model for the movement of a stock price
Solution 1
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.
Solution 2
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.
tamefoxes
Updated on August 01, 2022Comments
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tamefoxes over 1 year
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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tamefoxes about 11 yearsI 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
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Ross Millikan about 11 years@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.
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tamefoxes about 11 yearsWouldn't the mean be np?
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Ross Millikan about 11 years@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?