Question 6, Page 279

Let T be a random variable with range [0,\(\infty\)] and fT its density function. Find \(\mu\)(T) and \(\sigma^2\)(T) if both conditions are met:

1 – When t < 0, fT(t) = 0

2 – When t > 0, fT(t) =

  1. \(3e^{-3t}\)

  2. \(9te^{-3t}\)

  3. \(3/(1 + t)^4\)

To find \(\mu\)(T), you will find E(X), which is equal to the following:

\(\int_{0}^{\infty} t \, f(t) \, dt\)

To find \(\sigma^2\)(T), you will find \(E((X - \mu\)))^2$, which is equal to the following formula:

\(\int_{0}^{\infty} (t - \mu)^2 \, f(t) \, dt\)

Answer (a):

\(\mu\)(T) = 1/3

\(\sigma^2\)(T) = 1/9

integrate(function(t) {t * 3 * exp(-3 * t)}, lower = 0, upper = Inf)
## 0.3333333 with absolute error < 8.1e-08
integrate(function(t) {(t - 1/3)^2 * 3 * exp(-3 * t)}, lower = 0, upper = Inf)
## 0.1111111 with absolute error < 1.5e-06

Answer (b):

\(\mu\)(T) = 2/3

\(\sigma^2\)(T) = 2/9

integrate(function(t) {t * 9 * t * exp(-3 * t)}, lower = 0, upper = Inf)
## 0.6666667 with absolute error < 4.4e-06
integrate(function(t) {(t - 2/3)^2 * 9 * t * exp(-3 * t)}, lower = 0, upper = Inf)
## 0.2222222 with absolute error < 1.8e-05

Answer (c):

\(\mu\)(T) = 1/2

\(\sigma^2\)(T) = 3/4

integrate(function(t) {t * 3 / ((1 + t)^4)}, lower = 0, upper = Inf)
## 0.5 with absolute error < 5.6e-15
integrate(function(t) {(t - 0.5)^2 * 3 / ((1 + t)^4)}, lower = 0, upper = Inf)
## 0.75 with absolute error < 8.3e-15