ASSIGNMENT 8

IS 605 FUNDAMENTALS OF COMPUTATIONAL MATHEMATICS

  1. 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.)

Solution:

#Expected value for the first bulb to burn

E <- 1/(100/1000)
E
## [1] 10

Expected value for the first bulb to burn: 10

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

Solution:

\(f_Z(z) = (1/2)e^{-\lambda|z|}\) can be re-written as \(f_Z(z) = \begin{cases} (1/2)e^{-\lambda z}, & \mbox{if } z \ge 0, \\ (1/2)e^{\lambda z}, & \mbox{if }z <0. \end{cases}\)

Since \(X_1\) and \(X_2\) have exponential density, their PDF is

\(f_{X_1}(x)=f_{X_2}(x)=\begin{cases} \lambda e^{-\lambda x}, & \mbox{if } x\ge 0, \\ 0, & \mbox{ otherwise. }\end{cases}\)

\[ \begin{split} f_Z(z) &= f_{X_1+(-X_2)}(z) \\ &= \int_{-\infty}^{\infty} f_{-X_2}(z-x_1) f_{X_1}(x_1) dx_1 \\ &= \int_{-\infty}^{\infty} f_{X_2}(x_1-z) f_{X_1}(x_1) dx_1 \\ &= \int_{-\infty}^{\infty} \lambda e^{-\lambda(x_1-z)} \lambda e^{-\lambda x_1} dx_1 \\ &= \int_{-\infty}^{\infty} \lambda^2 e^{-\lambda x_1 + \lambda z} e^{-\lambda x_1} dx_1 \\ &= \int_{-\infty}^{\infty} \lambda^2 e^{\lambda z - \lambda x_1 - \lambda x_1} dx_1 \\ &= \int_{-\infty}^{\infty} \lambda^2 e^{\lambda(z-2x_1)} dx_1 \end{split} \]

Consider \(z=x_1-x_2\), then \(x_2=x_1-z\).

If \(z \ge 0\), then \(x_2=(x_1-z) \ge 0\), and \(x_1 \ge z\), and, using WolframAlpha, \(f_Z(z) = \int_{z}^{\infty} \lambda^2 e^{\lambda(z-2x_1)} dx_1 = \frac{1}{2} \lambda e^{-\lambda z}\).

If \(z < 0\), then \(x_2=(x_1-z) \ge 0\), and \(x_1 \ge 0\), and \(f_Z(z) = \int_{0}^{\infty} \lambda^2 e^{\lambda(z-2x_1)} dx_1 =\frac{1}{2} \lambda e^{\lambda z}\).

Combining two sides we get \(f_Z(z) = \begin{cases} (1/2)e^{-\lambda z}, & \mbox{if } z \ge 0, \\ (1/2)e^{\lambda z}, & \mbox{if }z <0. \end{cases}\)

so , \(f_Z(z) = (1/2)e^{-\lambda|z|}\)

  1. 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.
  1. \(P(|X - 10| \ge 2)\)

solution:

k <- (2/sqrt(100/3))
round(1/k^2,2) 
## [1] 8.33
  1. \(P(|X - 10| \ge 5)\)

solution:

k <- (5/sqrt(100/3))
round(1/k^2,2)
## [1] 1.33
  1. \(P(|X - 10| \ge 9)\)

solution:

k <- (9/sqrt(100/3))
round(1/k^2,2)
## [1] 0.41
  1. \(P(|X - 10| \ge 20)\)

solution:

k <- (20/sqrt(100/3))
round(1/k^2,2)
## [1] 0.08