Chapter 7 Page 313 - Exercise 11

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?

Answer:

We have \(E[X_{i}]= 1/\lambda_{i}\). Therefore \(\lambda_{i}=1/1000\) .

\(min X_{i} ∼ Exp(\lambda)\) where \(\lambda\) = \(\sum_{i=1}^{100} \lambda_{i} = 100 * (1/1000) = 1/10\) (since all lambdas are the same and we have 100 of them)

Finally: \(E[min X_{i}] = 1/\lambda\) where \(\lambda=\sum_{i=1}^{100} \lambda_{i}\)

               = 1/(1/10) 

               = 10

Therefore the expected time for the first bulb to burn out is 10 hours.

Chapter 7 Page 313 - Exercise 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 \(f_Z(z) = (1/2){\lambda}e^{-\lambda|z|}\).

Answer:

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

The densities of the three random variables \(X_1\), \(X_2\) and \(Z\) can be represented by \(f_{X_1}\), \(f_{X_2}\) and \(f_{Z}\) respectively.

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

Applying the convolution of \(X_1\) and \(X_2\), we will have:

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

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

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

Merging both equations , we will have: \(f_Z(z) = \begin{cases} (1/2){\lambda}e^{-\lambda z}, & \mbox{if } z \ge 0, \\ (1/2){\lambda}e^{\lambda z}, & \mbox{if }z <0. \end{cases}\)

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

Exercise 1 - Chapter 8.2 on page 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:

  1. \(P(|X - 10| \ge 2)\)
  2. \(P(|X - 10| \ge 5)\)
  3. \(P(|X - 10| \ge 9)\)
  4. \(P(|X - 10| \ge 20)\)

Answer:

Chebyshev Inequality: \(P(|X-\mu|\ge\epsilon) \le \frac{\sigma^2}{\epsilon^2}\) or, per example 8.4, \(P(|X-\mu|\ge k\sigma) \le \frac{1}{k^2}\).

For our problem, we have \(\mu=10\) and \(\sigma = \sqrt{\frac{100}{3}} = \frac{10}{\sqrt{3}}\).

If \(\epsilon = k\sigma\), then \(k=\frac{\epsilon}{\sigma} = \frac{\epsilon\sqrt{3}}{10}\).

Let \(u\) be upper bound in Chebyshev’s Inequality, then \(u = \frac{1}{k^2} = \frac{1}{(\epsilon\sqrt{3}/10)^2} = \frac{100}{3\epsilon^2}\).

  1. \(\epsilon=2\), the upper bound is \(u= \frac{100}{3\times 2^2} = \frac{25}{3} \approx 8.3333\). Since probability cannot be greater than \(1\), the upper bound is \(1\).
  2. \(\epsilon=5\), the upper bound is \(u= \frac{100}{3\times 5^2} = \frac{4}{3} \approx 1.3333\). Since probability cannot be greater than \(1\), the upper bound is \(1\).
  3. \(\epsilon=9\), the upper bound is \(u= \frac{100}{3\times 9^2} = \frac{100}{243} \approx 0.4115\).
  4. \(\epsilon=20\), the upper bound is \(u= \frac{100}{3\times 20^2} = \frac{1}{12} \approx 0.0833\).