Question

Suppose you own one share of a stock whose price changes according to a standard Brownian motion process. Suppose that you purchased the stock at a price \(b+c\), \(c \geq 0\), and the present price is \(b\). You have decided to sell the stock either when it reaches the price \(b+c\) or when an additional time \(t\) goes by (whichever occurs first). What is the probability that you do not recover your purchase price?

Short answer

The probability of not recovering the purchase price is given by: \[P =\frac{2}{\sqrt{2\pi t}} \int_{0}^{c} e^{- \frac{u^2}{2t}} du\]

Step-by-step solution

Define the Problem

We are given the problem of finding out the probability that we do not recover our purchase price. We define the following variables: - \(S(t)\): the price of the stock at time \(t\), which is a standard Brownian motion - \(T\): the first time when \(S(t) = b+c\) or when an additional time \(t\) goes by - \(P\): the probability of not recovering the purchase price, i.e., the probability that \(S(T) < b+c\). We want to find the value of \(P\).

Apply the Reflection Principle

The reflection principle states that, if \(Y(t)=b\) is the value of a standard Brownian motion at some time \(t\), and \(W(t)\) is another standard Brownian motion that starts at \(Y(t)\), then the joint probability distribution of \(Y(t)\) and the minimum of \(W(t)\) with respect to \(t\) is the same as the joint probability distribution of \(Y(t)\) and its reflected process. By applying the reflection principle, we can find the probability that \(S(T) < b+c\). We'll consider another Brownian motion process, \(\tilde{S}(t) = b-c - (S(t) - b) = 2b - c - S(t)\), which is the mirrored process of \(S(t)\) about the horizontal line at \(y = b\). Now, we reformulate the problem as finding the probability that the first time \(S(t)\) crosses \(b+c\), it does so from below (i.e., it moves from below the horizontal line at \(b+c\) to the line).

Compute the Probability

Consider the following set of events: - \(E_1\): \(S(t)\) first hits \(b+c\) from below - \(E_2\): \(\tilde{S}(t)\) is below \(b+c-2c\) at time \(t\) - \(E_3\): \(S(t)\) and \(\tilde{S}(t)\) have not yet hit \(b+c\) by time \(t\) We want to find \(P(\text{not } E_1| E_3)\), i.e., the probability that the stock price does not recover the purchase price. We have that \(P(E_2|E_3) = P(\text{not } E_1| E_3)\), since for each path \(S\) in \(E_3\), there exists a path \(\tilde{S}\) in \(E_2\). Now, the joint probability of \(E_1\cap E_3\) is equal to the joint probability of \(E_2\cap E_3\). So we have: \[P(\text{not } E_1| E_3) = P(E_2| E_3) = \frac{P(E_1\cap E_3)}{P(E_3)}\] As we know that the position of \(S(t)\) and \(\tilde{S}(t)\) at time \(t\) is distributed normally, we can directly compute the probability as follows: \[P(\text{not } E_1| E_3) = \frac{2}{\sqrt{2\pi t}} \int_{0}^{c} e^{- \frac{u^2}{2t}} du\]

Final Result

The probability of not recovering the purchase price is given by: \[P =\frac{2}{\sqrt{2\pi t}} \int_{0}^{c} e^{- \frac{u^2}{2t}} du\]