Problem 1:

Let \(X_1, X_2, \ldots, X_n\) be \(n\) mutually independent random variables, each uniformly distributed on the integers from 1 to \(k\). Let \(Y\) denote the minimum of the \(X_i\)’s. Find the distribution of Y

Calculate the Cumulative Distribution Function (CDF)

The CDF, \(F_Y(y)\), gives the probability that \(Y\) is less than or equal to \(y\), where \(Y\) is the minimum of a set of independent and identically distributed (i.i.d.) uniform random variables.

# Function to calculate the CDF of Y, the minimum of n i.i.d. uniform random variables
cdf_Y <- function(y, n, k) {
  if(y < 1 || y > k) {
    return(0)  # y is outside the range of the uniform distribution
  } else {
    return(1 - ((k - y) / k)^n)
  }
}

# Example: n = 5, k = 10, y = 3
n <- 5
k <- 10
y <- 3

# Calculate the CDF of Y at y
cdf_at_y <- cdf_Y(y, n, k)

cat("The probability that Y ≤", y, "when n =", n, "and k =", k, "is:", cdf_at_y, "\n")
## The probability that Y ≤ 3 when n = 5 and k = 10 is: 0.83193

Problem 2:

A machine has an expected lifetime of 10 years, implying an expected failure rate of once every ten years. Various probability models are used to analyze the probability of failure after 8 years.

Part a: geometric

What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a geometric. (Hint: the probability is equivalent to not failing during the first 8 years..)

p_fail <- 1 / 10  
# Probability of not failing for first 8 years
p_notfail_8year <- (1 - p_fail)^8

expected_value_geo <- 1 / p_fail
std_deviation_geo <- sqrt((1 - p_fail) / p_fail^2)

cat("Probability of not failing during the first 8 years (Geometric): ", p_notfail_8year)
## Probability of not failing during the first 8 years (Geometric):  0.4304672
cat("Expected value (Geometric): ", expected_value_geo)
## Expected value (Geometric):  10
cat("Standard deviation (Geometric): ", std_deviation_geo)
## Standard deviation (Geometric):  9.486833

Part b: exponential

What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as an exponential.

lambda <- 1 / 10
# 1-prob machine will fail within 8 years
p_fail_after_8 <- 1 - pexp(8, lambda) 

expected_value_exp <- 1 / lambda
std_deviation_exp <- 1 / lambda


cat("Probability of fail after 8 years (Exponential): ", p_fail_after_8)
## Probability of fail after 8 years (Exponential):  0.449329
cat("Expected value (Exponential): ", expected_value_exp)
## Expected value (Exponential):  10
cat("Standard deviation (Exponential): ", std_deviation_exp)
## Standard deviation (Exponential):  10

Part c: binomial

What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a binomial. (Hint: 0 success in 8 years)

n <- 8  
p <- 1/10  

# Probability of no failures in 8 years (0 successes)
p_no_failures <- dbinom(0, size = n, prob = p)


# Expected value and standard deviation
expected_value <- n * p
std_deviation <- sqrt(n * p * (1 - p))


cat("Probability of no failures in 8 years:", p_no_failures)
## Probability of no failures in 8 years: 0.4304672
cat("Expected value (number of failures in 8 years):", expected_value)
## Expected value (number of failures in 8 years): 0.8
cat("Standard deviation:", std_deviation)
## Standard deviation: 0.8485281

Part d: poisson

What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a Poisson.

#When modeling the failure of a machine using a Poisson distribution, we interpret the machine's expected lifetime of 10 years as an average failure rate of λ= 1/10
lambda_8_years <- 1/10*8

# Probability of the no failure in first 8 years
p_no_failures_8 <- dpois(0, lambda_8_years)


# Expected value and standard deviation
expected_value <- lambda_8_years
std_deviation <- sqrt(lambda_8_years)

# Display the results
cat("Probability of the machine no failure in first 8 years:", p_no_failures_8)
## Probability of the machine no failure in first 8 years: 0.449329
cat("Expected value :", expected_value)
## Expected value : 0.8
cat("Standard deviation:", std_deviation)
## Standard deviation: 0.8944272