Problem 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 \(X_i\)’s. Find the distribution of \(Y\) .

Simulation

I really don’t know the formula to determine the expected minimum of a random variable and don’t remember seeing it in the readings. So let’s try simulating this…

# k = range of possible values
# n = number of mutually independent random variables
# m = number of values in each of the n variables
 sim <- function(k, n, m){
     df <- data.frame(matrix(ncol = n, nrow = m))
     name_list <- c()
     Y <- c()
     for(i in seq(n)){
         name <- paste('X_', i, sep = "")
         name_list[i] <- name
         ran_var <- c(sample(1:k, m, replace=T))
         df[, i] <- ran_var
         Y[i] <- min(ran_var)
     }
     colnames(df) <- name_list
     print(df)
     print("Y =", Y)
     return(Y)
 }

If the number of values in each variable is >= k you would expect the min to be 1 or close to it. So let’s try our simulation function with 5 random variables, \(X_1, ...X_5\) with 12 value each in the range 1 through 10.

set.seed(10)
# k = 10
# n = 5
# m = 12
sim1 <- sim(10, 5, 12)
##    X_1 X_2 X_3 X_4 X_5
## 1    6   2   5   9   2
## 2    4   6   8  10   9
## 3    5   4   9   7   4
## 4    7   5   3   6  10
## 5    1   1   8   3   3
## 6    3   3   4   3   5
## 7    3   4   6   1   2
## 8    3   9   1   8   6
## 9    7   9   2   3   5
## 10   5   7   9   2   5
## 11   7   8   5   1   4
## 12   6   4   8   5   6
## [1] "Y ="
counts <- table(sim1)
barplot(counts)

As expected, most of our minimums are equal to 1 with only one of the minimum values equal to 2.

However, if the number of values in each variable is < k you would have a much different range of values in Y. Let’s try our simulation function again this time with 5 random variables, \(X_1, ...X_5\) with 12 values each in the range 1 through 100.

set.seed(10)
# k = 100
# n = 5
# m = 12
sim2 <- sim(100, 5, 12)
##    X_1 X_2 X_3 X_4 X_5
## 1   51  12  41  83  11
## 2   31  60  71  96  81
## 3   43  36  84  69  36
## 4   70  43  24  51  94
## 5    9   6  78  28  25
## 6   23  27  36  23  48
## 7   28  40  54   2  20
## 8   28  84  10  73  59
## 9   62  87  17  25  46
## 10  43  62  90  17  47
## 11  66  78  43   2  40
## 12  57  36  75  49  51
## [1] "Y ="
sim2 <- sort(sim2)
counts <- table(sim2)
barplot(counts)

Once again, as expected, this time we get 5 different minimum values within the range 2 through 11.

Math

Here’s a relatively simple way I found online (in the comments) of this post for computing the expected minimum of a set of independent identically distributed random variables (IID’s)…

Add another \(X_{n+1}\) random variable which is also uniformly distributed on the integers from 1 to \(k\) to the collection, let \(Y = min{X_1, ..., X_n}\), and compute \(P(X_{n+1} < Y)\) in two ways:

  1. First, \(P(X_{n+1} < Y)\) is just the probability that \(X_{n+1}\) is the smallest of the \(n+1\) IID uniformly random variables, and this is \(\frac{1}{(n+1)}\) by symmetry.
  2. Second, \(P(X_{n+1} < Y)\) is just \(E[Y]\), since \(X_{n+1}\) is a uniform random variable from 1 to k.

The first part makes saense to me, but the second part I am not so sure about… Also, the post was for uniform distributions from 0 to 1 so maybe that makes a difference?

Problem 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.).

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..)

\[ P(T > 8) = (1-0.1)^8 = 0.9^8 = 0.4304672\\ \]

Expected value, variance, and standard deviation.

\[ \mu = E[X] = \frac{1}{p} = \frac{1}{0.1} = 10\\ \sigma^2 = V[X] = \frac{1-p}{p^2} = \frac{1-0.1}{0.1^2} = 90\\ \sigma = \sqrt{90} = 9.486833 \]

(1-0.1)^8
## [1] 0.4304672
pgeom(0, 0.9^8)
## [1] 0.4304672

Part 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.

\[ P(T > 8) = 1-(1 - e^{- \lambda x}) = e^{- 0.1 \times 8}\\ \]

Expected value, variance, and standard deviation.

\[ \mu = E[X] = \frac{1}{\lambda} = \frac{1}{0.1} = 10\\ \sigma^2 = V[X] = \frac{1}{\lambda^2} = \frac{1}{0.1^2} = 100\\ \sigma = \sqrt{100} = 10 \]

exp(- 0.1*8)
## [1] 0.449329
pexp(8,0.1, lower.tail = FALSE)
## [1] 0.449329

Part 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)

\[ P(T > 8) = {8 \choose 0} \times 0.1^0 \times (1-0.1)^{8-0}= 1 \times 1 \times 0.9^8 = 0.4304672\\ \]

Expected value, variance, and standard deviation for an eight year period…

\[ \mu = E[X] = np = 8 \times 0.1 = 0.8\\ \sigma^2 = V[X] = np(1-p) = 8 \times 0.1 \times 0.9 = 0.72\\ \sigma = \sqrt{0.72} = 0.8485281 \]

choose(8,0)*0.1^0*(1-0.1)^8
## [1] 0.4304672
pbinom(0, 8, 0.1)
## [1] 0.4304672

Part 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.

\[ P(T > 8) = \frac{e^{-\lambda} \lambda^k}{k!} = \frac{e^{-(.1 \times 8)} (.1 \times 8)^0}{0!} = 0.449329 \]

Expected value, variance, and standard deviation for an eight year period.

\[ \mu = E[X] = \lambda = 0.1*8 = 0.8\\ \sigma^2 = V[X] = \lambda = 0.1*8 = 0.8\\ \sigma = \sqrt{\lambda} = \sqrt{0.1*8} = \sqrt{0.8} = 0.8944272 \]

Expected value, variance, and standard deviation for a one year period.

\[ \mu = E[X] = \lambda = 0.1 = 0.1\\ \sigma^2 = V[X] = \lambda = 0.1 = 0.1\\ \sigma = \sqrt{\lambda} = \sqrt{0.1} = 0.3162278 \]

# lambda = the probability in an 8 year period, 0.1 per year times 8 years
(exp(1)^(-(0.1*8))*(0.1*8)^0)/factorial(0)
## [1] 0.449329
ppois(0, 0.1*8)
## [1] 0.449329