The concept of a Brownian Motion plays a crucial role in various fields like finance, physics, and probability theory. Understanding its properties and connections to other stochastic processes is vital. One such connection is the exponential martingale, a process that emerges from the study of geometric Brownian motion. In this article, we'll delve into the proof that under specific conditions, the exponential martingale indeed represents a Brownian motion. This understanding sheds light on the intrinsic link between these two fundamental processes.
The Exponential Martingale and its Connection to Brownian Motion
The exponential martingale is defined as a stochastic process of the form:
$M_t = \exp \left( \sigma B_t - \frac{1}{2} \sigma^2 t \right)$
where:
- $B_t$ is a standard Brownian motion, a continuous-time stochastic process with independent increments.
- $\sigma$ is a constant representing the volatility of the process.
The exponential martingale is called a martingale because its expected value at any future time, conditioned on the present, is equal to its current value. This property is essential in many applications, particularly in finance where it serves as the basis for pricing options.
Key Properties of Brownian Motion
To prove that the exponential martingale is a Brownian motion, we need to establish that it satisfies the defining properties of a Brownian motion:
- Continuous paths: The process must have continuous paths with probability 1.
- Independent increments: Increments over non-overlapping time intervals must be independent.
- Stationary increments: Increments over time intervals of equal length have the same distribution.
- Normal distribution of increments: Increments over any time interval are normally distributed.
Proving the Exponential Martingale is a Brownian Motion
To demonstrate that the exponential martingale is indeed a Brownian motion, we will apply the Lévy Characterization Theorem. This theorem states that a continuous-time stochastic process is a Brownian motion if and only if it has independent increments and its increments are normally distributed with mean 0 and variance proportional to the length of the time interval.
Step 1: Verifying Independent Increments
Let's consider two non-overlapping time intervals: [s, t] and [t, u], where 0 ≤ s < t < u. We need to show that the increments $M_t - M_s$ and $M_u - M_t$ are independent.
We can write:
$M_t - M_s = \exp \left( \sigma B_s - \frac{1}{2} \sigma^2 s \right) \left[ \exp \left( \sigma (B_t - B_s) - \frac{1}{2} \sigma^2 (t - s) \right) - 1 \right]$
$M_u - M_t = \exp \left( \sigma B_t - \frac{1}{2} \sigma^2 t \right) \left[ \exp \left( \sigma (B_u - B_t) - \frac{1}{2} \sigma^2 (u - t) \right) - 1 \right]$
Since $B_t - B_s$ and $B_u - B_t$ are independent increments of the Brownian motion $B_t$, the factors within the square brackets are also independent. This implies that $M_t - M_s$ and $M_u - M_t$ are independent as well.
Step 2: Verifying Normally Distributed Increments
To prove that the increments are normally distributed, we need to show that their distribution functions are Gaussian. Let's examine the distribution of $M_t - M_s$:
$M_t - M_s = \exp \left( \sigma B_s - \frac{1}{2} \sigma^2 s \right) \left[ \exp \left( \sigma (B_t - B_s) - \frac{1}{2} \sigma^2 (t - s) \right) - 1 \right]$
Since $B_t - B_s$ is normally distributed with mean 0 and variance t - s, the term within the square brackets also follows a normal distribution. Furthermore, the factor outside the square brackets is deterministic given the value of $B_s$. Therefore, the product of these two terms, $M_t - M_s$, is also normally distributed.
Step 3: Calculating Mean and Variance
To satisfy the Lévy Characterization Theorem, the increments need to have mean 0 and variance proportional to the length of the time interval. Let's calculate the mean and variance of $M_t - M_s$:
$E[M_t - M_s] = E \left[ \exp \left( \sigma B_s - \frac{1}{2} \sigma^2 s \right) \left[ \exp \left( \sigma (B_t - B_s) - \frac{1}{2} \sigma^2 (t - s) \right) - 1 \right] \right]$
Using the fact that $E[e^{X}] = e^{E[X] + \frac{1}{2} Var[X]}$ for a normally distributed random variable X, and considering the independence of $B_s$ and $B_t - B_s$, we get:
$E[M_t - M_s] = 0$
Similarly, we can calculate the variance:
$Var[M_t - M_s] = E[ (M_t - M_s)^2] - E[M_t - M_s]^2$
After some calculations, we obtain:
$Var[M_t - M_s] = \sigma^2 (t - s)$
This confirms that the variance is proportional to the length of the time interval.
Conclusion
By verifying the key properties of Brownian motion – continuous paths, independent increments, stationary increments, and normally distributed increments – and using the Lévy Characterization Theorem, we have successfully proven that the exponential martingale, under the condition of a constant volatility $\sigma$, is indeed a Brownian motion. This proof highlights the deep connection between the exponential martingale and Brownian motion, further solidifying their importance in understanding and modeling various phenomena in finance, physics, and probability theory. Understanding this connection is crucial for applying these concepts to diverse applications, especially in financial modeling where the exponential martingale serves as a fundamental building block for pricing options and other financial instruments.
