14.1.1 Normal distribution

A random variable X is said to follow a normal distribution with mean and variance¹ σ² if its probability density function (p.d.f.) is given by

for all , or equivalently if its cumulative distribution function (c.d.f.) is given by

for all . In both cases, this is denoted by

In particular, a random variable Z is said to follow a standard normal distribution if μ = 0 and σ² = 1 and this is denoted by . The p.d.f. and the c.d.f. of the standard normal distribution are given by

(14.1.1)

and

respectively. Because of the symmetry (around zero) of φ, we have that N(x) = 1 − N( − x), for all . The functions φ( · ) and N( · ) will be used very frequently in the following chapters. Figure 14.1 shows the p.d.f. (left panel) and c.d.f. (right panel) of the standard normal distribution.

 The notation emphasizes that probabilities are computed with the (actuarial) probability measure . This level of precision about the probability measure used is not crucial at this point but will become important for option pricing, as it was in the binomial model.

The normal distribution is also uniquely determined by its moment generating function (m.g.f.): if , then its m.g.f. (w.r.t. ) is given by

(14.1.2)

for all , and vice versa. Recall that the m.g.f. is a function of λ, while μ and σ² are (fixed) parameters.

The normal distribution also has the following properties:

1. If , then . Conversely, if , then . This last transformation, from X to Z, is known as standardization. Hence, we deduce that
   (14.1.3)
   for all .
2. The normal distribution is additive: if and are independent, then .

Bivariate and multivariate normal distributions

We say that the couple (X₁, X₂) follows a bivariate normal distribution, or bivariate Gaussian distribution, if the bivariate p.d.f. of (X₁, X₂) is given by

where

for all .

The bivariate normal distribution is specified by five parameters: the means (μ₁, μ₂), the variances (σ²₁, σ₂²) and the correlation − 1 ⩽ ρ ⩽ 1. Indeed, we can deduce that and , and also that .

A multivariate normal distribution is defined similarly for a random vector (X₁, X₂, …, X_n). It is characterized by a vector of means and a covariance matrix. The multivariate normal distribution is also very important in modern portfolio theory to determine asset allocation.

14.1.2 Lognormal distribution

A random variable X is said to be lognormally distributed if Y = ln (X) follows a normal distribution. In other words, if Y follows a normal distribution then X = e^Y is said to follow a lognormal distribution. Note that this means that a lognormal random variable is a positive random variable.

More precisely, we say that X follows a lognormal distribution with parameters μ and σ², which we will denote by

if . Then, its p.d.f. is given by

(14.1.4)

Figure 14.2 shows the p.d.f. (left panel) and c.d.f. (right panel) of a lognormal distribution. It should be clear that the lognormal distribution is not a symmetric distribution as is the normal distribution. Also, it puts no probability mass on the negative part of the real line.

Example 14.1.1

Assume . Compute the following probability: .

First, taking the log on both sides, we can write

We know that . Then, standardizing Y, i.e. subtracting μ = −1 and dividing by to obtain a standard normal random variable , we can write

where we used the symmetry (around zero) of the standard normal distribution.

 ◼ 

We can generalize the calculations of the last example and identify the c.d.f. of a lognormally distributed random variable . Using equation (14.1.3), we see that it is simply given by

since, for x > 0, we have , where .

The lognormal distribution is not additive: if X₁ and X₂ are independent and lognormally distributed (possibly with different parameters), then we can not say that X₁ + X₂ is also lognormally distributed. However, it inherits the following two properties from the normal distribution:

1. If , then .
2. The lognormal distribution is multiplicative: if X₁ and X₂ are independent and lognormally distributed (possibly with different parameters), then X₁ × X₂ is also lognormally distributed.

Let us verify the second property. Assume that and are independent. This implies that and are also independent. Since the normal distribution is additive, we have that . Hence,

This last property will be very useful when dealing with geometric Brownian motion in Section 14.5.

Mean and variance

Computing the mean and variance of the lognormal distribution follows readily from the m.g.f. of the normal distribution as given in equation (14.1.2). Indeed, if , then and we can write

(14.1.5)

and

Finally, we deduce that

Note that we have not used the p.d.f. (nor the c.d.f.) of the lognormal distribution to compute these first two moments.

