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
model for the movement of a stock price
2 Answers
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.
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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0@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? – 2012-10-18