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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.

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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

1 Answer 1

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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.

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