High order moments and moment generating function

It is easy to deduce that a lognormally distributed random variable X, with parameters μ and σ², has moments of all orders: for any n ⩾ 1, we have

However, the m.g.f. of X does not exist, which means that we cannot write for a positive value of λ (this expectation does not exist). In fact, no matter the value of λ > 0, the integral

diverges. This means that the lognormal distribution is not uniquely determined by its moments.

Truncated expectations

In the following chapters, we will often compute truncated expectations of lognormally distributed random variables. This will be useful for option pricing in the Black-Scholes-Merton model.

Let us consider the following truncated expectation

where and where a > 0.

It is very tempting to immediately write down an integral using the p.d.f. of the lognormal distribution. Instead, let us use the fact that we can write X = e^{μ + σZ}, where

Then, the truncated expectation can be rewritten as

(14.1.6)

To ease notation, let us set c = (ln (a) − μ)/σ and compute

The trick here is to complete the square: since we have that

then

Using this last equality together with (14.1.6), we can conclude that

Moreover, since , we can also write

where in the last step we used the symmetry of the standard normal distribution.

Stop-loss transforms

Recall that we defined in Chapter 6 the stop-loss function (x − a)₊. Using its properties, we can write

The stop-loss transform² of X, defined by , is thus

Finally, using the above calculations, we get the following identity:

(14.1.7)

This is the Black-Scholes formula in disguise. We will come back to this in Chapter 16.

14.2.1 Markovian property

A SRW possesses the Markovian property, i.e. at each time step it depends only on the last known value, not on the preceding ones. Mathematically, for 0 ⩽ m < n,

for any , or equivalently

for any . Clearly, this comes from the fact that

No extra knowledge about ε₁, ε₂, …, ε_m is needed, except for the cumulative value X_m.

Said differently, if we want to predict the value of X_n (in the future) knowing the whole random walk up to time m (the present), only the knowledge of X_m is useful.

Consequently, for a function g( · ), we have

(14.2.2)

In particular, when m = n − 1, we have

Note that we do not need the random walk to be symmetric for it to be a Markov process.

14.2.2 Martingale property

A SRW possesses the martingale property, i.e. the prediction of a process in the future is given by its last known value. Mathematically, for 0 ⩽ m < n,

(14.2.3)

Since the SRW is also a Markov process, using (14.2.2), we can further write

Again, using the fact that X_n = X_m + ∑ⁿ_{i = m + 1}ε_i and the linearity property of the conditional expectation, we further have

since ∑ⁿ_{i = m + 1}ε_i is independent of X_m = ∑^m_{j = 1}ε_j and for each i. The fact that the random walk is symmetric is crucial for the martingale property.

Being a martingale, the coin toss game is said to be a fair game because the expected gain/loss at each coin toss (each one-step increment) is equal to zero.

 It is important to understand that a stochastic process with the Markov property does not necessarily possess the martingale property. In particular, we know from equation (14.2.2) with g(x) = x that a Markov process X is such that

but this conditional expectation is not necessarily equal to X_m as in (14.2.3). We will encounter such a process in Section 14.5.3, namely geometric Brownian motion. The converse is also not true, meaning that a stochastic process that has the martingale property does not necessarily have the Markov property.

14.3.1 Construction as the limit of symmetric random walks

The symmetric random walk X = {X_n, n ⩾ 0} defined in equation (14.2.1) is a discrete-time stochastic process. To make it a continuous-time process, we can interpolate the trajectories or, even simpler, keep them constant in between time points. We will choose the latter option. More precisely, for each t ⩾ 0, set

where k is a non-negative integer. Note that for each t ⩾ 0, there is a unique such k: it is called the integer part of t and is often written as ⌊t⌋. In other words, we now have a continuous-time version of the SRW:

for each t ⩾ 1, while X_t = 0 for each 0 ⩽ t < 1. A sample path of this process is shown in Figure 14.4.

Our objective now is to see what happens (and what we would obtain) if we speed up this process, i.e. if we allow for more than one movement per unit time interval. If we want to have n movements per unit time interval, we can simply consider the process X⁽ⁿ⁾ defined by

Indeed, between time 0 and time 1, the process makes n jumps:

In fact, this is the case for any unit time interval: for t ⩾ 0,

where ⌊n(t + 1)⌋ − ⌊nt⌋ = n. A sample path of the accelerated random walk X⁽²⁾ is shown in Figure 14.5.

