# model for the movement of a stock price

A model for the movement of a stock price supposes that if the present price is S then after one period, say one second, it will either go up to uS with probability p or go down to dS with probability q = 1 - p. Assuming that successive movements are independent,approximate the probability that the stock will be up by at least 5% after the next 1000 periods for u = 1.02, d = 0.95 and p = 0.6

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You need to find how many ups and downs are necessary.

Since $1.05^{723} \times 0.98^{277} \approx 1.115$ and $1.05^{722} \times 0.98^{278} \approx 1.039$, you need at least $723$ upticks, as you have already found. You need the number of upticks to be to be at least $\dfrac{\log(1.05/0.95^{1000})}{\log(1.02/0.95)}.$

Using the R code 1-pbinom(772,1000,0.6) this probability is about 2.929176e-16 which is extremely small. A normal approximation such as 1-pnorm(722.5,mean=600,sd=sqrt(240)) would give 1.332268e-15 which is also extremely small: as Ross says, you are almost eight standard deviations above the mean.

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Hint: The ending value is a binomial distribution. First calculate how many ups are required so you are up by 5%. Note that the order of ups and downs don't matter to your final result, just the number. Then calculate the chance that you have that many.

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I did calculate that and figured out that i need the random variable X=> 723. However, when I used a cumulative density function for binomials on many online calculators from i = 723 to 1000, I would get 0 because the probability is very small. This made me think that my method was incorrect – tamefoxes Oct 18 '12 at 3:42
@user43956: I didn't check 723, but it looks about right. A good check is how many standard deviations you need to be above the mean. If I recall correctly, the standard deviation is $\sqrt {npq} \approx 15$ (I used $p=.5, \sqrt {1000}\approx 30$ as I could do that easily), so you need to be 8 sd high which never happens. – Ross Millikan Oct 18 '12 at 3:48
Wouldn't the mean be np? – tamefoxes Oct 18 '12 at 4:18
@user43956: no, the mean is $u^{np}d^{nq}$ because on average you have $np$ ups and $nq$ downs. The asymmetry in the ups and downs is very important. What is the expected value of a single change? – Ross Millikan Oct 18 '12 at 4:23