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Chapter 10: Problem 17

Show that standard Brownian motion is a Martingale.

Short Answer

Expert verified
To show the standard Brownian motion is a Martingale, we need to prove it satisfies three properties: 1. The process is adapted to the filtration: \(W_t\) only depends on the information up to time t (\(\mathcal{F}_t\)). 2. The process has finite expectation: \(E[|W_t|] = \sqrt{\frac{2t}{\pi}}\), which is finite. 3. The conditional expectation satisfies: \(E[W_t|\mathcal{F}_s] = W_s\) for \(0 \le s \le t\). As all three properties are met, the standard Brownian motion is a Martingale.

Step by step solution

01

Adapted to the Filtration

For a standard Brownian motion \(W_t\), at any time t, the value \(W_t\) only depends on the information up to time t, namely \(\mathcal{F}_t\). Therefore, the process is adapted to the filtration \(\mathcal{F}_t\).
02

Finite Expectation

Recall that for a standard Brownian motion \(W_t\), \(W_t \sim \mathcal{N}(0, t)\). Thus, the expectation value of \(W_t\) is given by: \(E[|W_t|] = \int_{-\infty}^{\infty}|x|\frac{1}{\sqrt{2\pi t}}e^{-\frac{x^2}{2t}}dx\) Splitting the integral into the two intervals, where x is positive and negative: \(E[|W_t|] = 2\int_{0}^{\infty}x\frac{1}{\sqrt{2\pi t}}e^{-\frac{x^2}{2t}}dx\) Using integration by substitution, let \(u = \frac{x^2}{2t}\), we have: \(E[|W_t|] = 2\int_{0}^{\infty}\sqrt{2tu}\frac{1}{\sqrt{2\pi t}}e^{-u}d(\sqrt{2tu}) = \sqrt{\frac{2t}{\pi}}\) Since t is finite, \(E[|W_t|]\) must also be finite.
03

Conditional Expectation

Now let's compute the conditional expectation \(E[W_t|\mathcal{F}_s]\), where \(0 \le s \le t\): \(E[W_t|\mathcal{F}_s] = E[(W_t - W_s) + W_s|\mathcal{F}_s] = E[W_t - W_s|\mathcal{F}_s] + E[W_s|\mathcal{F}_s]\) Since \(W_t - W_s \sim \mathcal{N}(0, t-s) \) is independent of \(\mathcal{F}_s\), the term \(E[W_t - W_s|\mathcal{F}_s]\) becomes: \(E[W_t - W_s|\mathcal{F}_s] = E[W_t - W_s] = 0\) For the term \(E[W_s|\mathcal{F}_s]\), it is equal to \(W_s\) as the process is adapted and the value of \(W_s\) depends only on \(\mathcal{F}_s\). Thus, we have: \(E[W_t|\mathcal{F}_s] = 0 + W_s = W_s\) Since the three conditions of a Martingale process are satisfied, we can conclude that the standard Brownian motion is a Martingale.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Martingale
Martingale is a central concept in probability theory, particularly in stochastic processes. A process like this typically models a fair game where the expected value of the next observation, given all the past observations, is equal to the present observation. In simpler terms, future predictions are unbiased by past information.

For a process to be considered a martingale, the following three conditions must be met:
  • The process should be adapted to a given filtration, meaning each variable in the process only depends on available information up to that point.

  • The expectation of each variable in the process should be finite, ensuring that it can be properly managed and calculated.

  • The conditional expectation of the future value given the past values should be equal to the current value, indicating unbiased prediction.

Standard Brownian motion meets all these conditions, hence proving that it is indeed a martingale.
Filtration
Filtration is an essential part of understanding Brownian motion as a martingale. It refers to the increasing sequence of information available over time, representing all observations up to the present. This is critical because it ensures the process is appropriately updated with new information as time progresses.

In the context of a standard Brownian motion, filtration is denoted by \(\mathcal{F}_t\). This essentially means that at any given time \(t\), the path of the motion is adapted to, or influenced by, the filtration \(\mathcal{F}_t\).

This adaptability is achieved because each random variable in the process relies solely on previously accumulated information. Consequently, this ensures that predictions are anchored in reality, evolving based on the most current understanding of the process.
Conditional Expectation
Conditional expectation plays a crucial role in verifying whether a process is a martingale. It involves predicting the expected value of a future observation, given a certain amount of past information. Essentially, it's like asking, "Given what we know now, what do we expect will happen next?"

For standard Brownian motion, to establish it as a martingale, the conditional expectation formula \(E[W_t|\mathcal{F}_s] = W_s\) is used. Here, \(E[W_t - W_s|\mathcal{F}_s] = 0\) thus proving that the movement from time \(s\) to \(t\) is an unbiased random walk.
  • The term \(E[W_s|\mathcal{F}_s]\) represents the current value which is calculated based only on the information up to time \(s\).

  • Since the change \(W_t - W_s\) is independent of the past information \(\mathcal{F}_s\) and has an expectation of zero, it highlights the unbiased nature of future forecasts.

The conditional expectation confirms the unbiased progress of the motion, directly substantiating it as a martingale.

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Most popular questions from this chapter

The current price of a stock is 100 . Suppose that the logarithm of the price of the stock changes according to a Brownian motion with drift coefficient \(\mu=2\) and variance parameter \(\sigma^{2}=1 .\) Give the Black-Scholes cost of an option to buy the stock at time 10 for a cost of (a) 100 per unit. (b) 120 per unit. (c) 80 per unit.

Let \(\\{X(t),-\infty

Let \(X(t)=\sigma B(t)+\mu t\), and for given positive constants \(A\) and \(B\), let \(p\) denote the probability that \(\\{X(t), t \geqslant 0\\}\) hits \(A\) before it hits \(-B\). (a) Define the stopping time \(T\) to be the first time the process hits either \(A\) or \(-B\). Use this stopping time and the Martingale defined in Exercise 19 to show that $$ E\left[\exp \left\\{c(X(T)-\mu T) / \sigma-c^{2} T / 2\right\\}\right]=1 $$ (b) Let \(c=-2 \mu / \sigma\), and show that $$ E[\exp \\{-2 \mu X(T) / \sigma\\}]=1 $$ (c) Use part (b) and the definition of \(T\) to find \(p\). What are the possible values of \(\exp \left\\{-2 \mu X(T) / \sigma^{2}\right] ?\)

Show that \(\\{Y(t), t \geqslant 0\\}\) is a Martingale when $$ Y(t)=B^{2}(t)-t $$ What is \(E[Y(t)] ?\)

Let \(Y(t)=t B(1 / t), t>0\) and \(Y(0)=0 .\) (a) What is the distribution of \(Y(t) ?\) (b) Compare Cov \((Y(s), Y(t))\). (c) Argue that \(\\{Y(t), t \geqslant 0\\}\) is a standard Brownian motion process.
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