Problem One

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 .

\[For 1 \le j \le k\] \[P(Y =y) = \frac{(k - y + 1)^n - (y - k)^n}{k^n}\]

To get here: If we look at the probability of Y = 1

\[P(Y = 1) = k^n - (k - 1)^n\] where K^n is the possible number of assignments for the entire collection of random variables since k = the number of integers or buckets that each random variable can be, and n is the total number of random variables. (K-1)^n would represents all of the options where none of the Xi’s are equal to 1.

If we look at the probability of Y = 2 \[P(Y = 2) = k^n - (k - 2)^n-[k^n - (k - 1)^n]\] where K^n is the possible number of assignments and (K-2)^n represents the number of ways that we could assign X1, …,Xn with all values being greater than 2, and then we subtract the probabilty of Y=1 which we figured out above

we simplify \[P(Y = 2) = k^n - (k - 2)^n-[k^n - (k - 1)^n] -> (k - 1)^n - (k - 2)^n \]

and then we generalize the equation so that: \[P(Y =y) = \frac{(k - y + 1)^n - (y - k)^n}{k^n}\]

Problem Two

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.).