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$u=1.012,d=0.990,\text{and}p=.52.$

Short answer

The required probability is $0.9993$.

Step-by-step solution

Step 1. Given information.

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

After one period, the price will be either $us\text{withprobability}p$or role="math" localid="1646819419512" $ds$with probability $1-p$.

And$u=1.012,d=0.990,andp=.52$

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

role="math" localid="1646820170868" $s·u^X·d^{1000-X}=s·{\left(\frac{u}{d}\right)}^X·d^{1000}$

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

role="math" localid="1646820185164" $$\begin{gathered} \Rightarrow s·{\left(\frac{u}{d}\right)}^X·d^{1000}\ge 1.3s \\ \Rightarrow {\left(\frac{u}{d}\right)}^X·d^{1000}\ge 1.3 \end{gathered}$$

Taking log on the both sides, we get

role="math" localid="1646820396751" $$\begin{gathered} \log X\left(\frac{u}{d}\right)+1000\log d\ge \log 1.3 \\ \Rightarrow X\ge \frac{\log 1.3-1000\log d}{\log \left({\displaystyle \frac{u}{d}}\right)} \end{gathered}$$

As $u=1.012andd=0.990$

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

$\therefore X\ge 469.2089$

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$0.9993$.