Q. 5.29 - A First Course in Probability

Question

A model for the movement of a stock supposes that if the present price of the stock is $s$, then after one period, it will be either $us$ with probability $p$ or $ds$ with probability $1 - p$. Assuming that successive movements are independent, approximate the probability that the stock’s price will be up at least $30$ percent after the next $1000$ periods if $u = 1.012, d = 0.990, \text{and} p = .52.$

Step-by-Step Solution

Answer

The required probability is $0.9993$.

Step 1. Given information.

It is given that the present price of the stock $= s$

After one period, the price will be either $us \text{with probability} p$ or $ds$ with probability $1 - p$.

And $u = 1.012, d = 0.990, and p = .52$

Step 2. Find the number of periods in which the price of stock rises.

Let, 

Initial price of stock be $s$ and

The number of periods among $1000$ times period in which the stock increases be $X$.

Then the end price will be given by

$s·u^X·d^{1000-X}=s·{\left(\frac{u}{d}\right)}^X·d^{1000}$

In order to get $30\%$ up the end price be $1.3$ times of the initial price.

$$\begin{gathered} \Rightarrow s·{\left(\frac{u}{d}\right)}^X·d^{1000} \ge 1.3s \\ \Rightarrow {\left(\frac{u}{d}\right)}^X·d^{1000} \ge 1.3 \end{gathered}$$

Taking log on the both sides, we get

$$\begin{gathered} \log X \left(\frac{u}{d}\right)+1000 \log d \ge \log 1.3 \\ \Rightarrow X\ge \frac{\log 1.3-1000 \log d }{\log \left({\displaystyle \frac{u}{d}}\right)} \end{gathered}$$

As $u=1.012 and d=0.990$

$X\ge \frac{\log 1.3-1000 \log \left(0.990\right) }{\log \left({\displaystyle \frac{1.012}{0.990}}\right)}$

$\therefore X\ge 469.2089$

Step 3. Find the required probability.

The probability that at least $470$ time periods the stock will have to rise among $1000$ time periods.

Here, $X$ is a binomial with parameters

$n=1000 \text{and} p=0.52$

So,

$$\begin{gathered} \mu =np=1000\times 0.52=520 \\ \sigma =\sqrt{npq}=\sqrt{1000\times 0.52\times \left(1-0.52\right)}=15.7987 \end{gathered}$$

and

$P\left(X\ge 470\right)=P\left(X\ge 469.5\right)$ using continuity correction

$$\begin{gathered} P\left(X\ge 469.5\right)=P\left(\frac{X-\mu }{\sigma }\ge \frac{469.5-\mu }{\sigma }\right) \\ P\left(X\ge 469.5\right)=P\left(z\ge \frac{469.5-520}{15.7987}\right) \\ =1-P\left(z\ge -3.1964\right) \\ =1-0.0007 \\ =0.9993 \end{gathered}$$

Therefore, the required probability is $0.9993$.