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.
\(P(X = 1) =(k^n - (k - 1) ^n)/(k^n)\)
Continuing:
\(P(X = 3) =(k^n - (k - 3) ^n)/(k^n)\)
We generalize:
\(P(X = m) =(k^n - (k - m) ^n)/(k^n)\)
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.).
Using the geometric model: \(Pr(X = k) = (1-p)^{k-1}p\)
Let’s take the cumulative odds of the machine having failed in the first 8 years and then subtract it from 1. With the geometric distribution, the odds of a failure every year are lower due to the fact that they also include the odds that it did not fail the previous years.
1 - pgeom(8, (1/10))
## [1] 0.3874205
Expected value: 10 years - flip the fraction
1/(1/10)
## [1] 10
Standard Deviation:
For a geometric distribution, this is \(\sqrt{(1-p)} / p\)
sqrt(1-.1)/.1
## [1] 9.486833
Using the exponential model:
Exponential model: \(P(X \leqslant x) = 1 -e^{-mx}\)
We’ll get the complement.
1 - pexp(8, (1/10))
## [1] 0.449329
Expected value: Still 10
1/(1/10)
## [1] 10
Standard deviation: \(\sqrt{(1/p^2)}\)
sqrt(1/(1/10)^2)
## [1] 10
Modelling as a binomial: \(p^x(1-p)^{n-x}\)
We’ll look at getting no success outcomes in 8 years.
pbinom(0,8,.1)
## [1] 0.4304672
Expected value (first success): In this context, that means .8 expected failures in 8 years.
.1 * 8
## [1] 0.8
Standard deviation:
sqrt(8*.1*(1-.1))
## [1] 0.8485281
Let’s get Poisson: \((e^-{\lambda} \lambda^{x}) / x!\)
ppois(0, 8/10)
## [1] 0.449329
Expected value:
8/10
## [1] 0.8
Standard deviation:
sqrt(8/10)
## [1] 0.8944272