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Answers without the blur. Sign up and see all textbooks for free! Q. 8.11

Expert-verified Found in: Page 391 ### A First Course in Probability

Book edition 9th
Author(s) Sheldon M. Ross
Pages 432 pages
ISBN 9780321794772 # 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${\sigma }^{2}$. That is if Yn represents the price of the stock on the$n$ th day, then ${Y}_{n}={Y}_{n-1}+{X}_{n},n\ge 1$where ${X}_{1},{X}_{2},...$are independent and identically distributed random variables with mean $0$and variance${\sigma }_{2}$. Suppose that the stock’s price today is$100$. If${\sigma }_{2}=1$, what can you say about the probability that the stock’s price will exceed $105$after$10$ days?

The required probability is$\mathbf{.}\mathbf{0571}$.

See the step by step solution

## Step 1 Given Information.

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

## Step 2 Explanation.

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

${Y}_{n}={Y}_{n-1}+{X}_{n},,n\ge 1$

where ${X}_{1},{X}_{2},\dots$are independent and identically distributed random variables with mean $\mu =0$and variance${\sigma }^{2}$?

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 ${\sigma }^{2}=1$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 :${Y}_{9}={Y}_{8}+{X}_{9}=100+{X}_{1}+{X}_{2}+\cdots +{X}_{9}$

$10$th day :${Y}_{10}={Y}_{9}+{X}_{10}=100+{X}_{1}+{X}_{2}+\cdots +{X}_{9}+{X}_{10}$

## Step 3 Explanation.

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

role="math" ${Y}_{10}=100+{X}_{1}+{X}_{2}+\cdots +{X}_{9}+{X}_{10}$

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

$E\left[{Y}_{10}\right]=100+10\mu =100,\phantom{\rule{0ex}{0ex}}Var\left({Y}_{10}\right)=10{\sigma }^{2}=10$

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

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

$P\left\{{Y}_{10}>105\right\}=1-P\left\{{Y}_{10}\le 105\right\}\phantom{\rule{0ex}{0ex}}=1-P\left\{\frac{{Y}_{10}-E\left[{Y}_{10}\right]}{\sqrt{Var\left({Y}_{10}\right)}}\le \frac{105-E\left[{Y}_{10}\right]}{\sqrt{Var\left({Y}_{10}\right)}}\right\}\phantom{\rule{0ex}{0ex}}=1-P\left\{\frac{{Y}_{10}-100}{\sqrt{10}}\le \frac{105-100}{\sqrt{10}}\right\}\approx 1-\Phi \left(1.58\right)\phantom{\rule{0ex}{0ex}}\text{Table 5.1 (texthook, Chapter 5)}1-.9429=.\mathbf{0571}\mathbf{.}$ ### Want to see more solutions like these? 