Question 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

Let Xi be a discrete Uniform distribution on (1,k) \[ P(X_i = x) = \frac{1}{k}, x=1,…, k \\ P(X_i = x) \leq \sum_{y = 1}^{x}{\frac{1}{k}} = \frac{x}{k}, x=1,2,…, k \\ Hence, P(X_i \gt x) = 1 - \frac{x}{k} \\ P(Y \lt y) = P(min\{X_1, \dots, X_n\} \ge y) \\ = 1 - (P(X_i \ge y))^n \\ = 1 - (1 - \frac{y}{k})^n \\ P ( Y = y) = P ( Y \le y) - P ( Y \le y -1) \\ = 1 - (1 - \frac{y}{k})^n - (1 -(1 - \frac{y - 1}{k})^n) \\ = (1 - \frac{y - 1}{k})^n - (1 - \frac{y}{k})^n \\ = (\frac{k - y + 1}{k})^n - (\frac{k - y}{k})^n \\ = \frac{(k - y + 1)^n - (k - y)^n}{k^n}, y = 1,2,\dots,k \]

Question 2

Your organization owns a copier (future lawyers, etc.) or MRI (future doctors). This machine has a manufacturer’s expected lifetime of 10 years. This means that we expect one failure every ten years. (Include the probability statements and R Code for each part.).

  1. 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..)

Answer

Geometric Progression: P(X=n)=(1−p)^(n−1)∗p

n <- 8
p <- 1 / 10
geometric_prob_8_years <- ((1 - p)^ (n - 1) * p)
paste("P(X=8) = ", geometric_prob_8_years)
## [1] "P(X=8) =  0.04782969"

Expected Value: \[E[X] = \frac{1}{p} \\\]

geo_expected_value <- 1 / p
paste("E(X) = ", geo_expected_value)
## [1] "E(X) =  10"

Standard Deviation: \[\sigma = \sqrt \frac {1 - p}{p^2}\]

geo_standard_deviation <- sqrt((1 - p) / (p ^ 2))
paste("Standard Deviation = ", geo_standard_deviation)
## [1] "Standard Deviation =  9.48683298050514"
  1. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as an exponential.

Answer Exponential Distribution: \[P(X>=n)=e ^ {- \lambda n}\]

lambda <- 1 / 10
exponential_prob_8_years = exp(-lambda * n)
paste("P(X>=8) = ", exponential_prob_8_years)
## [1] "P(X>=8) =  0.449328964117222"

Expected Value: \[E[X] = \frac {1}{\lambda} \]

exp_expected_value <- 1 / lambda
paste("E(X) = ", exp_expected_value)
## [1] "E(X) =  10"

Stardard Deviation: \[\sigma = \sqrt \frac {1}{\lambda ^2 }\]

exp_standard_deviation <- sqrt(1 / (lambda ^ 2))
paste("Standard Deviation = ", exp_standard_deviation)
## [1] "Standard Deviation =  10"
  1. 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)

Answer

Binomial Distribution: \[P(X>=n) = {}_{n}C_{k}P^k(1−p)^{n−k}\]

k <- 0
# n and p are already defined under part (a)
bin_prob_8_years <- choose(n, k) * (p ^ k) * ((1 - p) ^ (n-k))
paste("P(X>=8) = ", bin_prob_8_years)
## [1] "P(X>=8) =  0.43046721"

Expected Value: \[E[X] = np\]

bin_expected_value <- n * p
paste("E(X) = ", bin_expected_value)
## [1] "E(X) =  0.8"

Stardard Deviation: \[\sigma = \sqrt {np(1-p)}\]

bin_standard_deviation <- sqrt(n * p * (1 - p))
paste("Standard Deviation = ", bin_standard_deviation)
## [1] "Standard Deviation =  0.848528137423857"
  1. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a Poisson.

Answer

Poisson Distribution: \[P(X>=n) = \frac {e^{-x} \lambda ^{x}}{x!}\]

lambda_poisson = n * (1 / 10)
poison_prob_8_years = (exp(- lambda_poisson) * lambda_poisson ^ 0) / factorial(0)
paste("P(X>=8) = ", poison_prob_8_years)
## [1] "P(X>=8) =  0.449328964117222"

Expected Value: \[E[X] = \lambda\]

poisson_expected_value <- lambda_poisson
paste("E(X) = ", poisson_expected_value)
## [1] "E(X) =  0.8"

Standard Deviation: \[\sigma = \sqrt \lambda\]

poisson_standard_deviation <- sqrt(lambda_poisson)
paste("Standard Deviation = ", poisson_standard_deviation)
## [1] "Standard Deviation =  0.894427190999916"