1. Let \(X_1, X_2, ... , X_n\) 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 \(Y\).
We need to find the probability that the smallest number among \(n\) random variables could be a certain value \(j\), or \(P(Y = j)\). The denominator of this probability distribution will be:
\[ k^n = \text{total number of options (ways) in which to assign values to}\ X_i \]
The numerator will be the number of ways of assigning values to \(X_i\) that result in \(j\) being the smallest value. If:
\[ \begin{align} (k-1)^n &= \text{all options in which none of the values equal 1 } \quad \text{and} \\ (k-2)^n &= \text{all options in which none of the values equal 2 }, \\ \end{align} \]
then imposing a restriction of values greater than \(2\) in \(X_i\) can be represented by subtracting options in which none of the values equal \(2\) OR equal \(1\) from \(k^n\).
\[ \begin{align} &[k^n - (k-2)^n] \ - \ [k^n - (k-1)^n], \quad \text{or simply} \\ &(k-1)^n - (k-2)^n \end{align} \]
More generally:
\[ \begin{align} (k-j)^n - (k-j+1)^n \qquad \text{number of options in which values equal } j \text{ or greater} \end{align} \]
And thus the probability distribution of Y can be represented by:
\[ P(Y=j) = \frac{(k-j)^n - (k-j+1)^n}{k^n} \]
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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.
a. What is the probability that the machine will fail after 8 years? Provide the expected value and standard deviation. Model as a geometric.
The probability using a geometric distribution is about 62% within 8 years, the expected value is 10 years, and the standard deviation is 0.72.
X <- 8
n <- 10
p_success <- 1/n # Probability that machine will fail within first ten years = success
p_fail <- 1-p_success # Probability that machine will NOT fail within first ten years = failure
pgeom(8, p_success)## [1] 0.6125795ev <- 1/p_success
ev## [1] 10sd <- (p_success*8)*p_fail
sd## [1] 0.72a. What is the probability that the machine will fail after 8 years? Provide the expected value and standard deviation. Model as an exponential.
The probability using an exponential distribution is about 55% within 8 years, the expected value is 10 years, and the standard deviation is about 0.85.
pexp(8, p_success)## [1] 0.550671ev <- 1/p_success
ev## [1] 10sd <- sqrt(X * p_success * p_fail)
sd## [1] 0.8485281a. What is the probability that the machine will fail after 8 years? Provide the expected value and standard deviation. Model as a binomial.
The probability using a binomial distribution is about 43% within 8 years, the expected value is 0.8, and the standard deviation is about 0.85.
pbinom(0, 8, p_success) #The probability of 0 machine failures within 8 years## [1] 0.4304672ev <- 8*p_success
ev## [1] 0.8sd <- sqrt(X * p_success * p_fail)
sd## [1] 0.8485281a. What is the probability that the machine will fail after 8 years? Provide the expected value and standard deviation. Model as a Poisson.
The probability using a Poisson distribution is about 45% within 8 years, the expected value is 0.8, and the standard deviation is about 0.89.
l <- 8 * p_success #lambda, or the mean number of occurrences in 8 years
ppois(0, l) #The probability of 0 or more machine failures in 8 years## [1] 0.449329ev <- l
ev## [1] 0.8sd <- sqrt(l)
sd## [1] 0.8944272