DATA605_HW7_DISTRIBUTIONS AND DENSITIES

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.

Solution 1

Assuming that each \(X_{i}\) has k possibilities, we will have \(k^{n}\) possible values in total

1 is the minimum of the Xi’s, the nymber of ways of getting Y.

\(P(x = 1) = \frac{k^{n} - (k-1)^{n}}{k^{n}}\)

\(P(x = 2) = \frac{(k-2+1)^{n} - (k-2)^{n}}{k^{n}}\)

For \(X_{1}, X_{2}, . . . , X_{n}\) mutually independent random variables, If Y = j then, the minimum value j is given as:

\(P(x = j) = \frac{(k-j+1)^{n} - (k-j)^{n}}{k^{n}}\)

Solution 2A (Geometric)

Geometric distribution is given as:

\(P_{r}(X=k) = (1−p)^{k-1}p\)

p <- 1 / 10
geometric_prob <- ((1 - p)^ (8 - 1) * p)
round(geometric_prob, 4)

## [1] 0.0478

Expected value

EV_geometric <- 1 / p
EV_geometric

## [1] 10

Standard Deviation

sd_geometric <- sqrt((1 - p) / (p ^ 2))
round(sd_geometric, 4)

## [1] 9.4868

2b. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as an exponential.

Solution 2B (Exponential)

The exponential distribution is given as:

\(f(x) = \lambda e^{\lambda x}\)

lambda <- 1 / 10
e <- exp(1)
exponential_prob <- lambda*e ^ (- lambda * 8)
round(exponential_prob, 4)

## [1] 0.0449

Expected value

EV_exponential <- 1 / lambda
EV_exponential

## [1] 10

Standard Deviation

sd_exponential <- sqrt(1 / (lambda ^ 2))
sd_exponential

## [1] 10

2c. 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)

Solution 2C (Binomial)

The Binomial distribution is given as:

\(f(r,n,p)=\binom{n}{r}p^{r}(1−p)^{n-r}\)

p <- 1 / 10
binomial_prob <- choose(8, 0) * (p ^ 0) * ((1 - p) ^ (8))
round(binomial_prob, 4)

## [1] 0.4305

expected_value_binomial <- p * 8
expected_value_binomial

## [1] 0.8

n = 8
sd_binomial <- sqrt(n * p  * (1 - p))
round(sd_binomial, 4)

## [1] 0.8485

2d. What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a Poisson.

Solution 2D (Poisson)

Poisson Distribution is given as:

\(P(x,\lambda)= \frac{e^{-\lambda}\lambda^{x}}{x!}\)

lambda_poisson = 8 * (1 / 10)
e <- exp(1)
poisson_prob <-(e ^ (- lambda_poisson) * lambda_poisson ^ 0) / factorial(0)
round(poisson_prob, 4)

## [1] 0.4493

Expected value

EV_poisson <- lambda_poisson
EV_poisson

## [1] 0.8

Standard Deviation

sd_poisson <- sqrt(lambda_poisson)
round(sd_poisson, 4) 

## [1] 0.8944