For 1 ≤ j ≤ k, m(j) = ((k−j+1)n−(k−j)n)/k^n
Since Y is the minimum of the Xs, then in order to find the distribution function m(j) = P(Y = j), we will need to count the number of ways that we can assign X1, X2, …, Xn to values between j and k with at least one Xi being assigned to j and divide by the total number of possible ways to assign X1, X2, …, Xn to values between 1 and k.
Using a geometric distribution:
f(x)=P(X=x)=(1−p)^x−1 * p P(X=8)=(9/10)^7 ∗ 1/10
dgeom(8,0.1)
## [1] 0.04304672
expected Value= 1/p
1/(1/10)
## [1] 10
standard deviation= √(1−p)/p^2
sqrt((1 - (1/10))/((1/10)^2))
## [1] 9.486833
Exponential Distribution
rate= 1/10
Failure after 8 years: 1-P(X<=P)
1-pexp(8, 0.1)
## [1] 0.449329
Expected value = 1/0.1
1/0.1
## [1] 10
Standard Deviation = 1/(0.1)^2
1/(0.1)^2
## [1] 100
Binomial Distribution
X ~ B(8, 0.1) P(X = 0)
dbinom(0,8,0.1)
## [1] 0.4304672
Expected Value = n * p
8*0.1
## [1] 0.8
Standard deviation = √(np∗(1−p)
sqrt((8 * 0.1) * (1 - 0.1))
## [1] 0.8485281
Poisson Distribution
X ~ P(0.8)
P(X=0)
ppois(0,0.8)
## [1] 0.449329
Expected value = 0.8
Standard Deviation = √(expected value)
sqrt(0.8)
## [1] 0.8944272