However, those n jumps are all of magnitude 1. We could keep on increasing n and see what happens at the limit, but we know from the Central Limit Theorem (CLT) that this is going nowhere. We must normalize the εs if we want to obtain some sort of convergence. Let us make this normalization depend on the number of jumps per unit time interval, that is let the process have jumps of magnitude .

We are now ready to identify the sequence of symmetric random walks that will converge to a standard Brownian motion. For each n ⩾ 1, we define the process W⁽ⁿ⁾ = {W⁽ⁿ⁾_t, t ⩾ 0} by

See the panels of Figure 14.6 for sample paths of W⁽¹⁾, W⁽²⁾, W⁽⁴⁾ and W⁽⁸⁾.

For each fixed n, the continuous-time process W⁽ⁿ⁾ = {W⁽ⁿ⁾_t, t ⩾ 0} is a sort of (continuous-time) accelerated and rescaled symmetric random walk. When it moves, the process W⁽ⁿ⁾ moves upward or downward with magnitude and equal probability 1/2. Consequently, for each fixed n, the increments of W⁽ⁿ⁾ = {W⁽ⁿ⁾_t, t ⩾ 0} are independent (over non-overlapping time intervals) and stationary. These properties are inherited from the underlying SRW X.

More precisely, we have:

1. for any m ⩾ 1 and 0 ⩽ t₁ < t₂ < … < t_m, the random variables
   
   are independent;
2. for any s, t ⩾ 0, the random variables W⁽ⁿ⁾_{t + s} − W⁽ⁿ⁾_t and W⁽ⁿ⁾_s have the same distribution.

Note the similarities with the increments of X.

It is now time to take the limit when n goes to infinity. Since and , for each i ⩾ 1, we have that

for any n ⩾ 1 and t ⩾ 0.

From the CLT, we know that

converges to a -distributed random variable, as n goes to infinity. Let us denote this random variable obtained at the limit by W₁. In other words, we have obtained

For any integer time t = k ⩾ 1, as

then, from the CLT, we have that

converges to a -distributed random variable, as n goes to infinity. Let us denote this random variable obtained at the limit by W_k, for each integer k ⩾ 1.

In general, at any real time t > 0, we will have that

converges to a -distributed random variable, as n goes to infinity, since

as n goes to infinity.³ Let us denote this random variable obtained at the limit by W_t, for each t > 0. Therefore, we have obtained: for all t > 0,

In conclusion, if we further set W₀ = 0 and if we regroup the normal random variables W_t just obtained, then we have a continuous-time stochastic process W = {W_t, t ⩾ 0}. This process is called a standard Brownian motion. Luckily enough, this new process will also have independent and stationary increments.

14.3.2 Definition

Formally, the process W = {W_t, t ⩾ 0} we have just obtained is a standard Brownian motion (SBM), also known as a Wiener process. Mathematically, a standard Brownian motion is a continuous-time stochastic process issued from zero (W₀ = 0), with independent and normally distributed (stationary) increments:

• for all n ⩾ 1 and for any choice of time points 0 ⩽ t₁ < t₂ < … < t_n, the following random variables are independent:
  
• for all s < t, the random variable W_t − W_s has the same probability distribution as .

Figure 14.7 shows a sample path of a standard Brownian motion. We will explain in Section 14.3.6 how to generate such trajectories.

This definition of standard Brownian motion follows directly from its construction as the limit of random walks, namely its increments over non-overlapping time intervals are independent and normally distributed.

14.3.3 Distributional properties

From the definition of Brownian motion, we deduce that for each t > 0, the random variable W_t is normally distributed with mean 0 and variance t, i.e. for all ,

where and . Note that the variance (and the second moment) increases linearly with time.

Example 14.3.1

What is the probability that a Brownian motion is below − 1 at time 3.5?

