1. Let X1, X2, . . . , Xn be n mutually independent random variables, each ofwhich is uniformly distributed on the integers from 1 to k. Let Y denote the minimum of the Xi’s. Find the distribution of Y .

Solution:

Given: X1, X2, . . . , Xn be n mutually independent random variables.

Each of which are uniformly distributed on the integers from 1 to k.

Y denote the minimum of the Xi’s.

To find the distribution of Y, we will find the probability that Y takes on each possible value from 1 to k,

where y is the range of all possible value from 1 to k.

For that we will’ calculate the cumulative distribution functions (CDFs) and probability mass functions (PMFs).

Let PMF of Y is P(Y=y) where y is the range of all possible value from 1 to k.

The probability that P(Y>y) is the probability that all Xi’s are greater than y.

Also, since Xi’s are are independent and uniformly distributed from 1 to k, the probability that any single Xi is greater than y is:

\[ = \frac{k-y}{y}\; \].

The probability of all Xi’s greater than y is:

\[ \left( \frac{k-y}{y}\; \right)^n\].

The CDF of Y:

where CDF is a function that describes the probability that a random variable (continuous or discrete) will take on a value less than or equal to a certain value.

\[F_{Y}(y) = P(Y \leq y) = 1 - P(Y > y) = 1 - \left( \frac{k-y}{k}\; \right)^n \]

Now to find the PMF of Y:

where PMF is a function that describes the probability of a discrete random variable taking on a certain value.

\[P(Y = y) = F_{Y}(y) - F_{Y}(y-1) \]

\[P(Y = y) = \left(1 - \left( \frac{k-y}{k}\; \right)^n \right) - \left(1 - \left( \frac{k-(y-1)}{k}\; \right)^n \right)\]

\[P(Y = y) = \left(\frac{k-(y-1)}{k}\; \right)^n - \left(\frac{k-y}{k}\; \right)^n \]

The PMF of Y where Xi’s is the minimum value and uniformly distributed from 1 to K.

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

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.).

Solution:

To model the time until the occurrence of a certain event, such as machine failure, we will use the exponential distribution.

The probability density function (PDF) of the exponential distribution is:

\[f(x) = \lambda * e^{-\lambda*x}\; \]

where λ is the rate parameter (1/mean)

x is the time until failure.

lambda <- 1/10

\[X \sim Geom(\frac{1}{10})\]

\[P(Y > n) = 1 - P(X \leq 8) = 1 - \sum_{i=1}^{8}(0.9)^{i-1}(0.1) \]

\[1 - 0.1(1+0.9+0.9^2+...+0.9^7)\]

\[E[X]= \frac{1}{\frac{1}{10}}= 10 \]

\[\sqrt{a}=b \]

\[\sigma = \sqrt{Var(X)}= \sqrt{90} = 3 \sqrt{10}\]

  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. (Hint: the probability is equivalent to not failing during the first 8 years..)

Probability that the machine will fail after 8 is 43%,

prob <- 1/10

year=8

round((1 - pgeom(year-1, prob)),4)
## [1] 0.4305

The Standard Deviation is 9.49.

std_dev <- round((sqrt((1-prob)/prob^2)),2)
std_dev
## [1] 9.49
  1. 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 that the machine will fail after 8 is 45%,

x <- 8
prob <- 1/10
round((1 - pexp(x, prob)), 4)
## [1] 0.4493

Standard deviation is 10.

sqrt(1/prob^2)
## [1] 10
  1. 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 that the machine will fail after 8 is 43%,

x <- 0
year <- 8
prob <- 1/10
round((dbinom(x, year, prob )), 2)
## [1] 0.43

The expected value is 0.80

expected_value <- year*prob
expected_value
## [1] 0.8

Standard deviation is 0.849.

std_dev <- sqrt(year*prob*(1-prob))
round(std_dev,3)
## [1] 0.849
  1. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a Poisson.

There is a 44.9% failure rate

x <- 0
year <- 8
prob <- 1/10
t <- 1 
round((dpois(x, year*prob/t)), 3)
## [1] 0.449

Expected Value is 0.8

exp_val <- year*prob/t

exp_val
## [1] 0.8

Standard Deviation is:

(year*prob/t)^.5
## [1] 0.8944272