11 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 µ = 0 and variance σ2 = 1/4.

If Y1 = 100, estimate the probability that Y365 is
(a) ≥ 100.
(b) ≥ 110.
(c) ≥ 120

# Parameters
mu_X <- 0  
sigma_X <- sqrt(1/4)  
Y1 <- 100 
n <- 364

# Mean and standard deviation for Y365
mu_Y365 <- Y1 + n * mu_X
sigma_Y365 <- sqrt(n) * sigma_X

# Probability calculations
# P(Y365 >= 100) is 0.5 due there's a 50-50 chance, 
# it's just as likely for the price to be near where it started as anywhere else. 
P_Y365_ge_100 <- 0.5  
P_Y365_ge_110 <- 1 - pnorm(110, mu_Y365, sigma_Y365)
P_Y365_ge_120 <- 1 - pnorm(120, mu_Y365, sigma_Y365)

# Output results
cat("P(Y365 >= 100):", P_Y365_ge_100, "\nP(Y365 >= 110):", P_Y365_ge_110, "\nP(Y365 >= 120):", P_Y365_ge_120)
## P(Y365 >= 100): 0.5 
## P(Y365 >= 110): 0.1472537 
## P(Y365 >= 120): 0.01801584

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

The moment generating function (MGF) for a binomial distribution with parameters \(n\) (number of trials) and \(p\) (probability of success) is:

\[ M_{X}(t) = (1 - p + pe^t)^n \]

Mean

The mean \(\mu\) is obtained by differentiating the MGF at \(t=0\):

\[ \begin{align*} \mu = M'_{X}(0) &= \frac{d}{dt}[(1 - p + pe^t)^n]_{t=0} \\ &= n(1 - p + pe^t)^{n-1} \cdot pe^t \bigg|_{t=0} \\ &= np \end{align*} \]

Variance

The variance \(\sigma^2\) involves the second derivative of the MGF at \(t=0\):

\[ \begin{align*} M''_{X}(0) &= \frac{d^2}{dt^2}[(1 - p + pe^t)^n]_{t=0} \\ &= n(n-1)(1 - p + pe^t)^{n-2} \cdot (pe^t)^2 + n(1 - p + pe^t)^{n-1} \cdot pe^t \bigg|_{t=0} \\ &= np(1-p) \end{align*} \]

Thus, \(\sigma^2 = np(1-p)\).

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

The MGF for an exponential distribution with rate \(\lambda\) is:

\[ M_{X}(t) = \frac{\lambda}{\lambda - t}, \text{ for } t < \lambda \]

Mean

The mean \(\mu\) is the first derivative of the MGF evaluated at \(t=0\):

\[ \begin{align*} \mu = M'_{X}(0) &= \frac{d}{dt}\left(\frac{\lambda}{\lambda - t}\right)\bigg|_{t=0} \\ &= \frac{\lambda}{(\lambda - 0)^2} \\ &= \frac{1}{\lambda} \end{align*} \]

Variance

The variance \(\sigma^2\) is the second derivative of the MGF at \(t=0\) minus the mean squared:

\[ \begin{align*} M''_{X}(0) &= \frac{d^2}{dt^2}\left(\frac{\lambda}{\lambda - t}\right)\bigg|_{t=0} \\ &= \frac{2\lambda}{(\lambda - 0)^3} \\ &= \frac{2}{\lambda^2} \end{align*} \]

Then, \(\sigma^2 = M''_{X}(0) - (M'_{X}(0))^2 = \frac{2}{\lambda^2} - \left(\frac{1}{\lambda}\right)^2 = \frac{1}{\lambda^2}\).