1 Exercise 11 page 363

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 \(X_n=Y_{n+1}-Y_n\) 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

a \(\geq100\)

b \(\geq110\)

c \(\geq120\)

Solution:

Our distribution is normal with mean \(\mu=n\mu=0\), variance \(\delta^2=n\delta^2=265*0.25=66.25\)

  1. 0.5

  2. 0.1096

  3. 0.0070

265*0.25
## [1] 66.25
pnorm(100,mean=100,sd=sqrt(66.25),lower.tail=FALSE)
## [1] 0.5
pnorm(110,mean=100,sd=sqrt(66.25),lower.tail=FALSE)
## [1] 0.1096127
pnorm(120,mean=100,sd=sqrt(66.25),lower.tail=FALSE)
## [1] 0.007001616
x <- seq(-4,4,length=100)*sqrt(66.25) + 100

hx <- dnorm(x,mean=100,sd=sqrt(66.25))

plot(x, hx)

2

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

Solution:

\(g(t)=\sum_{j=1}^{N} \binom{N}{j}p^jq^{N-j}e^{tj}\)

using binomial theorem \((a+b)^N=\sum_{j=0}^{N}\binom{N}{j}a^jb^{N-j}\), we get

\(g(t)=(pe^t+q)^N\)

\(\mu=g'(0)=Npe^t(pe^t+q)^{N-1}=Np\)

\(\mu^2=g''(0)=Npe^t(N-1)(pe^t+q)^{N-2}pe^t+Np=N(N-1)p^2+Np\)

\(\sigma^2=\mu_2-\mu^2=N(N-1)p^2+Np-(Np)^2=Np-Np^2\)

3

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

Solution:

\(g(t)=E(e^{tx})=\int_{0}^{\infty}e^{tx}\lambda e^{-\lambda x}=\lambda/(\lambda-t)\)

\(\mu=g'(0)=\lambda/(\lambda-t)^2=\lambda^{-1}\)

\(\mu^2=g''(0)=2\lambda/(\lambda-t)^3=2\lambda^{-2}\)

\(\sigma^2=\mu_2-\mu^2=2\lambda^{-2}+\lambda^{-2}=\lambda^{-2}\)