Page 303 Q. 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?

lambda <- 1/1000 #rate
bulbs <- 100  #num bulbs

first_bulb_time <- -log(1 - 1/100) / lambda
first_bulb_time
## [1] 10.05034

10 hours

Page 303 Q. 14

Assume that X1 and X2 are independent random variables, each having an exponential density with parameter λ. Show that Z = X1 - X2 has density fZ(z) = (1=2)^e**-λ|z|.

Since X1 and X2 are independent random variables with exponential densities, their PDFs are given by:

fX1(x) = λ * e^(-λx) fX2(x) = λ * e^(-λx)

Calculate the PDF of Z = X1 - X2:

fZ(z) = ∫[0,∞] fX1(x) * fX2(z + x) dx

Substitute the PDFs of X1 and X2:

fZ(z) = ∫[0,∞] (λ * e^(-λx)) * (λ * e^(-λ(z + x))) dx

Integrate

fZ(z) = λ^2 * ∫[0,∞] e^(-λx) * e^(-λ(z + x)) dx

fZ(z) = λ^2 * ∫[0,∞] e^(-λx) * e^(-λz) * e^(-λx) dx

fZ(z) = λ^2 * e^(-λz) * ∫[0,∞] e^(-2λx) dx

Integral and proof

fZ(z) = λ^2 * e^(-λz) * [-1/(2λ) * e^(-2λx)] [0,∞]

fZ(z) = λ^2 * e^(-λz) * (-1/(2λ)) * [0 - 1]

fZ(z) = (1/2) * λ * e^(-λz)

Page 320-21 Q. 1

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

  1. P(|X - 10| >= 2)
var_a<- 100/3
e_a <- 2
var_a/e_a^2
## [1] 8.333333
  1. P(|X - 10| >= 5)
var_b <- 100/3
e_b<- 5
var_b/(e_b^2)
## [1] 1.333333
  1. P(|X - 10| >= 9)
var_c <- 100/3
e_c <- 9
var_c/(e_c^2)
## [1] 0.4115226
  1. P(|X - 10| >= 20)
var_d <- 100/3
e_d <- 20
var_d/(e_d^2)
## [1] 0.08333333