Let \(X_1\), \(X_2\)…\(X_n\) 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 \(X_i\)s. Find the distribution of \(Y\).
Let \(j\) be a value within the range \([1,k]\).
We will define the probability of \(Y=j\) as: \[P(Y=j) = \frac{num\; ways \: event \; can \; happen}{total \; number \; of \; options}\]
We know that the total number of options is \(k^n\) because each of the \(n\) variables can take on \(k\) different values. Therefore, \(k^n\) will form the denominator in our above equation.
Intuitively, we know that as \(j\) increases, the probability of the minimum of the \(X_is\) being equal to \(j\) decreases. This is because the number of ways that \(j\) can be the minimum of the set of variables (our numerator in the probability equation above) decreases. In order to abstract this, let’s use an example.
Suppose \(k=4\) and \(n=2\). In order for \(Y=j\) to be true, we know that the set of values each variable can take on is as follows:
\(Y=j=1: \{1,2,3,4\}\)
\(Y=j=2: \{2,3,4\}\)
\(Y=j=3: \{3,4\}\)
\(Y=j=4: \{4\}\)
This means that in order for the minimum value to be 3, the variables cannot take on a value of 1 or 2. Similarly, in order for the minimum value to be 4, the variables can only take on a value of 4.
The number of ways the event can occur for each value of \(j\) can be modeled as:
\[number \; of \; values \; in \; the \;set^{number \; of \; variables}\]
Following our example:
\(Y=j=1: 4^2\)
\(Y=j=2: 3^2\)
\(Y=j=3: 2^2\)
\(Y=j=4: 1^2\)
We can abstract the number of times Y=j can occur as:
\[Y=j: (k-j+1)^n\] Putting this all together:
\[P(Y=j) = \frac{(k-j+1)^n}{k^n}\]
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).
We know that a “success” is the probability that the event will happen. In this instance, success is defined as the probability that a machine will fail. A failure is defined as the probability that the machine will last. For all of the following, let:
\[P(Success)=p = \frac{1}{10}\] \[P(Failure)=1-p = \frac{9}{10}\] \[X=number\; of\; trials\]
a. 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.)
Modeling as a geometric distribution, we know that: \[P(X>k)=(1-p)^k\]
where \(k\) is the total number of years.
Therefore, the probability of 0 successes (machine failures) in the first 8 years is:
p <- 1/10
k <- 8
geomDist <- (1-p)^k
geomDist## [1] 0.4304672The expected value is: \[E=\frac{1}{p}=10\]
The standard deviation is: \[\sigma = \sqrt{\frac{1-p}{p^2}}\]
geomStDev <- sqrt((1-p)/(p)^2)
geomStDev## [1] 9.486833b. What is the probability that the machine will fail after 8 years? Provide also the expected value and standard deviation. Model as an exponential.
Modeling as an exponential distribution, we know that:
\[P(X\leq k) = 1-e^{-k\lambda}\]
where \(k\) is the total number of years and \(\lambda\) is the average number of failures per year.
The complement of this is: \[P(X> k)=e^{-k\lambda}\]
Therefore, the probability of 0 successes (machine failures) in the first 8 years is:
k <- 8
expLambda <- 1/10
expDist <- exp(-k*expLambda)
expDist## [1] 0.449329The expected value is: \[E = \frac{1}{\lambda}\]
expExpVal <- 1/expLambda
expExpVal## [1] 10The standard deviation is:
\[\sigma = \sqrt{\frac{1}{\lambda^2}}\]
stDevExp <- sqrt(1/(expLambda^2))
stDevExp## [1] 10c. 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)
The binomial distribution is: \[b(n,p,k) = {{n}\choose{k}}p^k(1-p)^{n-k}\]
where \(n\) is the number of years and \(k\) is the number of successes.
Therefore, the probability of 0 successes (machine failures) in the first 8 years is:
n <- 8
k <- 0
p <- 1/10
choose(n,k)*p^k*(1-p)^(n-k)## [1] 0.4304672The expected value is:
expBinom <- n * p
expBinom## [1] 0.8The standard deviation is:
stDevBinom <- sqrt(n*p*(1-p))
stDevBinom## [1] 0.8485281d. What is the probability that the machine will fail after 8 years? Provide also the expected value and standard deviation. Model as a Poisson.
We know that the mean failure over the 8 year period is: \[\lambda = n * p\]
poissonLambda <- n * p
poissonLambda## [1] 0.8We can model this as a Poisson distribution:
\[P(X=n)= \frac{\lambda^ne^{-\lambda}}{n!}\]
where \(n\) is the number of failures in the first 8 years.
Therefore, the probability of 0 successes (machine failures) in the first 8 years is:
n <- 0
poissonDist <- ((poissonLambda^n)*exp(-poissonLambda))/factorial(n)
poissonDist## [1] 0.449329The expected value is:
\[E(X) = \lambda = 0.8\]
In a poisson distribution, we know that: \[E(X) = Var(X) = \lambda\]
Since \(Var(X) = \sqrt{\sigma}\), we know that the standard deviation is:
stDevPoisson <- sqrt(poissonLambda)
stDevPoisson## [1] 0.8944272