Q. 8.11 - A First Course in Probability

Question

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?

Step-by-Step Solution

Answer

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

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

$Y_{10}=100+X_1+X_2+\cdots +X_9+X_{10}$

Because of the independence of random variables$X_i$ and the corresponding properties of expectation and variance we get:

$$\begin{gathered} E\left[Y_{10}\right]=100+10\mu =100, \\ Var\left(Y_{10}\right)=10{\sigma }^2=10 \end{gathered}$$

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:

$$\begin{gathered} P\left\{Y_{10}>105\right\}=1-P\left\{Y_{10}\le 105\right\} \\ =1-P\left\{\frac{Y_{10}-E\left[Y_{10}\right]}{\sqrt{Var\left(Y_{10}\right)}}\le \frac{105-E\left[Y_{10}\right]}{\sqrt{Var\left(Y_{10}\right)}}\right\} \\ =1-P\left\{\frac{Y_{10}-100}{\sqrt{10}}\le \frac{105-100}{\sqrt{10}}\right\}\approx 1-\Phi (1.58) \\ \text{ Table 5.1 (texthook, Chapter 5) }1-.9429=.\mathbf{0571}\mathbf{.} \end{gathered}$$