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.

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

share|improve this question

2 Answers 2

up vote 1 down vote accepted

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.

share|improve this answer

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.

share|improve this answer
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

Your Answer


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.