Chapter 7

Problem 11

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

Given:

\(1/\lambda\) = 1000

\(\lambda\) = 1000

The Expected life is 1/100*\(\lambda\)

1/100 * 1/1000 = 10

The expected time for the first of the bulb to burnout is 10 hours.

Problem 14

Assume that \(X_1 and X_2\) are independent random variables, each having an exponential density with parameter \(\lambda\). Show that Z = \(X_1 − X_2\) has density

\(fZ(z)=(1/2)\lambda { e }^{ -\lambda |z| }\)

The probability density function for \(X_1 and X_2\) are

\(f(x_1)=\lambda { e }^{ -\lambda (x_1) }\) \(f(x_2)=\lambda { e }^{ -\lambda (x_2) }\)

The joint density of \(X_1 and X_2\) are \(\lambda { e }^{ -\lambda (x_1) }\) * \(\lambda { e }^{ -\lambda (x_2) }\) or \(\lambda^2 { e }^{ -\lambda (x_1 + x_2) }\)

When Z is negative, \(X_2\) = \(X_1\) - Z When Z is Positive, \(X_2\) > - Z and \(X_2\) will also be positive.

When Z is negeative,

\(\int_{-}^{\infty} z\lambda^2 { e }^{ -\lambda (z + 2x_2) }dx = {\lambda/2}{e}^{\lambda(z)}\)

When Z is positive,

\(\int_{0}^{\infty}{\lambda/2}{e}^{-\lambda(z+2x_2)}dx = {\lambda/2}e^{-\lambda|z|}\)

This becomes,

\(fZ(z)=(1/2)\lambda { e }^{ -\lambda |z| }\)

Chapter 8

Problem 1

Let X be a continuous random variable with mean μ = 10 and variance σ² = 100/3. Using Chebyshev’s Inequality, find an upper bound for the following probabilities.

  1. \(P(|X − 10| \ge 2)\).

kσ = 2

\(k = 2/\sqrt{100/3}\)

=\(1/k^2 = 8.3333\)

  1. \(P(|X − 10| \ge 5)\).

kσ = 5

\(k = 5/\sqrt{100/3}\)

=\(1/k^2 = 1.3333\)

  1. \(P(|X − 10| \ge 9)\).

kσ = 9

\(k = 9/\sqrt{100/3}\)

=\(1/k^2 = 0.4115\)

  1. \(P(|X − 10| \ge 20)\).

kσ = 20

\(k = 20/\sqrt{100/3}\)

=\(1/k^2 = 0.0833\)