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 .
Distribution of Y (Minimum of Uniformly Distributed Random Variables) Let \(X_1, X_2, \ldots, X_n\) be independent and uniformly distributed random variables from 1 to \(k\). The variable \(Y\) is defined as the minimum of these \(X_i\)’s, meaning \(Y = \min(X_1, X_2, \ldots, X_n)\). We aim to determine the probability distribution of \(Y\).
The key to finding the distribution of \(Y\) is to first consider the probability that \(Y\) is greater than some specific value \(m\), where \(1 \leq m \leq k\). If \(Y > m\), it implies that all \(X_i\) are greater than \(m\) because if even one of the \(X_i\) was less than or equal to \(m\), \(Y\) would be less than or equal to \(m\) (since \(Y\) is the minimum).
Since each \(X_i\) is uniformly distributed from 1 to \(k\), the probability that a single \(X_i\) is greater than \(m\) is:
\[ P(X_i > m) = \frac{k - m}{k} \]
This is because there are \(k - m\) favorable outcomes (selecting any of \(m+1, m+2, \ldots, k\)) out of \(k\) possible outcomes.
Given that the \(X_i\) are independent, the probability that all \(X_i\) are greater than \(m\) is the product of their individual probabilities:
\[ P(Y > m) = P(X_1 > m, X_2 > m, \ldots, X_n > m) = \left( \frac{k - m}{k} \right)^n \]
Now, to find the probability mass function of \(Y\), we calculate \(P(Y = m)\). This is the probability that \(Y\) is greater than \(m-1\) but not greater than \(m\). This can be expressed as:
\[ P(Y = m) = P(Y > m - 1) - P(Y > m) = \left( \frac{k - m + 1}{k} \right)^n - \left( \frac{k - m}{k} \right)^n \]
Here, \(P(Y > m - 1)\) gives the probability that all \(X_i\) are greater than \(m-1\), and \(P(Y > m)\) gives the probability that all \(X_i\) are greater than \(m\). The difference between these two probabilities gives the probability that \(Y\) equals \(m\).
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.
Geometric Distribution
What is the probability that the machine will fail after 8 years? Provide also the expected value and standard deviation. Model as a geometric distribution. (Hint: the probability is equivalent to not failing during the first 8 years.)
Probability statement
The geometric distribution is the probability of having the first
success on the \(x\)-th trial. Here,
success is the failure of the machine, and trials are years.
For a machine expected to fail once every ten years, \(p = \frac{1}{10}\). The probability of not failing for the first 8 years and then failing in the 9th year is:
\[ P(X = 9) = (1 - p)^8 \cdot p \]
Expected value \(E(X) = \frac{1}{p}\), standard deviation \(\sigma = \sqrt{\frac{1 - p}{p^2}}\).
# a. Geometric Distribution
p <- 1/10
prob_a <- (1 - p)^8 * p
expected_a <- 1/p
sd_a <- sqrt((1 - p) / p^2)
prob_a## [1] 0.04304672expected_a## [1] 10sd_a## [1] 9.486833Exponential Distribution
What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as an exponential.
Probability statement
The exponential distribution is suitable for modeling the time until an
event occurs.
For a rate \(\lambda = \frac{1}{10}\) (as one failure is expected every ten years):
\[ P(T > 8) = e^{-\lambda \cdot 8} \]
Expected value \(E(T) = \frac{1}{\lambda}\), standard deviation \(\sigma = \frac{1}{\lambda}\).
# b. Exponential Distribution
lambda <- 1/10
prob_b <- exp(-lambda * 8)
expected_b <- 1/lambda
sd_b <- 1/lambda
prob_b## [1] 0.449329expected_b## [1] 10sd_b## [1] 10Binomial Distribution
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)
Probability statement
Modeling this as a binomial with \(n = 8\) (years), and each year as a trial, the probability of success (failure) in any year is \(p = \frac{1}{10}\).
\[ P(X = 0) = \binom{8}{0} p^0 (1 - p)^8 = (1 - p)^8 \]
Expected value \(E(X) = np\), standard deviation \(\sigma = \sqrt{np(1 - p)}\).
# c. Binomial Distribution
n <- 8
p <- 1/10
prob_c <- dbinom(0, n, p)
expected_c <- n * p
sd_c <- sqrt(n * p * (1 - p))
p## [1] 0.1prob_c## [1] 0.4304672sd_c## [1] 0.8485281Poisson Distribution
What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a Poisson.
Probability statement
If we model the number of failures over time with a Poisson
distribution, with rate \(\lambda = 1\)
failure per 10 years, for 8 years \(\lambda_8
= 0.8\).
\[ P(X = 0) = e^{-0.8} \frac{0.8^0}{0!} \]
Expected value \(E(X) = \lambda_8\), standard deviation \(\sigma = \sqrt{\lambda_8}\).
# d. Poisson Distribution
lambda_8 <- 0.8
prob_d <- dpois(0, lambda_8)
expected_d <- lambda_8
sd_d <- sqrt(lambda_8)
prob_d## [1] 0.449329expected_d## [1] 0.8sd_d## [1] 0.8944272