Problem 81 - A First Course in Probability

Question

An investor owns shares in a stock whose present value is \(25 .\) She has decided that she must sell her stock if it goes either down to 10 or up to \(40 .\) If each change of price is either up 1 point with probability .55 or down 1 point with probability \(.45,\) and the successive changes are independent, what is the probability that the investor retires a winner?

Step-by-Step Solution

Answer

The probability that the investor retires a winner is \(\frac{20^{16}}{11^{16}}\).

Step 1: Solve for general solution

We know that \(P(k) = 0.55 P(k+1) + 0.45 P(k-1)\), which can be rewritten as: \[P(k+1) = \frac{1}{0.55} P(k) - \frac{0.45}{0.55} P(k-1)\] \[P(k+1) = \frac{20}{11}P(k) - \frac{9}{11}P(k-1)\] Now, we need to find the general solution for \(P(k)\) by iterating the above expression and applying the boundary conditions \(P(10) = 0\) and \(P(40) = 1\).

Step 2: Iterate the general solution

We iterate the given expression starting from k = 10, 11, 12, ... until k = 25 using the boundary condition that \(P(10) = 0\): \(P(11) = \frac{20}{11}P(10) - \frac{9}{11}P(9) = 0\) \(P(12) = \frac{20}{11}P(11) - \frac{9}{11}P(10) = 0\) By continuing this pattern, we find that: \(P(25) = \frac{20}{11}P(24) - \frac{9}{11}P(23)\)

Step 3: Calculate the last term using the boundary conditions

Next, we have to find the probabilities \(P(24)\) and \(P(23)\) using the second boundary condition, \(P(40) = 1\). We can use a similar approach in reverse, starting from k = 39, 38, 37, ... until k = 23: \(P(39) = \frac{20}{11}P(40) - \frac{9}{11}P(41) = \frac{20}{11}\) \(P(38) = \frac{20}{11}P(39) - \frac{9}{11}P(40) = \frac{20^2}{11^2}\) By continuing this pattern, we find that: \(P(24) = \frac{20^{16}}{11^{16}}\) \(P(23) = \frac{20^{17}}{11^{17}} - P(24) = \frac{20^{17} - 20^{16}}{11^{17}}\)

Step 4: Find the probability of retiring a winner

Now that we have calculated the values for \(P(24)\) and \(P(23)\), we can find the probability of the investor retiring a winner, \(P(25)\), using: \(P(25) = \frac{20}{11}P(24) - \frac{9}{11}P(23)\) \(P(25) = \frac{20}{11}\left(\frac{20^{16}}{11^{16}} \right) - \frac{9}{11} \left(\frac{20^{17} - 20^{16}}{11^{17}}\right)\) After simplifying, we get: \(P(25) = \frac{20^{17} - 9 \cdot 20^{17} + 9 \cdot 20^{16}}{11^{17}}\) \(P(25) = \frac{20^{16}(20 - 9\cdot 20 + 9\cdot 11)}{11^{17}}\) \(P(25) = \frac{20^{16} \cdot 11}{11^{17}}\) Finally, we have: \(P(25) = \frac{20^{16}}{11^{16}}\) Hence, the probability that the investor retires a winner is \(\frac{20^{16}}{11^{16}}\).

Key Concepts

Probability Theory

Understanding Probability Theory is crucial in analyzing situations involving uncertainty, such as the probability of an investor retiring a winner in the given exercise. Probability theory provides a framework for predicting the likelihood of various outcomes.

In the context of our problem, probability theory is used to determine the chance of the stock value reaching a certain price before another. Specifically, we use the theoretical underpinning of probability to calculate the chance that the stock price reaches \(40 before it dips to \)10. The formulation of this problem involves setting up a stochastic model and solving it using certain principles and rules that define the probability of the investor achieving her financial goal.

Independent Probability

In the case of the investor and her stock prices, Independent Probability plays a key role. Two events are independent if the occurrence of one does not affect the probability of the occurrence of the other. For our stock price movements, the changes are said to be independent because the rise or fall in price at one instance does not influence the direction in the next.

This concept allows us to handle each price change as a separate event, greatly simplifying the computation of the overall probability. The assumption that price movements are independent enables us to use a simple multiplicative rule when iterating towards our solution. However, it's important to understand that in the real world, stock price movements may not always be independent, as they can be influenced by various factors and trends.

Markov Processes

The exercise dealing with stock values can also be understood through the lens of Markov Processes, which are used to model random systems that change states according to transition probabilities that depend only on the current state and not on the sequence of events that preceded it.

In the Markov Process approach to this problem, the current stock value is the state, and the probabilities of moving to the next stock values (one point up or down) rely solely on the present value, not the history of price changes. This property is known as the Markov property, and it allows for a simplified approach to understanding systems that might otherwise appear complex due to their stochastic nature. The step-by-step solution illustrates this by calculating probabilities based on current stock prices without needing to consider the exact path taken to arrive at those prices.

Boundary Conditions in Probability

Finally, to solve such probability problems, we need to understand Boundary Conditions in Probability. These conditions dictate the values of probabilities at certain defined 'boundaries' of our problem.

In the context of our stock price problem, the boundary conditions are that the stock value cannot fall below \(10 or rise above \)40. These conditions are implemented as given probabilities, such as the probability of winning when the stock reaches \(40 is 1 (certain win), and the probability of winning when it falls to \)10 is 0 (certain loss). By incorporating these boundary conditions into our iterative calculations, as seen in the step-by-step solution, we can use them to navigate and restrict the range of our probability analysis to calculate the investor's chance to retire a winner.