# Normal Random Variable - uniform distribution

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.

• I'm guessing that you mean the colon to be a decimal point. Apr 30, 2012 at 1:52
• It's not at all clear why a uniform distribution is mentioned in this question. Apr 30, 2012 at 2:09

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.)