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

Response

\[ \begin{align} F_Y(y) = P(Y \leq y) &= 1 - P(Y>y) & \text{complement of max} \\ &= 1 - P(min(X_1,X_2 ... X_n) > y) & \text{minimum of the X's} \\ &= 1 - P(X_1 > y, X_2>y, ..., X_n > y) \\ &= 1 - P(X_1 >y) P(X_2>y)...P(X_n > y) & \text{due to mutually independent} \\ &= 1 - P(X_i >y)^n & \text{i.i.d} \\ &= 1 - (\frac{k-y}{k})^n & \\ \end{align} \]

k-y is the number of remaining outcomes divided by the total number of outcome which in this case is k.

This gives us the cumulative probability of Y less than or equal y. I think the question wants the probability of Y = y which is the probability density function. So, to remove the lower section of the cdf. We can substract \(P(Y \leq y -1 )\) from \(P(Y \leq y)\).

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

So, the distribution function is

\[ \begin{align} f_Y(y) &= P(Y = y) \\ &= P(Y \leq y) - P(Y \leq y-1) \\ &= 1 - (\frac{k-y}{k})^n - (1 - \frac{(k-y+1)^n}{k^n}) \\ &= \frac{(k-y+1)^n - (k-y)^n}{k^n} \end{align} \]

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. (Include the probability statements and R Code for each part.).

2.A

What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a geometric. (Hint: the probability is equivalent to not failing during the first 8 years..)

\[ X \sim Geom(\frac{1}{10}) \\ P(X > n) = 1 - P(X \leq 8) = 1 - \sum_{i=1}^{8} (0.9)^{i-1}(0.1) \\ \Rightarrow 1 - 0.1(1 + 0.9 + 0.9^2 + ... + 0.9^7) \\ E[X] = \frac{1}{1/10} = 10 \\ \sigma = \sqrt{Var(X)} = \sqrt{90} = 3\sqrt{10} \]

p <- 1/10  # given by problem
q <- 1 - p

no_failure_8_years <- 1 - pgeom(7, p)

no_failure_8_years
## [1] 0.4304672
# not failing during the first 8 years
q^8
## [1] 0.4304672

2.B

What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as an exponential.

\[ X \sim Exp(1/10) \\ f_X(8) = (0.1) e ^ {-(0.1) 8}, \ x \geq 0 \\ E[X] = \frac{1}{1/10} = 10 \\ \sigma = \sqrt{Var(X)} = \sqrt{\frac{1}{(1/10)^2}} = \sqrt{100} = 10 \]

exp_model <- pexp(8, p, lower.tail = FALSE)

exp_model
## [1] 0.449329

2.C

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)

\[ X \sim Binom(8,1/10) \\ P(X = 0) = {8 \choose 0} 0.1^0 (0.9)^8 \\ E[X] = np = 8(0.1) = 0.8 \\ \sigma = \sqrt{np(1-p)} = \sqrt{0.72} \]

n <- 8  # n trials
n_success <- 0  # n successes

binom_model <- pbinom(n_success, n, p)

binom_model
## [1] 0.4304672

2.D

What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a Poisson.

lamdba is 8/10 since out of the 10 years we want to know the probability it fail after 8 years.

\[ X \sim Pois(8/10) \\ P(X = 0 ) = e^{-0.8}\frac{0.8^0}{0!} = e^{-0.8} \\ E[X] = 0.8 \\ \sigma = \sqrt{0.8} \approx 0.8944 \]

pois_model <- ppois(0, 8/10)

pois_model
## [1] 0.449329