knitr::opts_chunk$set(echo = TRUE)
library(tinytex)
library(pracma)

Source files: [https://github.com/djlofland/DATA605_S2020/tree/master/]

Problem 11 (p 303)

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? (See Exercise 10.)

\[\text{from Ex 10: } M = min(X_j),\ \ \ mean = \mu/n \\ \mu = 1000, n = 100\]

n <- 100
mu <- 1000

(expected <- mu/n)
## [1] 10

Problem 14 (p 303)

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

\[f_Z(z) = (\frac{1}{2})\lambda e^{-\lambda|z|}\]

\[ f_{X}(x) = f_{Y}(x) = \left\{ \begin{array}\\ \lambda e^{-\lambda x}, & \text{if }x \ge 0,\\ 0, & \text{otherwise;}\\ \end{array} \right.\]

\[\text{Given: } Z = Y-X \text{, rearrange to get: } Y = X+Z\]

\[ P(Z \le z) = \int_{0}^{\infty}\lambda e^{-\lambda x} ( \int_{0}^{x+z} \lambda e^{-\lambda y} dy) dx\\ = \int_{0}^{\infty}\lambda e^{-\lambda x} ( 1 - e^{-\lambda (x+z)}) dx\\ = \int_{0}^{\infty}\lambda e^{-\lambda x} - \lambda e^{-\lambda x} e^{-\lambda (x+z)}) dx\\ = \int_{0}^{\infty}\lambda e^{-\lambda x} - \int_{0}^{\infty}\lambda e^{-\lambda x} e^{-\lambda (x+z)}) dx\\ = 1 - \lambda \int_{0}^{\infty} e^{(-2\lambda x - \lambda z)}) dx\\ P(Z \le z) = 1 - \lambda \frac{e^{-\lambda z}}{2\lambda} = 1 - \frac{1}{2}e^{-\lambda z}\\ \text{now differentiate with respect to Z to get the density function}\\ f_Z(z) = \frac{d}{dz}(1 - \frac{e^{-\lambda z}}{2}) = \frac{1}{2}\lambda e^{-\lambda z}\]

Problem 1 (p 320-321)

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

variance <- 100/3  # note this is sigma^2
mu <- 10
  1. \(P(|X − 10| \ge 2)\).
e <- 2
(upper <- variance / e^2)
## [1] 8.333333
  1. \(P(|X − 10| \ge 5)\)
e <- 5
(upper <- variance / e^2)
## [1] 1.333333
  1. \(P(|X − 10| \ge 9)\)
e <- 9
(upper <- variance / e^2)
## [1] 0.4115226
  1. \(P(|X − 10| \ge 20)\)
e <- 20
(upper <- variance / e^2)
## [1] 0.08333333