Homework 7

Exercise 1

Let X1, X2, . . . , Xn be n mutually independent random variables, each of which is uniformly distributed on the integers from 1 to k. Let Y denote the minimum of the Xi’s. Find the distribution of Y

Answer:

\[ \begin{align*} P(Y \leq y) & = 1 - P(Y > y) \\ & = 1 - P(\min\{X_1, X_2, \ldots, X_n\} > y) \\ & = 1 - P(X_1 > y, X_2 > y, \ldots, X_n > y) \\ & = 1 - \left(\frac{k - y}{k}\right)^n \\ \\ P(Y \leq y - 1) & = 1 - \left(\frac{k - y + 1}{k}\right)^n \\ \\ d(y) & = P(Y = y) = P(Y \leq y) - P(Y \leq y - 1) \\ & = \frac{(k - y + 1)^n - (k - y)^n}{k^n} \end{align*} \]

Exercice 2

In this section, I’ll answer the second exercice:

Question A

# Probability of failure after 8 years using geometric distribution
p_geometric <- 1/10
prob <- 1 - pgeom(8-1, p_geometric)

# Expected value and standard deviation for geometric distribution
mean <- 1 / p_geometric
sd <- sqrt((1 - p_geometric) / p_geometric^2)

prob

## [1] 0.4304672

mean

## [1] 10

sd

## [1] 9.486833

Question B

# Probability of failure after 8 years using exponential distribution
lambda <- 1 / 10
prob <- 1 - pexp(8, rate = lambda)

# Expected value and standard deviation for exponential distribution
mean <- 1 / lambda
sd <- 1 / lambda

prob

## [1] 0.449329

mean

## [1] 10

sd

## [1] 10

Question 3

# Probability of failure after 8 years using binomial distribution
n <- 8
p_binomial <- 1 / 10
prob <- dbinom(0, size = n, prob = p_binomial)

# Expected value and standard deviation for binomial distribution
mean <- n * p_binomial
sd <- sqrt(n * p_binomial * (1 - p_binomial))

prob

## [1] 0.4304672

mean

## [1] 0.8

sd

## [1] 0.8485281

Question 4

# Probability of failure after 8 years using Poisson distribution
lambda_poisson <- 8 / 10
prob <- ppois(0, lambda = lambda_poisson)

# Expected value and standard deviation for Poisson distribution
mean <- lambda_poisson
sd <- sqrt(lambda_poisson)

prob

## [1] 0.449329

mean

## [1] 0.8

sd

## [1] 0.8944272

…