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Q. 8.11

Expert-verified

A First Course in Probability

Found in: Page 391

Book edition 9th

Author(s) Sheldon M. Ross

Pages 432 pages

ISBN 9780321794772

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Short Answer

Many people believe that the daily change in the price of a company’s stock on the stock market is a random variable with a mean of $0$ and a variance of $σ^{2}$ . That is if Yn represents the price of the stock on the $n$ th day, then $Y_{n} = Y_{n - 1} + X_{n}, n \geq 1$ where $X_{1}, X_{2}, . . .$ are independent and identically distributed random variables with mean $0$ and variance $σ_{2}$ . Suppose that the stock’s price today is $100$ . If $σ_{2} = 1$ , what can you say about the probability that the stock’s price will exceed $105$ after $10$ days?

The required probability is $. 0571$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1 Given Information.

Let $Y_{n}$ represents the price of the stock on the $n$ th day, then $Y_{n} = Y_{n - 1} + X_{n}, n \geq 1$ where $X_{1}, X_{2}, . . .$ are independent and identically distributed random variables with mean $0$ and variance $σ^{2} .$ Suppose that the stock’s price today is $100$ and $σ_{2} = 1$ ,

Step 2 Explanation.

Let's $Y_{n}$ represent the price of the stock on the $n$ th day:

$Y_{n} = Y_{n - 1} + X_{n},, n \geq 1$

where $X_{1}, X_{2}, \dots$ are independent and identically distributed random variables with mean $μ = 0$ and variance $σ^{2}$ ?

Assume that today is $n^{*}$ th day. Additionally, assume that the stock's price today is $100$ :

$Y_{n^{*}} = Y_{n^{*} - 1} + X_{n^{*}} = 100 .$

Suppose that $σ^{2} = 1$ and let's consider the next $10$ few days. So,

today : $Y_{n^{v}} = Y_{0} = 100$

$1$ st day : $Y_{1} = Y_{0} + X_{1} = 100 + X_{1}$

$2$ nd day : $Y_{2} = Y_{1} + X_{2} = 100 + X_{1} + X_{2}$

$3$ rd day : $Y_{3} = Y_{2} + X_{3} = 100 + X_{1} + X_{2} + X_{3}$

$9$ th day : $Y_{9} = Y_{8} + X_{9} = 100 + X_{1} + X_{2} + \dots + X_{9}$

$10$ th day : $Y_{10} = Y_{9} + X_{10} = 100 + X_{1} + X_{2} + \dots + X_{9} + X_{10}$

Step 3 Explanation.

As we can see above, the price of the stock on the $10$ th day is

role="math" $Y_{10} = 100 + X_{1} + X_{2} + \dots + X_{9} + X_{10}$

Because of the independence of random variables $X_{i}$ and the corresponding properties of expectation and variance we get:

$E [Y_{10}] = 100 + 10 μ = 100, Var (Y_{10}) = 10 σ^{2} = 10$

The probability that the stock's price will exceed $105$ after $10$ days is $P \{Y_{10} > 105\}$ .

To approximate this probability we use the central limit theorem and in that case, we get:

$P \{Y_{10} > 105\} = 1 - P \{Y_{10} \leq 105\} = 1 - P \{\frac{Y_{10} - E [Y_{10}]}{\sqrt{Var (Y_{10})}} \leq \frac{105 - E [Y_{10}]}{\sqrt{Var (Y_{10})}}\} = 1 - P \{\frac{Y_{10} - 100}{\sqrt{10}} \leq \frac{105 - 100}{\sqrt{10}}\} \approx 1 - Φ (1.58) Table 5.1 (texthook, Chapter 5) 1 - . 9429 = . 0571 .$

Most popular questions for Math Textbooks

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