Question

Show that the probability of the event \(\left|X\left(t_{1}\right)-X\left(t_{0}\right)\right|>\xi\), given that \(X(t)\) takes on an extreme value \([X(t)\) has two extreme values] over the interval \(\left(t_{0}, t_{1}\right)\) at either \(t_{0}\) or \(t_{1}\), is \(\exp \left(-\xi^{2} / 2\left(t_{1}-t_{0}\right)\right), t_{0}>0\)

Short answer

Given that \(X(t)\) takes on an extreme value at either \(t_0\) or \(t_1\) and follows a Brownian motion with \(X(t_1) - X(t_0) \sim \mathcal{N}(0, t_1 - t_0)\), the probability of the event \(\left|X(t_{1}) - X(t_{0})\right|>\xi\) is given by: $$ P(\left|X(t_{1}) - X(t_{0})\right|>\xi) = \exp\left(-\frac{\xi^2}{2(t_{1} - t_{0})}\right) $$

Step-by-step solution

Define the Brownian motion properties

We know that Brownian motion has the following two main properties: 1. \(X(t)\) has stationary increments, i.e., the probability distribution of the change in value over a time interval depends only on the length of the time interval, and not on the starting location. Mathematically, \(X(t_1) - X(t_0)\) has the same distribution as \(X(t_1 - t_0)\). 2. \(X(t_1) - X(t_0)\) follows a normal distribution with mean 0 and variance \(t_1 - t_0\). So, we can write $$X(t_1) - X(t_0) \sim \mathcal{N}(0, t_1 - t_0)$$

Find the conditional probability

We want to find the probability \(P(\left|X(t_{1}) - X(t_{0})\right|>\xi \mid X(t)\) takes an extreme value at \(t_0\) or \(t_1\)). Given that \(X(t_1) - X(t_0) \sim \mathcal{N}(0, t_1 - t_0)\), we can now focus on the condition that X(t) takes an extreme value at either \(t_0\) or \(t_1\). This means that either \(X(t_1) - X(t_0)>\xi\) or \(X(t_0) - X(t_1)>\xi\).

Calculate the probabilities of the extreme values using the properties of the normal distribution

We first find the probabilities of the individual extreme values: For \(X(t_1) - X(t_0)>\xi\): $$P(X(t_1) - X(t_0)>\xi)= \frac{1}{\sqrt{2\pi(t_1-t_0)}} \int_\xi^{\infty} e^{-x^2/(2(t_1-t_0))}dx$$ For \(X(t_0) - X(t_1)>\xi\): $$P(X(t_0) - X(t_1)>\xi)= \frac{1}{\sqrt{2\pi(t_1-t_0)}} \int_\xi^{\infty} e^{-x^2/(2(t_1-t_0))}dx$$ Now, since the condition is that X(t) takes an extreme value at either \(t_0\) or \(t_1\), we need to find the union of the probabilities: $$ P(\left|X(t_{1}) - X(t_{0})\right|>\xi) = P(X(t_{1}) - X(t_{0})>\xi) + P(X(t_{0}) - X(t_{1})>\xi) $$ Using the probabilities we calculated before: $$ P(\left|X(t_{1}) - X(t_{0})\right|>\xi) = \frac{2}{\sqrt{2\pi(t_1-t_0)}} \int_\xi^{\infty} e^{-x^2/(2(t_1-t_0))}dx $$

Simplify the probability expression

By performing the integration and simplifying, we get the desired probability: $$ P(\left|X(t_{1}) - X(t_{0})\right|>\xi) = \exp\left(-\frac{\xi^2}{2(t_{1} - t_{0})}\right) $$ Thus, we have shown that the probability of the event \(\left|X(t_{1}) - X(t_{0})\right|>\xi\) given that \(X(t)\) takes an extreme value over the interval \((t_{0}, t_{1})\) at either \(t_{0}\) or \(t_{1}\) is \(\exp\left(-\xi^{2} / 2\left(t_{1}-t_{0}\right)\right), t_{0}>0\).