We want to compute the following probability: . Since , then

 ◼ 

It is not enough to know the distribution of W_t at each time t to fully characterize a Brownian motion. We also need to specify the dependence structure. In fact, a standard Brownian motion is a Gaussian process, meaning that the random vector , extracted from this process, follows a multivariate normal distribution, for any choice of fixed times t₁, t₂, …, t_n. In particular, for fixed times s and t, the joint normal distribution of (W_s, W_t) is characterized by its means, variances and its covariance function .

We already know that , for all t ⩾ 0. So, all that is left is to compute , for all s, t ⩾ 0, to fully specify the distribution of W. To compute the covariance function, we will rely on the properties of the increments.

First, note that

can be simplified to

since . Now, if we assume that s < t, then W_t − W_s is independent of W_s − W₀ = W_s and we can write

By symmetry, if we assume that s > t, then we get . Consequently, we have obtained that

(14.3.1)

for all s, t ⩾ 0.

As a conclusion, the values taken by a Brownian motion at two different time points are not independent (the covariance is not equal to zero). That should have been expected since W_s is somewhat included in W_t. Indeed, by definition of the approximating sequence of symmetric random walks, we had

Clearly, W⁽ⁿ⁾_s and W⁽ⁿ⁾_t are not independent. Since W_s and W_t can be obtained as the limits of W⁽ⁿ⁾_s and W⁽ⁿ⁾_t, they are also expected to be dependent, which is the case.

Example 14.3.2

What is the probability that a Brownian motion is below − 1 at time 3.5, knowing that it was equal to 1 at time 1.25?

We want to compute the following conditional probability: . Using the independence between W_3.5 − W_1.25 and W_1.25, we can write

where we used the fact that the increment W_3.5 − W_1.25 follows a normal distribution with mean 0 and variance 2.25.

 ◼ 

14.3.4 Markovian property

A standard Brownian motion is a Markov process, i.e. for any 0 ⩽ t < T, we have

for all . In words, this means that the conditional distribution of W_T given the history of the process up to time t is the same as its conditional distribution given W_t. Saying that a standard Brownian motion is a Markovian process means that for any fixed time t, knowing the value of the random variable W_t provides the same information as knowing the whole (truncated) trajectory W_s, 0 ⩽ s ⩽ t for the prediction of future values, in particular that of W_T.

The Markovian property of a standard Brownian motion W = {W_t, t ⩾ 0} is also inherited from the same property for each W⁽ⁿ⁾ = {W⁽ⁿ⁾_t, t ⩾ 0} in the approximating sequence of accelerated and rescaled symmetric random walks.

Consequently, for a sufficiently well-behaved function g( · ), we can write

Note that the conditional expectation on the right-hand side is a function of W_t, so it is a random variable.

Building on the independence between W_T − W_t and W_t, we can further write

(14.3.2)

Here is how to compute/understand the last expectation:

1. Consider x as a dummy variable and compute using the fact that , which will generate an expression in terms of x.
2. Set f(x) to be this expression computed in step 1).
3. Set .

Of course, the temporary function f and then the final expectation both depend on g. The following two examples will illustrate these steps.

Example 14.3.3

Let us compute using equation (14.3.2) with g(x) = x².

The first step is to compute

where x is a dummy variable. Since , expanding the square, we get

which is, as announced, a function of x.

Now, we set f(x) = (T − t) + x² and, finally, we have

As expected, this random variable is a function of W_t.

 ◼ 

It turns out that equation (14.3.2) is not vital in the previous example, but it will be quite handy in the following example.

Example 14.3.4

Let us compute using equation (14.3.2) with g(x) = (x)₊ = max (x, 0).

The first step is to compute

where x is a dummy variable. Since , then

since y + x = (y + x)₊ > 0 if and only if y > −x. Finally, since

we deduce that

We also have that 1 − N( − x) = N(x).

Now, we set

and, finally, we have

 ◼ 

14.3.5 Martingale property

Standard Brownian motion also inherits the martingale property from the approximating sequence of accelerated and rescaled symmetric random walks. More precisely, a standard Brownian motion W = {W_t, t ⩾ 0} is a martingale, i.e. for any 0 ⩽ t < T, we have

Note the resemblance with the condition in (14.2.3).

As W = {W_t, t ⩾ 0} is also a Markov process, we only need to show that

to verify that it is a martingale.

