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 .
In order to find the distribution function m(j) = P(Y = j), we will need to count the number of ways that we can assign X1, X2, …, Xn to values between j and k with at least one Xi being assigned to j and divide by the total number of possible ways to assign X1, X2, …, Xn to values between 1 and k.
k^n - the total possible number of assignments for the entire collection of random variables
(k −1)^n represents all of the options where none of the Xi’s are equal to 1.
k^n − (k − 1)^n is the number of ways of getting Y = 1
if Y = j then there are (k −j + 1)^n − (k − j)^n ways to assign X1, …, Xn so that the minimum value is j. Therefore, we should define m(j) to be (k − j + 1)^n − (k − j)^n
Source:
https://math.dartmouth.edu/archive/m20f10/public_html/HW5Solutions.pdf
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.).
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 (1 every 10)
Geometric distribution:
success = failure
P(x<=n) = (1-p)^(x)*p, n - expected number of failures before the first success
P(x>n) = 1 - P(x<=n)
n = 8
p = 0.1
q = 1-p
pgeom(n-1, p, lower.tail = F)## [1] 0.4304672# expected value
q = 1 - p
ex <-q/p
ex## [1] 9# standard deviation
std<-sqrt(q/p^2)
std## [1] 9.486833What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as an exponential.
Exponential distribution:
P(x<=n) = 1 - e−λx
P(x>n) = 1 - P(x<=n)
n = 8
1 - (1- exp(-0.1*n))## [1] 0.449329# checking with R
pexp(8, 0.1, lower.tail = F)## [1] 0.449329# expected value
ex <- 1/0.1
ex## [1] 10# standard deviation
std<-sqrt(1/0.1^2)
std## [1] 10What 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)
Binomial distribution:
P(X=0) = p^k * q^(n-k)
n = 8
k = 0
p^k * q^(n-k)## [1] 0.4304672# checking with R
pbinom(0,8,0.1, lower.tail = T)## [1] 0.4304672# expected value E(X) = n*p
ex<-n*p
ex## [1] 0.8std<-sqrt(n*p*q)
std## [1] 0.8485281What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a Poisson.
Poisson distribution:
P(X=k)^8 = ((e^-lambda*(lambda^k)/k!
average number of failures in 8 years will be: lambda = 8/10 = 0.8
ppois(0, 0.8, lower.tail = T)## [1] 0.449329lambda = 0.8
# expected value
ex<- lambda
ex## [1] 0.8std<-sqrt(lambda)
std## [1] 0.8944272