Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

So here's the question I'm trying to solve:

A stock price movement model supposes that if the current stock price is s, then, after one period, the stock price will be $us$ with probability $p$ and $ds$ with probability $1 - p$. Assuming that successive movements of the stock are independent, approximate the probability that the stock’s price will be up at least $50\%$ after the next $1000$ periods if $u = 1.1$, $d = 0.95$, and $p = 0.49$. What is the minimal value of $p$ to guarantee that the stock price will be over $60\%$ up after $1000$ periods?

And I don't even know where to start. Any suggestions as to how I should approach this problem?

thank you.

share|improve this question
    
I'm guessing that you mean the colon to be a decimal point. –  Michael Hardy Apr 30 '12 at 1:52
    
It's not at all clear why a uniform distribution is mentioned in this question. –  Michael Hardy Apr 30 '12 at 2:09
add comment

1 Answer

up vote 0 down vote accepted

You're multiplying 1000 times by something that's equal to either $u$ or $d$ each time. That means you're adding something equal to $\log u$ or $\log d$ 1000 times. The central limit theorem applies to independent random variables that are added; that's why we're taking logarithms.

We have $$ \mathbb{E}\left.\begin{cases} \log 1.1 & \text{with probability } 0.49 \\ \log 0.95 & \text{with probability } 0.51 \end{cases}\right\} = 0.49\log 1.1 + 0.51\log 0.95, $$ and $$ \operatorname{var}\left.\begin{cases} \log 1.1 & \text{with probability } 0.49 \\ \log 0.95 & \text{with probability } 0.51 \end{cases}\right\} = (0.49)(0.51)(\log 1.1 - \log 0.95)^2. $$ Hence $$ \frac{\left.\begin{cases} \log 1.1 & \text{with probability } 0.49 \\ \log 0.95 & \text{with probability } 0.51 \end{cases}\right\} - (0.49 \log1.1+0.51\log 0.95)}{\sqrt{(0.49)(.051)(\log1.1-\log0.95)}} $$ has expected value $0$ and standard deviation $1$.

Adding up $1000$ independent copies of this, we get $$ \frac{\text{sum}-1000(0.49\log1.1+0.51\log0.95)}{\sqrt{1000(0.49(0.51)(\log1.1-\log0.95)}} $$ has an approximately standard normal distribution. The problem is: what is the probability that this is more than $$ \frac{\log 1.5 - 1000(0.49\log1.1+0.51\log0.95)}{\sqrt{1000(0.49(0.51)(\log1.1-\log0.95)}}. $$ (I.e. going up at least $50\%$ is the same as multiplying by $1.5$ or more.)

So that last item is what you put into the table of values of the c.d.f. of the standard normal distribution.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.