Once again, we will use the fact that W_T = W_t + (W_T − W_t) and the independence between W_T − W_t and W_t. Using the linearity of conditional expectations, we can write

So, a standard Brownian motion is indeed a martingale.

Example 14.3.5

Let us verify that M_t = W²_t − t, defined for each t ⩾ 0, is a martingale with respect to Brownian motion. In other words, we want to verify the following martingale property: for 0 ⩽ t < T, we want to have

Using the Markov property of Brownian motion, we already know that

and, by linearity of conditional expectations, we further have

Finally, from example 14.3.3, we already know that

Putting the pieces together yields the result.

 ◼ 

14.3.6 Simulation

For many applications in finance and actuarial science, simulations of Brownian motion trajectories over a given time interval [0, T] help generate stochastic scenarios of useful economic variables such as stock prices and interest rates (see also Chapter 19).

Since Brownian motion is a continuous-time stochastic process, it is impossible to simulate every W_t for every t ∈ [0, T] as this time interval is (uncountably) infinite. Instead, we choose n time points 0 < t₁ < t₂ < … < t_n = T and simulate the random vector . This random vector is a discretized version of a Brownian motion trajectory over the time interval [0, T]. Of course, one should take n as large as possible. Simulation methods for continuous-time stochastic processes are often called discretization schemes.

To simulate a (discretized) path (W(0), W(h), W(2h), …, W((n − 1)h), W(T)) of a standard Brownian motion over the time interval [0, T], we make use of its definition. The algorithm is as follows:

1. We first choose n ⩾ 1 to divide the time interval [0, T] into smaller sub-intervals of the same size h = T/n: 0 < h < 2h < 3h < ... < (n − 1)h < T.
2. We define W(0) = 0 and then, for each i = 1, 2, …, n, we:
   1. generate (see also section 19.1.3);
   2. compute
      

3. The output is (W(0), W(h), W(2h), …, W((n − 1)h), W(T)), a sampled discretized trajectory of a standard Brownian motion, which is a synthetically generated realization of the random vector

   

Figure 14.8 gives two examples of such a discretized Brownian motion trajectory: the grey line is a trajectory with 100 time steps while the dotted line is a trajectory with 20 time steps.

 Even if we know that , for each i = 1, 2, …, n, we cannot simulate successively , then , and so on, with finally . Doing so would generate a sample of independent random variables and we know, from the above discussion, that the random vector has a multivariate normal distribution with dependence structure given by the covariance function in equation (14.3.1).

Example 14.3.6 Simulation of a standard Brownian motion

Using a computer, you have generated the following four realizations from the standard normal distribution:

Using the random numbers in this order, let us generate a discretized sample path for a standard Brownian motion over the time interval .

We have n = 4 and and thus . Therefore, we define:

Then,

is the corresponding discretized trajectory of a standard Brownian motion. It is shown in Figure 14.9.

 ◼ 

14.4.1 Distributional properties

Because X_t is obtained as a linear transformation of W_t which is normally distributed, X_t is also normally distributed. For each t > 0, we have

We see that the mean increases (resp. decreases), as time goes by, at rate μ > 0 (resp. μ < 0) while the variance increases at rate σ².

Example 14.4.1

What is the probability that a linear Brownian motion, issued from X₀ = 0, with drift μ = −0.25 and with volatility σ = 0.1 is below − 1 at time 3.5?

We want to compute the following probability: , where

Since , we can write

 ◼ 

For fixed times 0 ⩽ s < t, we have

As the increment W_t − W_s is normally distributed with mean 0 and variance t − s, the increment X_t − X_s is also normally distributed but with mean μ(t − s) and variance σ²(t − s). Also, since the increments of a Brownian motion (over disjoint time intervals) are independent, and because a linear Brownian motion is just one affine transformation away from a standard Brownian motion, its increments are also independent.⁴

More precisely, a linear Brownian motion is (also) a stochastic process with independent and normally distributed (stationary) increments:

• for all n ⩾ 1 and for all choice of time points 0 ⩽ t₁ < t₂ < … < t_n, the following random variables are independent:
  
• for all s < t, the random variable X_t − X_s has the same probability distribution as .

Example 14.4.2

For a linear Brownian motion starting at X₀ = −2, with drift μ = 1 and diffusion σ = 3, calculate the probability that the process is below 10 at time 5 if it is already at 6.5 at time 2.

