HW7_605

jbrnbrg

October 9, 2017

1)

Let \(X_1, X_2, . . . , X_n\) be \(n\) mutually independent random variables, each of which is uniformly distributed on the integers \(\{1,2,3,...,k\}\). Let \(Y\) denote the minimum of the \(X_i\)’s. Find the distribution of \(Y\).

1) Answer:

Here, \(Y\) is the minimum of all values \(\{X_1, X_2, . . . , X_n\}\). I want to find \(f_y(x)=P(Y=min(X_i))\) where \(min(X_i)=j\), the minimum value of all the \(X_i\)’s. To be more specific, the values of \(Y\) will take on some value within \(\{j,...,k\}\) with at least a single \(X_i\) being assigned to this \(j^{th}\) value where \(1 \leq j \leq k\).

As \(Y\) is a discrete distribution containing \(n\) elements, I’ll need to define the event space. Since each of the \(X_i\)’s are uniformly distributed on the integers \({1,2,3,...,k}\), the event space is permutations with repetion where \(n\) is the number of \(X_i\)’s to choose from and each \(X_i\) is a integer up to \(k\) yeilding: \(k^n\) different possibilities.

If we let \(Y=1\), the number of ways it’s possible to get this is \(k^n-(k-1)^n\) ways where \((k-1)^n\) is the total number of ways that none of the \(X_i\)’s are equal to \(1\). If \(Y=2\), we’d see that \((k-2)^n\) is the number of ways we could arrange the \(X_i\) such that their values are greater than 2. Then \(k^n-(k-2)^n- (k^n-(k-1))^n=-(k-2)^{n}+(k-1)^{n}\). If I continue along this same route, I note that if we set \(Y=j\), then there are \(-(k-2)^{n}+(k-1)^{n}\) ways to assign a min. value of \(j\) which leads me to:

\[(-(k-j)^{n}+(k-j+1)^{n})/k^{n}\]

for \(1 \leq j \leq k\).

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).

  1. 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)

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

  3. 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)

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

2a) Answer:

Geometric \(p=\frac{1}{10}\), \(E[X]=\frac{1-p}{p}\), \(\sigma = \sqrt{\frac{1-p}{p^2}}\)

\(P(X > 8)\):

lam <- 10
x <- 8

pgeom(x, prob = 1/lam, lower = FALSE)          # a) Geometric
## [1] 0.3874205

2b) Answer:

Exponential \(E[X]=\frac{1}{\lambda}=\frac{1}{10} \implies \sigma_k = \frac{1}{\lambda}= \frac{1}{10}\)

\(P(X > 8)\):

pexp(x, rate = 1/lam, lower = FALSE)           # b) Exponential
## [1] 0.449329

2c) Answer:

Binomial \(E[X]=np = 8/10\), \(\sigma = \sqrt{np(1-p)} \approx .849\)

\(P(X > 8)\): The probability of not getting a failure until after 8 years.

pbinom(0, size=8, prob = 1/10)                 # c) Binomial
## [1] 0.4304672

2d) Answer:

Poisson \(E[X] = Var[X]=10, \implies \sigma_k = \sqrt{\lambda}=\sqrt{10}\)

\(P(X > 8)\):

ppois(0 , lambda = x/lam)                      # d) Poisson
## [1] 0.449329