#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 .
Ans: Let, Y is equal to some value y. This occurs when all n random variables X1, X2, …, Xn are greater than or equal to y, and the probability of this event is:
P(Y = y) = P(X1 >= y, X2 >= y, …, Xn >= y)
Since the random variables are uniformly distributed from 1 to k, the probability that any one of them is less than y is (y - 1) / k. Therefore, the probability that all n of them are greater than or equal to y is:
P(Y = y) = (k - y + 1)^n / k^n
So, the distribution of Y is a discrete distribution on the integers from 1 to k, with probabilities given by the above formula.
To verify that this indeed forms a probability mass function, we need to show that the sum of probabilities over all possible values of Y is 1:
Sum(P(Y = y)) = Sum((k - y + 1)^n / k^n (for y = 1 to k))
Using the identity Sum i^m = (m+1)B_{m+1} where B_{m+1} is the (m+1)-th Bernoulli number, we can simplify the above expression to:
Sum(P(Y = y)) = (B_{n+1}(k+1) - B_{n+1}(1)) / k^n
Since B_{m+1} is 0 for odd m > 1, this expression reduces to a simple closed-form expression for the sum:
Sum(P(Y = y)) = (k+1)/((n+1)k^n)
Therefore, the distribution of Y is indeed a valid probability mass function.
In summary, the distribution of Y is a discrete distribution on the integers from 1 to k, with probabilities given by:
P(Y = y) = (k - y + 1)^n / k^n (for y = 1 to k)
#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.).
#a. 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..)
Ans: The probability that the machine will fail after 8 years is equivalent to the probability of not failing during the first 8 years, which can be calculated as:
P(X > 8) = (1 - P(X <= 8)) = (1 - (1 - p)^8)
The expected value and standard deviation of a geometric distribution are given by:
E(X) = 1/p
SD(X) = sqrt((1-p)/p^2)
# Probability of not failing in the first 8 years
p <- 1/10
prob_not_fail <- (1-(1 - p)^8)
# Probability of failing after 8 years
prob_fail_after_8_years <- 1 - prob_not_fail
prob_fail_after_8_years## [1] 0.4304672# Expected value
expected_value <- 1/p
expected_value## [1] 10#standard deviation
standard_deviation <- sqrt((1-p)/p^2)
standard_deviation## [1] 9.486833#b. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as an exponential.
Ans: If we model the failure rate of the machine as an exponential distribution, the probability of the machine failing after 8 years can be calculated as follows:
P(X > 8) = e^(-lambda*8), where lambda is the failure rate parameter.
Since the manufacturer’s expected lifetime of the machine is 10 years, we know that the failure rate parameter lambda is equal to 1/10 per year.
Therefore, the probability of the machine failing after 8 years can be calculated as:
P(X > 8) = e^(-8/10)
The expected value and standard deviation of an exponential distribution are given by:
E(X) = 1/lambda
SD(X) = 1/lambda
# Probability of failing after 8 years
lambda <- 1/10
prob_fail_after_8_years <- exp(-lambda*8)
prob_fail_after_8_years## [1] 0.449329# Expected value
expected_value <- 1/lambda
expected_value## [1] 10# Standard deviation
standard_deviation <- 1/lambda
standard_deviation## [1] 10#c. 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)
Ans: If we model the failure rate of the machine as a binomial distribution, where the number of trials is the number of years the machine has been in operation (n = 8), the probability of success (failure) is p = 1/10, and we are interested in the probability of zero successes (i.e., zero failures) in 8 years, we can calculate the probability as follows:
P(X = 0) = (n choose 0) * p^0 * (1-p)^(n-0)
The expected value and standard deviation of a binomial distribution are given by:
E(X) = np
SD(X) = sqrt(np*(1-p))
# Probability of zero failures in 8 years
n <- 8
p <- 1/10
prob_zero_failures <- choose(n, 0) * p^0 * (1-p)^(n-0)
prob_zero_failures## [1] 0.4304672# Expected value
expected_value <- n*p
expected_value## [1] 0.8# Standard deviation
standard_deviation <- sqrt(n*p*(1-p))
standard_deviation## [1] 0.8485281#d. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a Poisson.
Ans: If we model the failure rate of the machine as a Poisson distribution, where the rate parameter lambda is equal to 1/10 per year (the manufacturer’s expected lifetime of the machine), the probability of the machine failing after 8 years can be calculated as follows:
P(X > 0) = 1 - P(X = 0) = 1 - e^(-lambda*n)
The expected value and standard deviation of a Poisson distribution are given by:
E(X) = lambda*n
SD(X) = sqrt(lambda*n)
lambda <- 1/10
n <- 8
prob_fail_after_8_years <- 1 - exp(-lambda*n)
prob_fail_after_8_years## [1] 0.550671# Expected value
expected_value <- lambda*n
expected_value## [1] 0.8# Standard deviation
standard_deviation <- sqrt(lambda*n)
standard_deviation## [1] 0.8944272#Reference: https://rpubs.com/mrcuny/1012832