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 .
ANS:
For \[ 1 \leq j \leq k, m(j) = (k-j+1)^n - (k - j)^n / k^n \]
Since \(Y\) is the minimum value of \(Xi\) over all of the \(Xi'\)s, 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.
First, suppose that each \(Xi\) has k possibilities: 1, 2,….k. Then, 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 probability distribution function.
Now, the number of ways of getting \(Y = 1\) is \[ k^n -(k-1)^n \]
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.
If \(Y = 2\), then there are \[ k^n -(k-2)^n - [ k^n -(k-1)^n ] \] different options for the collection of $ Xi $ ’s, where the \[k^n\] represents the total number of options (with no restrictions), \[(k-2)^n\] represents the number of ways that we could assign \(X1,X2......Xn\) with all of their values being greater than 2, and, as we showed above, \[[k^n -(k-1)^n]\] represents the number of ways that we could assign \(X1,X2......Xn\) and have at least one of them equal 1 (i.e. Y = 1). We can simplify above equation to get \[ (k - 1)^n - (k - 2)^n \]
In general, if Y = j then there are \[ ( k - j + 1)^n -(k-j)^n \] ways to assign \(X1,X2......Xn\) so that minimumvalue is j.
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.).
E(X)=ns=10 Probability that machine fail once every 10 years
s is the probability that the machine fail
q is the probability that the machine does not fail
s=1/10
s## [1] 0.1q=1-s
q## [1] 0.9#ns=10
n=10/s
n## [1] 100#v(x)=nsq
n*s*q## [1] 9# Probability that machine will fail after 8 years
#using R function pgeom
p <- pgeom(8, s, lower.tail = F)
p## [1] 0.3874205#probability will copier will fail
p = 1 -p
p## [1] 0.6125795#expected value:mean
#s is probabity of copier failure
m <- round(1/s,2)
m## [1] 10#standard deviation
sd <- round(sqrt((1-s)/(s^2)),2)
sd## [1] 9.49\[ P(X>=9) = P( (X - EX(X))/\sqrt{(V(X)})\]
Using Exponential Distribution \(p(n) = e^{{-\lambda}x}\), probability that copier will be working or fail after \(8\) years is,
Expected value ns =10 Standard deviation = \(\sqrt{(V(X)})\)
#number of trials
i = 8
#failure rate of copier
s = 1/10
#copier working
f = 1 - s
#using R function pexp
p <- pexp(8, s, lower.tail = F)
p## [1] 0.449329#probability will copier will fail
p = 1 -p
p## [1] 0.550671#expected value: mean
#s is probabity of copier failure, lambda
m <- round(1/s,2)
m## [1] 10#standard deviation
#in expontential distribution standard deviation = mean
sd <- m
sd## [1] 10P(X = X) = \[ {{8}\choose{x}} (0.10)^x (0.90)^{8-x} \]
P(X = 8) = \[ {{8}\choose{8}} (0.10)^8 (0.90)^{8-8} \]
=0.43
#number of trials
i = 8
#probability of failure rate of copier
s = 1/10
#copier working
d = 1 - s
#using R function pbinom
p <- pbinom(0, size=8, prob=0.1)
p## [1] 0.4304672#probability will copier will fail
p = 1 -p
p## [1] 0.5695328#expected value:mean
#s is probabity of failure
#i is number of years
m <- round(i * s,2)
m## [1] 0.8#standard deviation
#square root of mean * (1-p)
sd <- sqrt(m * d)
sd## [1] 0.8485281#number of trials
t = 8
#mean or average failure rate of copier(success) in 8 years
k = 0
#probability of failure
p = 1/10
#lambda = years * probability of failure
l = t * p
#using R function ppois
pp <- ppois(0, lambda = l)
pp## [1] 0.449329#probability will copier will fail
pp = 1 -pp
pp## [1] 0.550671#expected value: mean
#s is probabity of failure(p)
#i is number of years(n)
m <- round(l,2)
m## [1] 0.8#standard deviation
#square root of mean
sd <- sqrt(m)
sd## [1] 0.8944272