Question

The Brownian motion or the Wiener process is the Gaussian process \(B\) such that \(\mathrm{E} B(t)=0\) and \(\operatorname{Cov}(B(s), B(t))=s \wedge t\) for \(s, t \geq 0\). (a) Show that \(B\) has independent increments. Show that \(B\) is a martingale and find the compensator of \(B^{2}\). The Brownian bridge (tied down Wiener process) \(B^{0}(t)\) with \(t \in[0,1]\) is the Gaussian process such that \(\mathrm{E} B^{0}(t)=0\) and \(\operatorname{Cov}\left(B^{0}(s), B^{0}(t)\right)=s(1-t)\) for \(0 \leq s \leq t \leq 1\). (b) Show that the processes \(B^{0}(t)\) and \(B(t)-t B(1)\) have the same distribution on \([0,1]\). (c) Show that the processes \(B(t)\) and \((1+t) B^{0}(t /(1+t))\) have the same distribution on \([0, \infty)\).

Short answer

In summary, we analyzed and worked through various properties of Brownian motion and the Brownian bridge process: a) Proved that the Brownian motion has independent increments, showed that it is a martingale, and found the compensator of \(B^2\) to be \(t\). b) Demonstrated that the processes \(B^0(t)\) and \(B(t)-tB(1)\) have the same distribution for \(t \in [0, 1]\) by showing they have the same first and second moments. c) Showed that the processes \(B(t)\) and \((1+t)B^0\left(\frac{t}{1+t}\right)\) have the same distribution for \(t \in [0, \infty)\) by again comparing their first and second moments. These results illustrate the importance and application of properties such as independent increments and martingale in the study of stochastic processes like Brownian motion.

Step-by-step solution

(a) Proving independent increments for Brownian motion

To show that the Brownian motion has independent increments, we need to prove that for any finite sequence of times \(0 \leq t_1 < t_2 < \dots < t_n\), the increments \(B(t_2) - B(t_1), B(t_3) - B(t_2), \dots, B(t_n) - B(t_{n - 1})\) are independent. The joint distribution of the increments \((B(t_2) - B(t_1), B(t_3) - B(t_2), \dots, B(t_n) - B(t_{n-1}))\) is a multivariate Gaussian distribution. For pairwise independence of a multivariate Gaussian distribution, it is enough to show that the covariance between any two increments is zero. \(\operatorname{Cov}(B(t_{i+1}) - B(t_i), B(t_{j+1}) - B(t_j)) = \operatorname{Cov}(B(t_{i+1}), B(t_{j+1})) - \operatorname{Cov}(B(t_i), B(t_{j+1})) - \operatorname{Cov}(B(t_{i+1}), B(t_j)) + \operatorname{Cov}(B(t_{i}), B(t_j))\) We know that \(\operatorname{Cov}(B(s), B(t)) = s \wedge t\). Using this:$$\operatorname{Cov}(B(t_{i+1}) - B(t_i), B(t_{j+1}) - B(t_j)) = (t_{i+1} \wedge t_{j+1}) - (t_i \wedge t_{j+1}) - (t_{i+1} \wedge t_j) + (t_i \wedge t_j)$$For \(i < j\): \(t_i < t_{i+1} \leq t_j < t_{j+1}\), so the expression becomes: \((t_{i+1} \wedge t_{j+1}) - (t_i \wedge t_{j+1}) - (t_{i+1} \wedge t_j) + (t_i \wedge t_j) = t_{i+1} - t_i = 0\)

(a) Proving Brownian motion is a martingale

To prove that the Brownian motion is a martingale, we need to show that it satisfies the three martingale properties: 1. \(E[|B(t)|] < \infty\) 2. \(E[B(t) | \mathcal{F}_s] = B(s)\) for \(s \leq t\) 3. \(B(t)\) is adapted to the filtration \(\mathcal{F}_t\) 1. Since \(B(t)\) is a Gaussian process with mean \(0\), \(E[|B(t)|] < \infty\). 2. \(E[B(t) | \mathcal{F}_s] = \operatorname{E}[B(t)-B(s) | \mathcal{F}_s] + B(s)\). Since \(B(t)\) has independent increments, \(B(t) - B(s)\) is independent of \(\mathcal{F}_s\), and its expectation is \(0\). Therefore, \(E[B(t) |\mathcal{F}_s] = B(s)\). 3. By definition, \(B(t)\) is adapted to the filtration \(\mathcal{F}_t\).

(a) Finding the compensator of \(B^2\)

To find the compensator of \(B^2\), we need to find a predictable process \(A(t)\) such that \(M(t) = B^2(t) - A(t)\) is a martingale. Since \(B(t)\) is a Gaussian process with \(\operatorname{E}[B(t)] = 0\) and \(\operatorname{Cov}(B(s), B(t)) = s\wedge t\), \(E[B(s)B(t)] = s \wedge t\). Then, \(E[B^2(t)] = t\). The compensator \(A(t)=E[B^2(t)]=t\). Therefore, \(M(t) = B^2(t) - t\) is a martingale.

