 4.10.

4.3 Exercises

Exercise 4.1 Use Ito’s Lemma to derive the stochastic differential equation for \(S_t^2\). Show that \(S(t)^2\) is a geometric Brownian motion if \(\mu\) and \(\sigma\) are constants and find \(\mathbb{E}[S(t)^2]\).

Exercise 4.2 Let \[\begin{align*} \frac{\,\mathrm{d}S_1}{S_1} &= \mu_1\,\mathrm{d}t + \sigma_1 \,\mathrm{d}B_1\\ \frac{\,\mathrm{d}S_2}{S_2} &= \mu_1\,\mathrm{d}t + \sigma_2 \,\mathrm{d}B_2 \end{align*}\] where the \(\mu_i\) and \(\sigma_i\) are constants and \(B_1\) and \(B_2\) are Brownian motions with constant correlation \(\rho\).

  1. Define \(Y=S_1S_2\). Show that \(Y\) is a geometric Brownian motion and calculate its drift and volatility.
  2. Repeat for \(Y=S_1/S_2\).

Hint: use the facts \(\mathrm{e}^{x+y}=\mathrm{e}^x \times \mathrm{e}^y\) and \(\mathrm{e}^x/\mathrm{e}^y = \mathrm{e}^{x-y}\).

:

  1. The reason for considering this concept is that compound returns like \((1+r_1)(1+r_2)\) are simpler to analyze in some contexts as \(\mathrm{e}^{x_1+x_2}\).↩︎