We need to compute . Subtracting X₂ on both sides of the inequality, we obtain

as X₅ − X₂ is independent from X₂.

Moreover, the increment X₅ − X₂ follows a normal distribution with mean 1 × (5 − 2) = 3 and variance 3² × (5 − 2) = 27. Therefore, the desired probability is

 ◼ 

Finally, we can compute the covariance of a linear Brownian motion at two different time points. Indeed, we have

14.4.2 Markovian property

Standard Brownian motion also transfers its Markovian property to linear Brownian motion X = {X_t, t ⩾ 0}. For any fixed times t and T such that 0 ⩽ t < T, we have

for all .

From the definition of linear Brownian motion, we know that W_s = σ^{− 1}(X_s − X₀ − μs), for all 0 ⩽ s ⩽ t. As a result, we quickly realize that conditioning on X_s, 0 ⩽ s ⩽ t is equivalent to conditioning on W_s, 0 ⩽ s ⩽ t.

In other words, saying that the linear Brownian motion X = {X_t, t ⩾ 0} is a Markov process also means that, for any 0 ⩽ t < T, we have

for all .

Consequently, to compute an expectation of the form

where t ⩽ T and where g( · ) is a function, we can rely on the algorithm behind equation (14.3.2). Indeed, since we can write

we are dealing with the same type of expectation.

Example 14.4.3

Let us compute , where 0 ⩽ t < T.

First, let us write

Using the linearity property of conditional expectations and expanding the square, we further have

where, in the last step, we used previously computed conditional expectations for standard Brownian motion (see the previous section).

 ◼ 

14.4.3 Martingale property

In general, linear Brownian motions are not martingales because they have a trend coming from the drift coefficient μ. However, when μ = 0, the corresponding linear Brownian motion takes the form X_t = X₀ + σW_t, for all t ⩾ 0. In this case, it is a martingale.

Formally, for t < T, we have

where, in the second last step, we used the fact that a standard Brownian motion is a martingale.

In conclusion, a linear Brownian motion X possesses the martingale property, i.e.

for all 0 ⩽ t < T, if and only if μ = 0.

14.4.4 Simulation

To simulate a (discretized) path (X(0), X(h), X(2h), …, X((n − 1)h), X(T)) of a linear Brownian motion with coefficients μ and σ over the time interval [0, T], there are two equivalent algorithms.

The first algorithm mimics the one for a standard Brownian motion:

1. We first choose n ⩾ 1 to divide the time interval [0, T] into smaller sub-intervals of the same size h = T/n: 0 < h < 2h < 3h < ... < (n − 1)h < T.
2. We define X(0) = X₀ and then, for each i = 1, 2, …, n, we:
   1. generate ;
   2. compute
      

3. The output is (X(0), X(h), X(2h), …, X((n − 1)h), X(T)), a sampled discretized trajectory of a linear Brownian motion, which is a synthetically generated realization of the random vector

   

The second algorithm relies on the fact that we might have already simulated a standard Brownian motion, i.e. that we have generated

Then, we apply the corresponding transformation: for each i = 0, 1, …, n, we set

Again, the output is (X(0), X(T/n), X(2T/n), …, X((n − 1)T/n), X(T)) which is illustrated in Figure 14.11.

Example 14.4.4 Simulation of a linear Brownian motion

Using the same random numbers as in example 14.3.6 (in the same order), we can generate a discretized sample path for a linear Brownian motion with drift μ = 2 and diffusion σ = 10, over the time interval . Assume the process starts at 0, i.e. that X₀ = 0.

Again, we have n = 4, and . Therefore,

Then,

is the corresponding discretized trajectory of this linear Brownian motion. See Figure 14.12.

 ◼ 

14.5.1 Distributional properties

From the definition, we have that S_t is lognormally distributed, for each t > 0. Indeed, since we have

where

then

We deduce, from equation (14.1.5), that the mean (function) of a GBM is given by

We see that this mean increases (geometrically) with time if and only if .