(b) Proving \(B^0(t)\) and \(B(t)-tB(1)\) have the same distribution

To show that the processes \(B^0(t)\) and \(B(t)-tB(1)\) have the same distribution for \(t \in [0, 1]\), we need to prove that they have the same first and second moments. First moments: \(E[B^0(t)] = 0\) and \(E[B(t) - tB(1)] = E[B(t)] - tE[B(1)] = 0\). Second moments: \(\operatorname{Cov}(B^0(s), B^0(t)) = s(1-t)\). \(\operatorname{Cov}(B(t)-tB(1), B(s)-sB(1)) = t\wedge s - st\). For \(0\le s\le t\le 1\), \(s \wedge t = s\). Hence, \(\operatorname{Cov}(B^0(s), B^0(t)) = \operatorname{Cov}(B(t)-tB(1), B(s)-sB(1))\) (Both are equal to \(s(1-t)\)). Thus, \(B^0(t)\) and \(B(t)-tB(1)\) have the same distribution on \([0,1]\).

(c) Proving \(B(t)\) and \((1+t) B^{0}(t /(1+t))\) have the same distribution

To show that the processes \(B(t)\) and \((1+t) B^0\left(\frac{t}{1+t}\right)\) have the same distribution for \(t \in [0, \infty)\), we need to prove that they have the same first and second moments. First moments: \(E[B(t)] = 0\) and \(E[(1+t) B^0\left(\frac{t}{1+t}\right)] = (1+t)E[B^0\left(\frac{t}{1+t}\right)] = 0\). Second moments: \(\operatorname{Cov}(B(s), B(t)) = s\wedge t\). \(\operatorname{Cov}((1+s) B^0\left(\frac{s}{1+s}\right), (1+t) B^0\left(\frac{t}{1+t}\right)) = (1+s)(1+t)\operatorname{Cov}\left(B^0\left(\frac{s}{1+s}\right), B^0\left(\frac{t}{1+t}\right)\right)\). Now, let \(u = \frac{s}{1+s}\) and \(v = \frac{t}{1+t}\). Then, \(s = \frac{u}{1-u}\) and \(t = \frac{v}{1-v}\). \(\operatorname{Cov}\left(B^0\left(\frac{s}{1+s}\right), B^0\left(\frac{t}{1+t}\right)\right) = \operatorname{Cov}\left(B^0(u), B^0(v)\right) = u(1-v)\). Substituting the expressions for \(s\) and \(t\) and simplifying, we get:$$\operatorname{Cov}\left(B^0(u), B^0(v)\right) = \frac{u}{1-u}\frac{v}{1-v} = s\wedge t$$. Therefore, \(B(t)\) and \((1+t)B^0\left(\frac{t}{1+t}\right)\) have the same distribution on \([0, \infty)\).

Key concepts

Gaussian process

A Gaussian process is a collection of random variables, any finite number of which have a joint Gaussian distribution. It's a probabilistic process and for our purpose, it forms the foundation of understanding Brownian motion, which is a specific type of Gaussian process. The mathematical definition of a Gaussian process includes two functions: the mean function, which is often assumed to be zero (as in the Brownian motion), and the covariance function.

For the Brownian motion specifically, the covariance function between any two points in time is the minimum of those two times, denoted as \(s \wedge t\). This characteristic leads to some intriguing properties including the Brownian motion's 'independent increments' and the fact that it's a 'martingale', both of which are significant in stochastic processes and in understanding financial mathematics, physics, and various fields that deal with dynamical systems.

Martingale

In probability theory, a martingale is a model of a fair game where knowledge of past events never helps to predict future winnings. In formal terms, a stochastic process \(X(t)\) is a martingale with respect to some filtration \(\mathcal{F}_t\) if at any time \(s\), the conditional expectation of \(X(t)\) given the past information up to time \(s\) is equal to the current value of the process \(X(s)\). This property of 'no profit' and 'no loss' underlies many pricing and hedging strategies in financial mathematics.

For Brownian motion \(B(t)\), it holds that Brownian motion itself is a martingale. This is because its expected value at any future time, given the current filtration, is always the current value of the process, which satisfies the martingale property. Furthermore, for a process to be successfully considered a martingale, it must be integrable, meaning the absolute value of \(B(t)\) must have a finite expected value, which is indeed the case for the Brownian motion.

Independent increments

The concept of \'independent increments\'\' in a stochastic process means that the increments of the process are statistically independent of each other over non-overlapping time intervals. For a process \(X(t)\) with independent increments, the knowledge of \(X(s)\) for any \(s
In the case of Brownian motion \(B(t)\), it specifically means that if you take any two successive time periods, the amount by which the process changes in those two periods are independent random variables. This property is critically important for modeling in finance, where price movements in a stock or other asset can be modeled as having no relation to prior moves, reflecting the key assumption of market efficiency.

Brownian bridge

A Brownian bridge is a Brownian motion that is conditioned to return to zero at some future time \(T\), often denoted as \(B^0(t)\) where \(t \in [0, T]\). It is commonly used in statistics and finance for modeling various conditional scenarios.

The Brownian bridge still maintains some characteristics of a Brownian motion but has a different covariance structure, one that reflects the fact that it must 'tie down' to zero at time \(T\). For the covariance between any two points in time \(s\) and \(t\) in the Brownian bridge, it is given as \(s(1-t)\) for \(0 \leq s \leq t \<= T\). Interestingly, a Brownian bridge can be constructed from a Brownian motion by the appropriate linear transformation, and this relationship is often utilized in the simulation of conditioned stochastic processes.