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.

I’ll use an example where \({X1, X2, X3, X4}\) will be the number of independent random variables, \(n = 4\), uniformly distributed on integers from 1 to 6, \(k=6\).

To calculate the number of ways \(n\) is uniformly distributed on \(k\), we’ll do \(6^4\), which gives us 1,296. This number will be the denominator, \(k^n\).

When \(Y=1\), the number of combinations where \(X1\) to \(X4\) are greater than 1 is \(5^4\). The total number of combinations is \(6^4\), so therefore to find the distribution when \(Y=1\) is \(6^4 - 5^4\), which can be expressed as \(k^n - (k-1)^n\).

When \(Y=2\), the number of combinations where \(X1\) to \(X4\) are greater than 2 is \(4^4\). Using the methods from the previous calculations, we can express the distribution when \(Y=2\) as \(6^4 - (6^4 - 5^4) - 4^4\). This can be expressed as \(k^n - (k^n - ((k-1)^n) - (k-2)^n\), simplified as \((k-1)^n - (k-2)^n\) when \(Y=2\).

Where \(P(Y=y)\), we can therefore state that when \(y=2\), \((k-2+1)^n - (k-2)^n)\), with \((k-2+1)^n\) simplified as \((k-1)^n\). The final distribution of \(Y\) is:

\((k-y+1)^n - (k-y)^n) / k^n\)

Reference: https://math.dartmouth.edu/archive/m20f10/public_html/HW5Solutions.pdf

  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.)
# Probability of Failure in 10 years
p <- 1/10

# Geometric Model for machine failing after 8 years
geom_model <- 1 - pgeom(8, p)
geom_model
## [1] 0.3874205

Expected Value:

geom_expected_value <- 1/p
geom_expected_value
## [1] 10

Standard Deviation:

sd_geom <- sqrt(1-p)/p
sd_geom
## [1] 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.
# rate of machine failing after 10 years
lambda <- 1/10
exp_model <- 1 - pexp(8, lambda)
exp_model
## [1] 0.449329

Expected Value:

exp_expected_value <- 1/lambda
exp_expected_value
## [1] 10

Standard Deviation:

sd_exp <- sqrt(1/lambda^2)
sd_exp
## [1] 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)
# Number of trials (using number of years machine will fail)
n <- 8

# Probability of 1 failure after 8 years
binom_model <- choose(n, 0) * p^0 * (1-p)^n
binom_model
## [1] 0.4304672

Expected Value:

# p = 0.1
binom_expected_value <- n * p
binom_expected_value
## [1] 0.8

Standard Deviation:

sd_binom <- sqrt(binom_expected_value*(1-p))
sd_binom
## [1] 0.8485281
  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.
n <- 8
p <- 0.1
lambda_rate <- n * p
pois_model <- ppois(0,lambda_rate)
pois_model
## [1] 0.449329

Expected Value:

pois_expected_value <- lambda_rate
pois_expected_value
## [1] 0.8

Standard Deviation:

sd_pois <- sqrt(lambda_rate)
sd_pois
## [1] 0.8944272