Hazal Gunduz
1. Let \(X_{1}\), \(X_{2}\), . . . , \(X_{n}\) 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 \(X_{i}\)’s. Find the distribution of Y.
Found this helpful [link] (“https://math.dartmouth.edu/archive/m20f10/public_html/HW5Solutions.pdf”) for understanding the way of solution.
Y denote the minimum of the \(X_{i}\)’s and each independent random variable \(X_{i}\) has k possibilities.
Since Y is the minimum value of \(X_{i}\) over all of the\(X_{i}\)’s, then to find the distribution function m(j) = P(Y = j), we will need to count the number of ways that we can assign \(X_{1}\), \(X_{2}\), …, \(X_{n}\) to values between j and k with at least one \(X_{i}\) being assigned to j and divide by the total number of possible ways to assign \(X_{1}\), \(X_{2}\), …, \(X_{n}\) to values between 1 and k.
\(X_{i}\) has k possibilities: 1, 2, …, k. Then, the total possible number of assignments for the entire collection \(X_{1}\), \(X_{2}\), …, \(X_{n}\) will be \(k^{n}\)
The number of ways of getting Y = 1 is \(k^{n}\) − \((k − 1)^n\), since \(k^{n}\) represents the total number of options and \((k-1)^{n}\) represents all of the options where none of the \(X_{i}\)’s are equal to 1.
Therefore the solution is: 1 ≤ j ≤ k, m(j) = \(\frac{(k - j + 1)^n - (k - j)^n}{k^n}\)
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..)
p <- 0.1
pgeom(8, p, lower.tail = F)## [1] 0.3874205expected <- 1/p
expected## [1] 10sd <- sqrt(((1-p) / p^2))
sd## [1] 9.486833b. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as an exponential.
pexp(8, p, lower.tail = F)## [1] 0.449329expected <- 1/p
expected## [1] 10sd <- sqrt(p^-2)
sd## [1] 10c. 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)
pbinom(0, 8, p)## [1] 0.4304672expected <- 8 * p
expected## [1] 0.8sd <- sqrt(8 * p * (1-p))
sd## [1] 0.8485281d. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a Poisson.
ppois(0, 8/10)## [1] 0.449329expected <- 8/10
expected## [1] 0.8sd <- sqrt(expected)
sd## [1] 0.8944272