The price of one share of stock in the Pilsdorff Beer Company (see Exercise 8.2.12) is given by Yn on the nth day of the year. Finn observes that the differences Xn = Yn+1 ??? Yn appear to be independent random variables with a common distribution having mean mu = 0 and variance sigma^2 = 1/4. If Y1 = 100, estimate the probability that Y365 is:

  1. \(\ge\) 100.

This problem requires estimating the probability that the expected value of the sum of 365 independent, identically distributed (iid) random variables is 0. The day-over-day difference is Yn+1 - Yn. Starting with Y1=100 and adding the daily difference for 364 days would get us to Y365. Y365-Y1=sum of 364 iid random variables with mean 0 and variance = 1/4 (std deviation = 1/2).

Bringing in the CLT, the standardized sum of these daily differences can be expressed as (see Page 351 of Introduction to Probability by Grinstead): \((S_{n}-n\mu)/(n\sigma^{2})^{1/2}\)

The problem being posed here is to find the probability that this standardized sum lies between the starting value of 100 (day 1) and ending value of 100 (day 365) i.e. that it is equal to 0. We solve this below.

n<-364
mu<-0
mean<-n*mu
sigma<-sqrt(n)/2
Sn<-100-100
cat("Mean is",mean,"\n")
## Mean is 0
cat("Sigma is",sigma,"\n")
## Sigma is 9.539392
prob.100<-pnorm(Sn,mean,sigma,lower.tail=FALSE)
cat("Probability that stock price on day 364 is greater than or equal to 100 is",prob.100,"\n")
## Probability that stock price on day 364 is greater than or equal to 100 is 0.5
  1. \(\ge\) 110.

The problem being posed here is to find the probability that this standardized sum lies between the starting value of 100 (day 1) and ending value of 110 (day 365) i.e. that it is equal to 0. We solve this below.

Sn<-110-100
prob.110<-pnorm(Sn,mean,sigma,lower.tail=FALSE)
cat("Probability that stock price on day 364 is greater than or equal to 110 is",prob.110,"\n")
## Probability that stock price on day 364 is greater than or equal to 110 is 0.1472537
  1. \(\ge\) 120.

The problem being posed here is to find the probability that this standardized sum lies between the starting value of 100 (day 1) and ending value of 120 (day 365) i.e. that it is equal to 0. We solve this below.

Sn<-120-100
prob.120<-pnorm(Sn,mean,sigma,lower.tail=FALSE)
cat("Probability that stock price on day 364 is greater than or equal to 120 is",prob.120,"\n")
## Probability that stock price on day 364 is greater than or equal to 120 is 0.01801584

2. Calculate the expected value and variance of the binomial distribution using the moment generating function.

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3. Calculate the expected value and variance of the exponential distribution using the moment generating function.

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