Homework 6

Question 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 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 $m$.

Question 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.

Geometric Distribution

1. 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] 0.04304672

expected_a

## [1] 10

sd_a

## [1] 9.486833

Exponential Distribution

2. 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] 0.449329

expected_b

## [1] 10

sd_b

## [1] 10

Binomial Distribution

3. 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] 0.1

prob_c

## [1] 0.4304672

sd_c

## [1] 0.8485281

Poisson Distribution

4. 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] 0.449329

expected_d

## [1] 0.8

sd_d

## [1] 0.8944272