DATA 605 HW 9

11. The price of one share of stock in the Pilsdorff Beer Company (see Exercise 8.2.12) is given by Yn on the n th 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 μ=0 and variance σ^2=1/4. If Y1=100, estimate the probability that Y365 is

    Since Xn=Yn+1−Yn, we can also say Yn+1=Xn+Yn, or in the case of year 1 and year 2, Y2=X1+Y1.

    Y365=Y1+X1+X2+X3…X364.

    Because Y1=100 and X1,X2,X3…X365 have μ=0, we know that the expected value of Y1 is 100+0+0+0…0=100.

    The variance of Y1,X1,X2,X3…X364 is 1/4 for each item. Therefore, the variance for Y365=365×1/4=91.25. Therefore, the standard deviation =91.25−−−−√=9.552. Using μ=0 and σ=91.25−−−−√

1. ≥ 100.

pnorm(100 - 100, mean = 0, sd = sqrt(91.25), lower.tail = FALSE)

## [1] 0.5

2. ≥ 110.

pnorm(110 - 100, mean = 0, sd = sqrt(91.25), lower.tail = FALSE)

## [1] 0.1475849

3. ≥ 120.

pnorm(120 - 100, mean = 0, sd = sqrt(91.25), lower.tail = FALSE)

## [1] 0.01814355

2. Calculate the expected value and variance of the binomial distribution using the moment generating function. The probability mass function for the binomial distribution is:

(nk)p^k(1−p)n−k

Therefore, the moment generating function for the binomial distribution is:

g(t)=n∑j=1(n|k)p^k(1−p)(n−k)e^(tj) =(1−p+pe^t)n

Calculating the first derivative:

g′(t)=npe^t(pet−p+1)^(n−1)

Evaluating for t = 0 to determine the mean μ1:

g′(0)=n×p×e^0(p × e^0 − p + 1)^(n−1)=np

Calculating the second derivative:

g′′(t)=npe^t(pet−p+1)^(n−2)(npet−p+1)

Evaluating for t = 0 to determine μ2:

g′′(0)=npe^0(pe0−p+1)^(n−2)(npe0 − p + 1)=np(np+1−p)

The variance is μ2−μ^2|1 = np(np+1−p)np=n^2p2(np+1−p)

3. Calculate the expected value and variance of the exponential distribution using the moment generating function. The probability density function for the exponential distribution is:

λe^−λx

Therefore, the moment generating function for the binomial distribution is:

g(t)=λ/λ−t for t<λ

Calculating the first derivative:

g′(t)=λ/(λ−t)2

Evaluating for t = 0 to determine the mean μ1:

g′(0)=1/λ

Calculating the second derivative:

g′′(t)=2λ(/λ−t)^3

Evaluating for t = 0 to determine μ2:

g′′(0)=2/λ2

The variance is μ2−μ2|1 = 2/λ^2−(1/λ)2=2/λ^2−1/λ2=1/λ2