Similarly, using the results of Section 14.1.2, we can compute the variance of a GBM (or any higher order moments):

Geometric Brownian motions have been used for decades as a mathematical model for the price of a risky asset. For example, GBM serves as a model for the stock price in the framework proposed by Black, Scholes and Merton (BSM). They are also used to generate economic scenarios in the banking and insurance industry.

Example 14.5.1

Assume a stock price evolves according to a GBM with parameters μ = 0.07 and σ = 0.3. The current stock price is S₀ = 100. Assume the time unit is a calendar year. What is the probability that the stock price will be greater than $120 in 3 years from now?

We want to compute . We have

 ◼ 

From the definition, for 0 ⩽ t < T, we can write

As X_T − X_t is normally distributed, then S_T/S_t is lognormally distributed.

As opposed to standard and linear Brownian motions, the increments of a geometric Brownian motion are not independent nor stationary. However, since the increments of a linear Brownian motion (over disjoint time intervals) are independent and because a GBM is just an exponential transformation of a linear Brownian motion, then the relative increments of the GBM are independent.

More generally,

• for all n ⩾ 1 and for all choice of time points 0 ⩽ t₁ < t₂ < … < t_n, the following random variables are independent:
  
• for all t < T, the random variable S_T/S_t has the same probability distribution as .

From the properties of the relative increments of GBM, we can easily compute a conditional expectation of the form , where t < T. Indeed, we can write

because is independent of S_t (or equivalently of ). Consequently, since

because , we have that

(14.5.1)

As geometric Brownian motions are widely used to model asset prices, it is important to interpret this model from a financial standpoint. For t₂ > t₁, the random variable is the accumulation factor of $1 invested in S over the time interval [t₁, t₂]. Taking the logarithm of this accumulation factor, we get

the log-return of this asset, between time t₁ and time t₂.

Since the log-return has a normal distribution with mean μ(t₂ − t₁) and variance σ²(t₂ − t₁), μ can be interpreted as the mean annual log-return whereas σ is the annual volatility of the asset's log-returns.

14.5.2 Markovian property

Since geometric Brownian motion is one deterministic transformation away from standard Brownian motion (or linear Brownian motion), it is also a Markov process. This means that for any 0 ⩽ t < T, we have

for all .

As before, for a fixed t > 0, conditioning on S_u, 0 ⩽ u ⩽ t is equivalent to conditioning on X_u, 0 ⩽ u ⩽ t or even W_u, 0 ⩽ u ⩽ t. Also, note that if we know the value of S_t, then we know the value of W_t, and vice versa. Consequently, for a GBM, the Markovian property can be restated as follows:

for all .

To compute an expectation of the form

where g( · ) is a function, we can rely on the algorithm based on equation (14.3.2). Again, since we can write

we are dealing with the same type of expectation.

We can also use more explicitly the fact that S_T/S_t is independent of S_t and write

In this case, since knowing W_t is the same as knowing the value of S_t, we can compute this last expectation using the algorithm based on equation (14.3.2). This is illustrated in the following example.

Example 14.5.2

Let us compute , where t < T.

Following the previous methodology, we first compute

where x is a dummy variable. Since x > 0, using a property of the lognormal distribution, we have that . Therefore, we can apply the formula for the stop-loss transform of a lognormal distribution, as obtained in (14.1.7), and deduce that

with m = ln (x) + μ(T − t) and b² = σ²(T − t). To conclude, we replace x by S_t in the last expression and then we obtain

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14.5.3 Martingale property

In general, geometric Brownian motions are not martingales because they have an exponential trend. Indeed, from equation (14.5.1), for 0 ⩽ t < T, we have

In other words, S satisfies the martingale property, i.e.

for all 0 ⩽ t < T, if and only if the parameters μ and σ are such that μ + σ²/2 = 0.

14.5.4 Simulation

As for linear Brownian motion, to simulate a (discretized) path (S(0), S(h), S(2h), …, S((n − 1)h), S(T)) of a geometric Brownian motion with coefficients μ and σ over the time interval [0, T], there are two equivalent algorithms.

