Chapter 7.2 Exercise 11 of Probability Text

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: Let \(X_{1}, X_{2},...,X_{100}\) be 100 lightbulbs with exponential lifetime \(\mu=1000\). The first lightbulb will fail at minimum value of \(X_{j}\) suppose that is M. The denisity of M is exponential with mean \(\mu/100\)

mu <- 1000
n <- 100
mu/n
## [1] 10

The expected time for the first bulb to fail out is 10 hours with out disregading other factors that cam also inflinece like electric grid and spikes.

Chapter 7.2 Exercise 14 of the Probability Text

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

Answer

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

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

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

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 \(f_Z(z) = \int_{0}^{\infty} \lambda^2 e^{\lambda(z-2x_1)} dx_1 =\frac{1}{2} \lambda e^{\lambda z}\).

through merging the two we will have \(f_Z(z) = \begin{cases} (1/2)e^{-\lambda z}, & \mbox{if } z \ge 0, \\ (1/2)e^{\lambda z}, & \mbox{if }z <0. \end{cases}\)

Chapter 8.2 Exercise 1 of the Probability Text

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, \(P(|X-\mu|\ge k\sigma) \le \frac{1}{k^2}\).

requiring a solution , \(\mu=10\) 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 limit is \(u= \frac{100}{3\times 2^2} = \frac{25}{3} \approx 8.3333\). The upper bound is \(1\) due to the fact that probability will not be greater than \(1\).
  2. \(\epsilon=5\), the upper bound is \(u= \frac{100}{3\times 5^2} = \frac{4}{3} \approx 1.3333\). Due to the fact that probability cannot be greater than \(1\), the upper bound is \(1\).
  3. \(\epsilon=9\), the value of the upper bound is \(u= \frac{100}{3\times 9^2} = \frac{100}{243} \approx 0.4115\).
  4. \(\epsilon=20\), the upper bound is \(u= \frac{100}{3\times 20^2} = \frac{1}{12} \approx 0.0833\).