/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 81 An investor owns shares in a sto... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 81

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?

Short Answer

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

Step by step solution

01

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 \(P(40) = 1\).
02

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: \(P(25) = \frac{20}{11}P(24) - \frac{9}{11}P(23)\)
03

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: \(P(24) = \frac{20^{16}}{11^{16}}\) \(P(23) = \frac{20^{17}}{11^{17}} - P(24) = \frac{20^{17} - 20^{16}}{11^{17}}\)
04

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 \(\frac{20^{16}}{11^{16}}\).

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Most popular questions from this chapter

Die \(A\) has 4 red and 2 white faces, whereas die \(B\) has 2 red and 4 white faces. A fair coin is flipped once. If it lands on heads, the game continues with dic \(A ;\) if it lands on tails, then dic \(B\) is to be used. (a) Show that the probability of red at any throw is \(\frac{1}{2}.\) (b) If the first two throws result in red, what is the probability of red at the third throw? (c) If red turns up at the first two throws, what is the probability that it is die \(A\) that is being used?

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