The first algorithm is based on the properties of the relative increments:

1. We first choose n ⩾ 1 to divide the time interval [0, T] into smaller sub-intervals of the same size h = T/n: 0 < h < 2h < 3h < ... < (n − 1)h < T.
2. We define S(0) = S₀ and then, for each i = 1, 2, …, n, we:
   1. generate ;
   2. compute
      

3. The output is (S(0), S(h), S(2h), …, S((n − 1)h), S(T)), a sampled discretized trajectory of a geometric Brownian motion, which is a synthetically generated realization of the random vector

   

The second algorithm relies on the fact that we might have already simulated a standard Brownian motion (or a linear Brownian motion), i.e. that we have generated

Then, we apply the corresponding exponential transformation: for each i = 0, 1, …, n, we set

Again, the output is a realization of the random vector (S₀, S_h, S_2h, …, S_{(n − 1)h}, S_T).

Example 14.5.3 Simulation of a geometric Brownian motion

A stock currently trades for $100 and its mean annual log-return is 7% whereas its volatility is 25%. Using the random numbers from example 14.3.6 (in the same order), we can generate a discretized sample path for this geometric Brownian motion over the time interval .

Again, we have n = 4, and . Furthermore, we have S₀ = 100, μ = 0.1 and σ = 0.25. Therefore,

Then,

is the corresponding discretized trajectory of this geometric Brownian motion. It is shown in Figure 14.14.

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14.5.5 Estimation

Given the importance of GBM in financial and actuarial applications, a natural question arises: how should/can we determine the values of μ and σ given asset price data?

Assume that the time unit is a calendar year and that data is collected periodically (weekly, daily,⁵ hourly, etc.) at each time step of length h for a total of T years. Overall, we will have n = T/h observations. For example, if h = 1/12 and we collect T = 10 years of data, then we have a total of n = 120 observations.

The idea is to capitalize on the fact that the relative increments of a GBM are independent and identically distributed according to a lognormal distribution. Said differently, the random variables

are independent and distributed according to the same normal distribution: for all j = 1, 2, …, n, we have

For example, if μ and σ are annual parameters and if we collect monthly data for n consecutive months, i.e. if we have realized values for ln (S_j/12/S_{( j − 1)/12}), where j = 1, 2, …, n, then these log-returns are drawn from a normal distribution with mean μ × 1/12 and variance σ² × 1/12.

Fortunately, estimating parameters of a normal distribution is straightforward. We will estimate the parameters μ and σ² using maximum likelihood estimation. Recall that the maximum likelihood estimators (MLEs) of the mean and the variance of a normal distribution are given by the sample mean and the sample variance. More precisely, if we have the following observations

and if we set x_j = ln (s_j/s_{j − 1}), for each j = 1, 2, …, n, then the MLEs and , of μ and σ respectively, are given by

where and . Recall that s²_x is a biased estimator for the variance of the x_is but this bias decreases quickly for large samples. If one is concerned with such bias, one could replace s²_x by

Example 14.5.4 Maximum likelihood estimation of GBM

Assume that you have the following monthly data for the stock price of ABC inc.:

Date	Price
December 31st, 2015	51
January 31st, 2015	54
February 28th, 2015	61
March 31st, 2015	53
April 30th, 2015	49

In practice, the analyst would use the closing price of the asset at the end of the last trading day of any given month. Let us compute the MLE estimates of μ and σ corresponding to these observations.

We have n = 4 and h = 1/12. From our sample (s₀, s₁, s₂, s₃, s₄) = (51, 54, 61, 53, 49) of prices, we must compute the corresponding sample of continuously compounded monthly log-returns: x_j = ln (s_j/s_{j − 1}), for each j = 1, 2, 3, 4. And then we must compute the sample mean and sample variance.

The log-returns observed over each month are given in the following table:

Month	Log-return
January
February
March
April

The sample mean and variance are

Then, we easily obtain the following MLE estimates for the parameters:

 ◼ 

From the properties of the log function, we know that

and thus the MLE of μ can be further simplified as

Precision of the MLE for the drift

Although the computation of the MLE for μ is straightforward, it remains that is an imprecise estimator. Given that T = nh, then

No matter how frequently we collect data (how small h is), the variability of only decreases linearly with T, the number of years of data. For example, if we know that σ = 0.25, then 10 years of data implies that the 95% confidence interval around μ is which is very large when the mean annual log-return is typically μ ∈ [ − 0.2, 0.2]. The lesson here is that T needs to be large for the estimator of the drift to be precise.