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 .

This [source] (https://math.dartmouth.edu/archive/m20f10/public_html/HW5Solutions.pdf) was very helpful in understanding the way to approach this problem:

The total possible combinations of Y that can be obtained is \(k^n\) given they are independent random variables and can have n total permutations. This total value will be important given that that there is a uniform distribution. If the range of the uniform distribution of Xi’s begins at 1, then \((k-1)\) would be the polynomial term representing the values where y is greater than 1 as otherwise it would be below the minimum of the range as specified. By using the complement to identify where Y=1 then \(k^n - (k-1)^n\) should represent all of the potential options where X does equal 1. If the same logic is applied to find the potential values for Y where it is greater than 2 \((k-2)\) then \((k-2)^n\) would represent all of the possible permutations of the random variable for this value of X. The complement for the value of X=2 can be obtained by \(k^n - (k-2)^n\) and it could also be identified for X=3 by taking \(k^n - (k-3)^n\)

When taking the difference between these two values similar to using integration to capture the area of this function between potential values:

\(k^n - (k-3)^n - (k^n - (k-2)^n)\)

Simplifying the terms gets:

\((k-2)^n - (k-3)^n\)

Using the same process between 2 and 1:

\(k^n - (k-2)^n - (k^n - (k-1)^n)\) becomes \((k-1)^n - (k-2)^n\)

This process could be repeated for all values through \(k\) but it looks as though it will lead to the same simplified term $(k - x +1)^n -(k - x)^n $

Given that this is a uniform distribution, the same probability is needed for all potential values in this distribution, which as previously derived above is \(k^n\) so the probability distribution of Y should be:

\(\frac{(k- x +1)^n -(k - x)^n}{k^n}\)

  1. 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..)

\(P(X=x) = p(1-p)^{n-1}\)

p <- 1/10
geo_prob <- (1-p)**(8-1)*p**0
sprintf('The probability that the machine fails modeled using the geometric model is: %f',geo_prob)
## [1] "The probability that the machine fails modeled using the geometric model is: 0.478297"
geo_fx <- 1/p
sprintf('The expected value for the geometric distribution is %f',geo_fx)
## [1] "The expected value for the geometric distribution is 10.000000"
std_geo <- sqrt(1-p)/p
sprintf('The standard deviation in the geometric model for this example is: %f',std_geo)
## [1] "The standard deviation in the geometric model for this example is: 9.486833"
  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.

\(\lambda e^{-\lambda t}\)

mu_ex <- 10
ex_lambda <- 1/mu_ex
t <- 8
p_exp <- ex_lambda * exp(ex_lambda*t)

sprintf('The probability that the machine fails modeled using the exponential model is: %f',p_exp)
## [1] "The probability that the machine fails modeled using the exponential model is: 0.222554"
sprintf('The expected value for the exponential distribution is %f',1/ex_lambda)
## [1] "The expected value for the exponential distribution is 10.000000"
std_ex <- sqrt(1/ex_lambda**2)
sprintf('The standard deviation in the exponential model for this example is: %f',std_ex)
## [1] "The standard deviation in the exponential model for this example is: 10.000000"
  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)

\(\binom{n}{k}(1-p)^{n-k}p^k\)

b_prob <- 1/10
failures <- 8
bino <- choose(8,0)*(1-b_prob)**8*b_prob**0

sprintf('The probability that the machine will fail after 8 years for the Binomial distribution is %f',bino)
## [1] "The probability that the machine will fail after 8 years for the Binomial distribution is 0.430467"
ex_bino <- failures *  b_prob
print(ex_bino)
## [1] 0.8
sprintf('The expected value in the binomial model for this example is: %f',ex_bino)
## [1] "The expected value in the binomial model for this example is: 0.800000"
std_bino <- sqrt(failures * b_prob * (1-b_prob))
sprintf('The standard deviation in the binomial model for this example is: %f',std_bino)
## [1] "The standard deviation in the binomial model for this example is: 0.848528"
  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.

\(\frac{\lambda^k}{k!} e^{-\lambda}\)

lambda <- 10
poisson_prob <- lambda ** 8 * exp(-lambda) / factorial(8)


sprintf('The probability that the machine will fail after 8 years for the Poisson distribution is %f',poisson_prob)
## [1] "The probability that the machine will fail after 8 years for the Poisson distribution is 0.112599"
sprintf('The expected value for the Poisson distribution is %f',lambda)
## [1] "The expected value for the Poisson distribution is 10.000000"
sprintf('The standard deviation for the Poisson distribution is %f',sqrt(lambda))
## [1] "The standard deviation for the Poisson distribution is 3.162278"