Question

Consider the random walk that in each \(\Delta t\) time unit either goes up or down the amount \(\sqrt{\Delta t}\) with respective probabilities \(p\) and \(1-p\), where \(p=\frac{1}{2}(1+\mu \sqrt{\Delta t})\). (a) Argue that as \(\Delta t \rightarrow 0\) the resulting limiting process is a Brownian motion process with drift rate \(\mu\). (b) Using part (a) and the results of the gambler's ruin problem (Section 4.5.1), compute the probability that a Brownian motion process with drift rate \(\mu\) goes up \(A\) before going down \(B, A>0, B>0\)

Short answer

In the given random walk, as \(\Delta t \rightarrow 0\), it converges to a Brownian motion process with drift rate \(\mu\). The probability that the Brownian motion process with drift rate \(\mu\) goes up A before going down B is given by: $$ P(T_A < T_B) = \frac{\exp(\frac{2\mu A}{\Delta t}) - 1}{\exp(\frac{2\mu(A + B)}{\Delta t}) - 1} $$

Step-by-step solution

Part (a): Show the limiting process is a Brownian motion process with drift rate \(\mu\)

We are given a random walk with the following parameters: - In each time step \(\Delta t\), it either goes up or down the amount \(\sqrt{\Delta t}\); - The probability of going up is \(p = \frac{1}{2}(1 + \mu \sqrt{\Delta t})\); - The probability of going down is \(1 - p = \frac{1}{2}(1 - \mu \sqrt{\Delta t})\). As \(\Delta t\) goes to zero, the changes in the random walk become smaller and the process approaches a continuous time stochastic process. Define the increment of the process as: $$ X(\Delta t) = \begin{cases} \sqrt{\Delta t}, &\text{ with probability } p = \frac{1}{2}(1 + \mu \sqrt{\Delta t})\\ -\sqrt{\Delta t}, &\text{ with probability } 1-p = \frac{1}{2}(1 - \mu \sqrt{\Delta t}) \end{cases} $$ The expectation and variance of the increment are given by: $$ E[X(\Delta t)] = p\sqrt{\Delta t} - (1-p)\sqrt{\Delta t} = \mu \Delta t \\ Var[X(\Delta t)] = E[X^2(\Delta t)] - E[X(\Delta t)]^2 = \Delta t $$ As \(\Delta t \rightarrow 0\), the increments \(X(\Delta t)\) become small and frequent, and the limiting process becomes a continuous time stochastic process. Since the increments have a continuous distribution, are independent, and have a normal mean and variance, the limiting process is a Brownian motion process with drift rate \(\mu\).

Part (b): Compute the probability that Brownian motion goes up A before going down B

Based on part (a) and the gambler's ruin problem results, let's first denote: - \(T_A\): the first time that the Brownian motion process goes up A; - \(T_B\): the first time that the Brownian motion process goes down B. We want to compute the probability \(P(T_A < T_B)\). The gambler's ruin problem teaches us that for a random walk, the probability of reaching a certain level before another is given by: $$ P(T_A < T_B) = \frac{\exp(\frac{2\mu A}{\sigma^2}) - 1}{\exp(\frac{2\mu(A + B)}{\sigma^2}) - 1} $$ where \(\sigma\) is the standard deviation of the increments. In our case, the standard deviation \(\sigma = \sqrt{Var[X(\Delta t)]} = \sqrt{\Delta t}\). As \(\Delta t \rightarrow 0\), the continuous-time limit of the above expression is: $$ \lim_{\Delta t\to0} P(T_A < T_B) = \frac{\exp(\frac{2\mu A}{\Delta t}) - 1}{\exp(\frac{2\mu(A + B)}{\Delta t}) - 1} $$ Since the limiting process is a Brownian motion with drift rate \(\mu\), the probability that it goes up A before going down B is: $$ P(T_A < T_B) = \frac{\exp(\frac{2\mu A}{\Delta t}) - 1}{\exp(\frac{2\mu(A + B)}{\Delta t}) - 1} $$ Now we have the probability of Brownian motion with drift rate \(\mu\) going up A before going down B.