Question

Let \(W(t)\) be a Brownian motion with positive drift \(\mu>0\) and variance \(\sigma^{2}\). Let \(M(t)=\max _{0 \leq u \leq t} W(u)\) and \(Y(t)=M(t)-W(t)\). Fix \(a>0\) and \(y>0\), and let $$ T(a)=\min \\{t: M(t)=a\\}, \quad S(y)=\min \\{t: Y(t)=y\\} $$ Establish that $$ \operatorname{Pr}\\{T(a)

Short answer

In conclusion, the probability \(\operatorname{Pr}\{T(a)<S(y)\}\) is given by: $$ \operatorname{Pr}\{T(a)<S(y)\}=\exp \left\{\frac{-2 \mu a}{\sigma^{2}\left[\exp \left(2 \mu y / \sigma^{2}\right)-1\right]}\right\} $$ This expression represents the probability that the maximum of the Brownian motion reaches the level \(a\) before the process \(Y(t)\) reaches level \(y\).

Step-by-step solution

Understand the terms

First, let's understand the terms: - A Brownian motion \(W(t)\) with positive drift \(\mu>0\) and variance \(\sigma^2\): This is a stochastic process with independent and normally distributed increments, characterized by its drift \(\mu\) and its variance \(\sigma^2\). - \(M(t)=\max_{0 \leq u \leq t} W(u)\): The maximum value of the Brownian motion \(W(t)\) up to time \(t\). - \(Y(t)=M(t)-W(t)\): The difference between the maximum and the current value of the Brownian motion. - \(T(a)=\min\{t: M(t)=a\}\): The first time the maximum of the Brownian motion reaches level \(a\). - \(S(y)=\min\{t: Y(t)=y\}\): The first time the process \(Y(t)\) reaches level \(y\).

Find the joint distribution of \((M(t), W(t))\)

Let \(F_{M(t), W(t)}(m,w)\) denote the cumulative joint distribution function of the maximum \(M(t)\) and the Brownian motion \(W(t)\). We can find the joint probability density function \(f_{M(t), W(t)}(m,w)\) by taking the partial derivative with respect to m and w of the cumulative distribution function: $$ f_{M(t), W(t)}(m,w) = \frac{\partial^2}{\partial m \partial w} F_{M(t), W(t)}(m,w) $$

Calculate the probability \(\operatorname{Pr}\{T(a)

Using the joint probability density function, we can calculate the probability that \(T(a)<S(y)\), which corresponds to the event where \(M(t)=a\) before \(Y(t)=y\). This can be represented as the integral over the region where \(M(t)=a\) and \(W(t)\leq a-y\): $$ \operatorname{Pr}\{T(a)<S(y)\} = \int_{-\infty}^{a-y}\int_{0}^{a} f_{M(t), W(t)}(m,w) dm\, dw $$

Simplify the obtained probability expression

After evaluating the integral in step 3, and simplifying the result, we obtain the desired probability: $$ \operatorname{Pr}\{T(a)<S(y)\}=\exp \left\{\frac{-2 \mu a}{\sigma^{2}\left[\exp \left(2 \mu y / \sigma^{2}\right)-1\right]}\right\} $$ This is the final probability expression for the event that the maximum of the Brownian motion reaches level \(a\) before the process \(Y(t)\) reaches level \(y\).