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? (Page 303).
\(mu = 1000\) (the expected lifetime of a bulb is 1000 hours) \(n = 100\) so the expected time for the first of the bulbs to burn out is \(E(M)=\mu/n=1000/100=10\).
Therefore, the expected time for the first of these bulbs to burn out is 10 hours.
14 Assume that X1 and X2 are independent random variables, each having an exponential density with parameter ??. Show that Z = X1 ??? X2 has density fZ(z) = (1/2)??e?????|z| (Page 303).
Because \(f_Z(z) = (1/2)e^{-\lambda|z|}\), so it can be re-written as \(f_Z(z) = \begin{cases} (1/2)e^{-\lambda z}, & \mbox{if } z \ge 0, \\ (1/2)e^{\lambda z}, & \mbox{if }z <0. \end{cases}\)
Because \(X_1\) and \(X_2\) have exponential density, their PDF is
\[ \begin{split} f_Z(z) &= f_{X_1+(-X_2)}(z) \\ &= \int_{-\infty}^{\infty} f_{-X_2}(z-x_1) f_{X_1}(x_1) dx_1 \\ &= \int_{-\infty}^{\infty} f_{X_2}(x_1-z) f_{X_1}(x_1) dx_1 \\ &= \int_{-\infty}^{\infty} \lambda e^{-\lambda(x_1-z)} \lambda e^{-\lambda x_1} dx_1 \\ &= \int_{-\infty}^{\infty} \lambda^2 e^{-\lambda x_1 + \lambda z} e^{-\lambda x_1} dx_1 \\ &= \int_{-\infty}^{\infty} \lambda^2 e^{\lambda z - \lambda x_1 - \lambda x_1} dx_1 \\ &= \int_{-\infty}^{\infty} \lambda^2 e^{\lambda(z-2x_1)} dx_1 \end{split} \]
Because \(z=x_1-x_2\), we write the formula to \(x_2=x_1-z\).
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 \(f_Z(z) = \begin{cases} (1/2)e^{-\lambda z}, & \mbox{if } z \ge 0, \\ (1/2)e^{\lambda z}, & \mbox{if }z <0. \end{cases}\)
1 Let X be a continuous random variable with mean µ = 10 and variance ?? 2 = 100/3. Using Chebyshev’s Inequality, find an upper bound for the following probabilities. (a) P(|X ??? 10| ??? 2). (b) P(|X ??? 10| ??? 5). (c) P(|X ??? 10| ??? 9). (d) P(|X ??? 10| ??? 20). (Page 320).
(Chebyshev Inequality) Let X be a continuous random variable
with density function \(f(x)\). Suppose X has a finite expected value \(\mu = E(X)\) and
finite variance \(\sigma^2 = V(X)\). Then for any positive number \(\epsilon > 0\) we have
\(P(|X - \mu| \geq \epsilon) \leq \sigma^2/\epsilon^2\)
\(\epsilon = 2\) and the upper bound is \(\sigma^2/\epsilon^2 = (100/3)/4 = 25/3\)
\(\epsilon = 5\) and the upper bound is \(\sigma^2/\epsilon^2 = (100/3)/25 = 4/3\)
For a) and b), the upper bound cannot be greater than 1, therefore the upper bounds are 1 for both a) and b).
\(\epsilon = 9\) and the upper bound is \(\sigma^2/\epsilon^2 = (100/3)/81 = 100/243\)
\(\epsilon = 20\) and the upper bound is \(\sigma^2/\epsilon^2 = (100/3)/400 = 1/12\)