1. Page- 363, Ex- 11

The price of one share of stock in the Pilsdor 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

Ans:

Let Xn = Yn+1 - Yn, then we have X1, X2, …, X364 as independent random variables with the same distribution such that E(Xn) = Mu = 0 and Var(Xn) = sigma^2 = 1/4.

Using the property of linearity of expectation, we have:

E(Y365) = E(Y1 + X1 + X2 + … + X364) = E(Y1) + E(X1) + E(X2) + … + E(X364) = 100 + 0 + 0 + … + 0 = 100.

Also, using the property of the variance of a sum of independent random variables, we have:

Var(Y365) = Var(Y1 + X1 + X2 + … + X364) = Var(Y1) + Var(X1) + Var(X2) + … + Var(X364) = 0 + 1/4 + 1/4 + … + 1/4 = 91.

Therefore, Y365 follows a normal distribution with mean 100 and variance 91, i.e., Y365 ~ N(100, 91).

  1. Let Z = (Y365 - E(Y365))/sqrt(91) be the standard normal variable.

Then, P(Y365 >= 100) = P(Z >= (110 - 100)/sqrt(91)) = P(Z >= 0) = 0.5, where Z is the standard normal variable.

  1. Let Z = (Y365 - E(Y365))/sqrt(91) be the standard normal variable.

Then, P(Y365 >= 110) = P(Z >= (110 - 100)/sqrt(91)) = P(Z >= 1.05) = 0.1472537

  1. Let Z = (Y365 - E(Y365))/sqrt(91) be the standard normal variable.

Then, P(Y365 >= 120) = P(Z >= (120 - 100)/sqrt(91)) = P(Z >= 2.09) = 0.01801584

# Given values
Y1 <- 100
mu <- 0
sigma2 <- 1/4
n <- 365

# Calculate mean and variance of Yn
mean_Yn <- Y1
var_Yn <- (n-1)*sigma2

# Calculate the standard deviation of Yn
sd_Yn <- sqrt(var_Yn)

# (a) Probability that Y365 >= 100
p_a <- pnorm((100  - mean_Yn) / sd_Yn, lower.tail = FALSE)
cat("Probability that Y365 >= 100: ", p_a, "\n")
## Probability that Y365 >= 100:  0.5
# (b) Probability that Y365 >= 110
p_b <- pnorm((110 - mean_Yn) / sd_Yn, lower.tail = FALSE)
cat("Probability that Y365 >= 110: ", p_b, "\n")
## Probability that Y365 >= 110:  0.1472537
# (c) Probability that Y365 >= 120
p_c <- pnorm((120 - mean_Yn) / sd_Yn, lower.tail = FALSE)
cat("Probability that Y365 >= 120: ", p_c, "\n")
## Probability that Y365 >= 120:  0.01801584

Additional:

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

Ans:

The moment generating function (MGF) of the binomial distribution Bin(n, p) is given by:

M(t) = (pe^t+ 1 - p)^n

To find the expected value, we take the first derivative of the MGF with respect to t and evaluate it at t = 0:

E(X) = M’(0) = np

To find the variance, we take the second derivative of the MGF with respect to t and evaluate it at t = 0:

Var(X) = M’’(0) + M’(0) - [M’(0)]^2 = np(1-p)

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

Ans:

The moment generating function (MGF) of the exponential distribution with parameter lambda is given by:

M(t)=1/(1-lambda*t)

To find the expected value, we take the first derivative of the MGF with respect to t and evaluate it at t = 0:

E(X) = M’(0) = 1/lambda

To find the variance, we take the second derivative of the MGF with respect to t and evaluate it at t = 0:

Var(X) = M’’(0) + M’(0) - [M’(0)]^2 = 1/lambda^2