Question 11, page 303

A company buys 100 light bulbs, each of which has an exponential lifetime of 1000 hours. What is the expected time for the 1st of these bulbs to burn out?

\(\lambda = 1000\) \(n = 100\)

Since the expected time has an exponential distribution we can calculate the mean or expected value using \(E(x) = \frac {\mu}{n}\)

The expected time for the 1st of these bulbs to burn out is \(= \frac {1000}{100} = 10\) hours

mu <- 1000
n <- 100

mu/n
## [1] 10

Question 14, page 303

Assume the X1 and X2 are independent random variables, each having an exponential density with parameter \(\lambda\).

Show that Z = X1 - X2 has a density \(fz(z) = (1/2)\lambda e^{-\lambda|z|}\)

Using the convolution formula we can write \[f_Z(z) = \int^{\infty}_{-\infty} f_{X_1}(x) f_{-X_2}(z-x)dx \\ f_{-X_2}(z-x) = f_{X_2}(x-z) \\ \int^{\infty}_{0}f_{X_1}(x) f_{X_2}(x-z) dx\\ z < 0 \\ \int^{\infty}_{0} \lambda e^{-\lambda x} \lambda e^{-\lambda (x-z)}dx \\ \lambda e^{\lambda z} \int^{\infty}_{0} \lambda e^{-2 \lambda x}dx \\ \lambda e^{\lambda z} \left (-\frac{1}{2} e^{-2 \lambda x} {|}\begin{array}{} \infty \\ 0 \end{array} \right) \\ = \frac{\lambda}{2}e^{\lambda z} ~;{~when~ z < 0} \\ ~The ~distribution ~of ~z ~must ~be ~symmetric ~around ~0 ~so,\\ f_Z(z) = \frac{\lambda}{2}e^{\lambda |z|} , ~ if ~z ~ > 0 \]

Question 1, 320

Let X be a continuous variable with mean \(\mu\) = 10 and the variance \(\sigma^2\) = 100/3. Using Chebyshev’s Inequality, find an upper bound for the following probabilities.

Chebyshev’s Inequality

\[P(|X-\mu| \geq \epsilon) \leq \frac{\sigma^2}{\epsilon^2}\]

variance <- 100/3
chebs_inequality <- function(var, k){
  var / (k **2)
}

\((a)~ P(|X - 10| \geq 2)\)

k <- 2
chebs_inequality(variance, 2)
## [1] 8.333333

P(|X - 10| \(\geq\) 2) = 8.3333333

The upper bound is 8.3333333

\((b)~ P(|X - 10| \geq 5)\)

k <- 5
chebs_inequality(variance, 5)
## [1] 1.333333

P(|X - 10| \(\geq\) 5) = 1.3333333

The upper bound is 1.3333333

\((c)~ P(|X - 10| \geq 9)\)

k <- 9
chebs_inequality(variance, 9)
## [1] 0.4115226

P(|X - 10| \(\geq\) 5) = 0.4115226

The upper bound is 0.4115226

\((d)~ P(|X - 10| \geq 20)\)

k <- 20
chebs_inequality(variance, 20)
## [1] 0.08333333

P(|X - 10| \(\geq\) 5) = 0.0833333

The upper bound is 0.0833333