91Ó°ÊÓ

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

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" $$\begin{gathered} ⇒s·{\left(\frac{u}{d}\right)}^X·d^{1000}â‰\yen 1.3s \\ ⇒{\left(\frac{u}{d}\right)}^X·d^{1000}â‰\yen 1.3 \end{gathered}$$

Taking log on the both sides, we get

role="math" localid="1646820396751" $$\begin{gathered} \log X\left(\frac{u}{d}\right)+1000\log dâ‰\yen \log 1.3 \\ ⇒Xâ‰\yen \frac{\log 1.3-1000\log d}{\log \left({\displaystyle \frac{u}{d}}\right)} \end{gathered}$$

As $u=1.012andd=0.990$

$Xâ‰\yen \frac{\log 1.3-1000\log \left(0.990\right)}{\log \left({\displaystyle \frac{1.012}{0.990}}\right)}$

$∴Xâ‰\yen 469.2089$

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,

$$\begin{gathered} \mathrm{μ}=np=1000×0.52=520 \\ \mathrm{σ}=\sqrt{npq}=\sqrt{1000×0.52×\left(1-0.52\right)}=15.7987 \end{gathered}$$

and

$$\begin{gathered} P\left(Xâ‰\yen 469.5\right)=P\left(\frac{X-\mathrm{μ}}{\mathrm{σ}}â‰\yen \frac{469.5-\mathrm{μ}}{\mathrm{σ}}\right) \\ P\left(Xâ‰\yen 469.5\right)=P\left(zâ‰\yen \frac{469.5-520}{15.7987}\right) \\ =1-P\left(zâ‰\yen -3.1964\right) \\ =1-0.0007 \\ =0.9993 \end{gathered}$$

Therefore, the required probability is$0.9993$.