Important distributions

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.

Y denotes the minimum of the \(Xi\)’s, such that each independent random variable \(Xi\) has \(k\) possibilities.

Suppose that each \(Xi\) has \(k\) possibilities: 1, 2, …, \(k\). The total possible number of assignments for the entire collection of random variables X1, X2, …, Xn is \(k^n\) . This will form the denominator for our probability distribution function.

The number of ways of getting \(Y\) = 1 is \(k^n\) - \((k - 1)^n / k^n\), since \(k^n\) represents the total number of options and \((k-1)^n\) represents all of the options where none of the \(Xi\)’s are equal to 1.

When X = 1: \(P(X=1)\) = \(\frac{k^n - (k-1)^n}{k^n}\)

Similarly When X= 2 & 3: \(P(X=2)\) = \(\frac{(k-2+1)^n - (k-2)^n}{k^n}\) \(P(X=3)\) = \(\frac{(k-3+1)^n - (k-3)^n}{k^n}\)

Generalization this for (X=m): \(P(X=m)\) = \(\frac{(k-m+1)^n - (k-m)^n}{k^n}\)

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.).

• 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..)

We can use the Geometric Distribution formula:

\[P(T=n) = q^{n-1}p\]

# set var
p_fail <- 1/10
n <- 8
p_not_fail <- 1-p_fail

# compute
(p_fail_after_8_geo <- 1-pgeom(n-1, p_fail))

## [1] 0.4304672

# std dev
(sd_geom <- sqrt(1-p_fail)/p_fail)

## [1] 9.486833

# expected value
(exp_val_geom <- 1/p_fail)

## [1] 10

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

\[P(X \leq k)=\left\{\begin{array}{ll} 1-e^{-k / \mu}, & k \geq 0, \\ 0, & \underline{k<0 .} \end{array} \quad\left(\text { where } \mu=\frac{1}{\lambda}\right)\right.\]

# compute
(p_fail_after_8_exp <- 1-pexp(n, p_fail))

## [1] 0.449329

#std dev
(sd_exp <- sqrt(1/(p_fail)^2))

## [1] 10

# exp value
(exp_val_exp <- 1/p_fail)

## [1] 10

• 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)

\[\left(\begin{array}{l} n \\ j \end{array}\right)=\frac{n !}{j !(n-j) !}\]

# compute
(p_fail_after_8_bi <- pbinom(0,n,p_fail))

## [1] 0.4304672

# std dev
(sd_bi <- sqrt(n*p_not_fail*p_fail))

## [1] 0.8485281

# exp value
(exp_val_bi <- n * p_fail)

## [1] 0.8

• What is the probability that the machine will fail after 8 years?. Provide also the expected value and standard deviation. Model as a Poisson. \[P(X=x) = \frac{\lambda^x e^{-\lambda}}{x !}\]

# set var
lambda <- 8/10

# compute
(p_fail_after_8_pois <- ppois(0, lambda))

## [1] 0.449329

# std dev
(sd_pois <- sqrt(lambda))

## [1] 0.8944272

# exp val
(exp_val_pois <- lambda)

## [1] 0.8