Question

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 \(p\). Hint: What are the possible values of \(\exp \left\\{-2 \mu X(T) / \sigma^{2}\right\\} ?\)

Short answer

The probability \(p\) that the process hits \(A\) before it hits \(-B\) can be expressed as: $$ p = \frac{e^{\frac{2\mu B}{\sigma^2}} - 1}{e^{\frac{2\mu B}{\sigma^2}} - e^{\frac{-2\mu A}{\sigma^2}}}. $$

Step-by-step solution

Define the stopping time T

Let's define the stopping time \(T\) as $$ T = \inf \{t \geq 0: X(t) = A \text{ or } X(t) = -B\}. $$ This represents the first time when the process \(X(t)\) hits either the value \(A\) or the value \(-B\).

Find the expected value using the Martingale property

Using the Martingale property from Exercise 19, we have $$ E\left[e^{\frac{c(X{(T)}-\mu T)}{\sigma}-\frac{c^2T}{2}}\right] = 1. $$

Substitute c = -2μ/σ

Now, let's substitute \(c = -\frac{2\mu}{\sigma}\): $$ E\left[e^{\frac{-2\mu(X{(T)}-\mu T)}{\sigma^2}+\frac{2\mu^2T}{\sigma^2}}\right] = 1. $$ Simplifying the exponent, we get: $$ E\left[e^{\frac{-2\mu X{(T)}}{\sigma^2}}\right] = 1. $$

Find the probability p

From the definition of stopping time \(T\), we know that when \(T\) is reached, there are only two possible values for \(X(T)\): \(A\) or \(-B\). So, we can consider two cases: 1. If \(X(T) = A\), then \(e^{\frac{-2\mu X{(T)}}{\sigma^2}} = e^{\frac{-2\mu A}{\sigma^2}}\). 2. If \(X(T) = -B\), then \(e^{\frac{-2\mu X{(T)}}{\sigma^2}} = e^{\frac{2\mu B}{\sigma^2}}\). We can now rewrite the expected value expression using these two cases and the fact that \(p\) is the probability of reaching \(A\) before \(-B\): $$ E\left[e^{\frac{-2\mu X{(T)}}{\sigma^2}}\right] = p e^{\frac{-2\mu A}{\sigma^2}} + (1-p) e^{\frac{2\mu B}{\sigma^2}} = 1. $$ Rewriting the expression for \(p\), we have: $$ p = \frac{e^{\frac{2\mu B}{\sigma^2}} - 1}{e^{\frac{2\mu B}{\sigma^2}} - e^{\frac{-2\mu A}{\sigma^2}}}. $$ So, the probability \(p\) that the process hits \(A\) before it hits \(-B\) can be expressed as: $$ p = \frac{e^{\frac{2\mu B}{\sigma^2}} - 1}{e^{\frac{2\mu B}{\sigma^2}} - e^{\frac{-2\mu A}{\sigma^2}}}. $$

Key concepts

Stochastic Processes

Imagine flipping a coin. Each flip is random, and predicting the outcome on the next flip is uncertain. Now, think of something more complex, like the stock market, where prices change randomly over time. This randomness over time can be described using mathematical structures known as stochastic processes.\

A stochastic process is a collection of random variables representing the evolution of some system of random values over time. In our textbook exercise, the stochastic process is given by the formula \(X(t)=\sigma B(t)+\mu t\), which models a continuous-time process that consists of a scaled Brownian motion, \(\sigma B(t)\), and a drift component, \(\mu t\). This reflects how an asset might behave in financial markets, factoring in both random fluctuation (the Brownian motion) and a general trend (the drift term).\

Understanding stochastic processes is crucial for fields like finance, physics, and engineering, as they provide insights into how systems behave over time under uncertainty. This foundation is essential when diving into more complex concepts like stopping times and martingales.

Stopping Time

Timing can be everything - it's true in life, and it's certainly true in the world of stochastic processes. In our exercise, we are introduced to a concept called stopping time, which is a random time that depends on the stochastic process itself.\

Technically, stopping time is a variable that signifies the point at which a certain event happens for the first time. For our process \(X(t)\), we define the stopping time \(T\) as the first instance when \(X(t)\) reaches either \(A\) or \(-B\). This is like waiting for a stock to hit a certain price before selling it.\

Stopping times are very useful in financial mathematics, particularly in option pricing and risk management, and are crucial for the derivation of various properties, such as the martingale property used later in our solution.

Martingales

The concept of a martingale is quite fascinating. It originates from a class of betting strategies, where future winnings are solely based on past events. In the realm of mathematics, particularly probability theory, a martingale is a stochastic process that represents a fair game where, on average, the expected value of the next observation is equal to the present observation.\

This means that no matter how the process evolves, the expected future value, given all past values, always equals the current value. For instance, consider a gambler with a fair coin; if the gambler has no past information that can predict future outcomes, then the sequence of coin flips constitutes a martingale because each flip's expected value is zero (fair game).\

In our textbook problem, the martingale property is used to show that the expected exponential function of the process at stopping time \(T\) equals one, thus satisfying a particular martingale equality. This property is not just important in the context of gambling or mathematical theory, but also in financial modeling, where it's used to price derivatives and assess risk.

Expected Value

The average outcome is often what we're after. That's the idea behind the expected value, which is a fundamental concept in probability and statistics. Expected value is the long-run average value of repetitions of the experiment it represents.\

For a simple example, consider rolling a six-sided die. The expected value of a roll is 3.5 because, over many rolls, the average result converges to this number, despite the die landing on whole numbers between 1 and 6.\

In the context of our exercise, the expected value is a tool used to solve for the probability \(p\). It represents what outcome you would expect on average if you could repeat the stochastic process an infinite number of times. The beauty of expected value lies in its ability to provide a single summary number for a complex random phenomenon. In financial terms, it's akin to forecasting the average return on investment over a long period. This simplicity in the heart of complexity is what makes expected value such a powerful and widely-used concept in various fields, from economics to engineering.