Homework Week 9

Exercise 1

Y1 <- 100  
n <- 365  
mu <- 0  
sigma <- sqrt(1/4)

# Probabilities
prob_Y365_100 <- 1 - pnorm((100 - Y1) / sqrt(n * sigma^2))
prob_Y365_110 <- 1 - pnorm((110 - Y1) / sqrt(n * sigma^2))
prob_Y365_120 <- 1 - pnorm((120 - Y1) / sqrt(n * sigma^2))

# Output
cat("Estimated probabilities:\n")

## Estimated probabilities:

cat("Y365 >= 100:", prob_Y365_100, "\n")

## Y365 >= 100: 0.5

cat("Y365 >= 110:", prob_Y365_110, "\n")

## Y365 >= 110: 0.1475849

cat("Y365 >= 120:", prob_Y365_120, "\n")

## Y365 >= 120: 0.01814355

Exercise 2

Binomial Distribution

The moment generating function (MGF) of a binomial distribution with parameters \(n\) (number of trials) and \(p\) (probability of success in each trial) is given by:

\[ M_X(t) = (1 - p + pe^t)^n \]

Expected Value

To find the expected value \(E(X)\), we take the first derivative of the MGF with respect to \(t\), and then evaluate it at \(t = 0\):

\[ E(X) = M'_X(0) = \frac{d}{dt}M_X(t)\bigg|_{t=0} \]

\[ E(X) = np \]

Variance

For the variance \(Var(X)\), we take the second derivative of the MGF with respect to \(t\), and then evaluate it at \(t = 0\):

\[ Var(X) = M''_X(0) + [M'_X(0)]^2 = \frac{d^2}{dt^2}M_X(t)\bigg|_{t=0} + \left(\frac{d}{dt}M_X(t)\bigg|_{t=0}\right)^2 \]

\[ Var(X) = np(1-p) \]

Exponential Distribution

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

\[ M_X(t) = \frac{\lambda}{\lambda - t} \]

Expected Value

To find the expected value \(E(X)\), we take the first derivative of the MGF with respect to \(t\), and then evaluate it at \(t = 0\):

\[ E(X) = M'_X(0) = \frac{d}{dt}M_X(t)\bigg|_{t=0} \]

\[ E(X) = \frac{1}{\lambda} \]

Variance

For the variance \(Var(X)\), we take the second derivative of the MGF with respect to \(t\), and then evaluate it at \(t = 0\):

\[ Var(X) = M''_X(0) + [M'_X(0)]^2 = \frac{d^2}{dt^2}M_X(t)\bigg|_{t=0} + \left(\frac{d}{dt}M_X(t)\bigg|_{t=0}\right)^2 \]

\[ Var(X) = \frac{1}{\lambda^2} \]

…