Chapter 7.2 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?
Solution: Assume X1,X2,… X100 be 100 lightbulbs has exponential lifetime μ=1000. The expected time for the first light bulb to burn out is at minimum value of Xj then denisity of minimum value is exponential mean μ/100
Exp_x <- 1000
n <- 100
Exp_x/n## [1] 10Answer: Tthe expected time for the first of these bulbs to burn out is 10 hours.
Chapter 7.2 Exercise 14
Assume that X1 and X2 are independent random variables, each having an exponential density with parameter Lamda. Show that Z = X1 − X2 has density \(f_Z(z) = (1/2)e^{-\lambda|z|}\).
\(f_Z(z) = (1/2)e^{-\lambda|z|}\) is written as \(f_Z(z) = \begin{cases} (1/2)e^{-\lambda z}, & \mbox{if } z \ge 0, \\ (1/2)e^{\lambda z}, & \mbox{if }z <0. \end{cases}\)
As \(X_1\) and \(X_2\) has exponential density
\[ \begin{split} f_Z(z) &= f_{X_1+(-X_2)}(z) \\ &= \int_{-\infty}^{\infty} f_{-X_2}(z-x_1) f_{X_1}(x_1) dx_1 \\ &= \int_{-\infty}^{\infty} f_{X_2}(x_1-z) f_{X_1}(x_1) dx_1 \\ &= \int_{-\infty}^{\infty} \lambda e^{-\lambda(x_1-z)} \lambda e^{-\lambda x_1} dx_1 \\ &= \int_{-\infty}^{\infty} \lambda^2 e^{-\lambda x_1 + \lambda z} e^{-\lambda x_1} dx_1 \\ &= \int_{-\infty}^{\infty} \lambda^2 e^{\lambda z - \lambda x_1 - \lambda x_1} dx_1 \\ &= \int_{-\infty}^{\infty} \lambda^2 e^{\lambda(z-2x_1)} dx_1 \end{split} \]
Let us assume \(z=x_1-x_2\), then \(x_2=x_1-z\).
Incase \(z \ge 0\), then \(x_2=(x_1-z) \ge 0\), and \(x_1 \ge z\), and \(f_Z(z) = \int_{z}^{\infty} \lambda^2 e^{\lambda(z-2x_1)} dx_1 = \frac{1}{2} \lambda e^{-\lambda z}\).
Incase \(z < 0\), then \(x_2=(x_1-z) \ge 0\), and \(x_1 \ge 0\), and \(f_Z(z) = \int_{0}^{\infty} \lambda^2 e^{\lambda(z-2x_1)} dx_1 =\frac{1}{2} \lambda e^{\lambda z}\).
after merging the two we can see \(f_Z(z) = \begin{cases} (1/2)e^{-\lambda z}, & \mbox{if } z \ge 0, \\ (1/2)e^{\lambda z}, & \mbox{if }z <0. \end{cases}\)
Chapter 8.2 Exercise 1
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.
Chebyshev Inequality is \(P(|X-\mu|\ge\epsilon) \le \frac{\sigma^2}{\epsilon^2}\) as per example 8.4, \(P(|X-\mu|\ge k\sigma) \le \frac{1}{k^2}\).
solution is \(\mu=10\) and \(\sigma = \sqrt{\frac{100}{3}} = \frac{10}{\sqrt{3}}\).
In case \(\epsilon = k\sigma\), then \(k=\frac{\epsilon}{\sigma} = \frac{\epsilon\sqrt{3}}{10}\).
Lets assume \(u\) is upper bound in Chebyshev’s Inequality then \(u = \frac{1}{k^2} = \frac{1}{(\epsilon\sqrt{3}/10)^2} = \frac{100}{3\epsilon^2}\).