Chapter 7: Page 303

Problem 11. A company buys 100 lightbulbs, each of which has an exponential lifetime of 1000 hours. What is the expected time for the ???rst of these bulbs to burn out? (See Exercise 10.)

(Ex 10): X1, X2, …, Xn be n independent random variables each of which has an exponential density with mean µ. Let M be the minimum value of the Xj (µ/n)
Xi \(\sim\) Exp(\(\lambda\)i), then min Xi \(\sim\) Exp(\(\Sigma\) \(\lambda\)i)
Exp|X| = 1000; \(\lambda\)i=1/1000; \(\Sigma\) \(\lambda\)i=1/10 (n*\(\lambda\))
Exponential Distribution \(P(X \leq x) = 1 - e^{-\lambda x}\)
First bulb to burn out: \(P(min Xi) = 1 - e ^{-\lambda x} = 1 - e^{-1/10} = 10hrs\)

Problem 14. Assume that X1 and X2 are independent random variables, each having an exponential density with parameter \(\lambda\). Show that Z = X1-X2 has density

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

\[ fX_{1}(x) = fX_{2}(x) = \{ \lambda e^{-\lambda x} x \geq 0 \] \[ fz(z) = \int_{-\infty}^{\infty} f(x_{1})(x)f(x_{2})(x-z)dx \] \[ \int_{0}^{\infty}\lambda e^{-\lambda x}\lambda e^{-\lambda (x-z)}dx \] \[ \int_{0}^{\infty}\lambda^{2} e^{-2\lambda x + \lambda z}dx \] \[ \lambda e^{\lambda z}\int_{0}^{\infty} e^{-2\lambda x}dx \] \[ (1/2)\lambda e ^ {- \lambda |z|}\]

Chapter 8: Page 303

Problem 1. Let X be a continuous random variable with mean µ = 10 and variance \(\sigma^{2}\) = 100/3. Using Chebyshev’s Inequality, find an upper bound for the following probabilities.

Chebyshev’s Inequality:

\[ P(|X-\mu|\geq k\sigma) \leq 1/k^{2} \]

(a) P(|X-10|\(\geq\) 2)
k*sd = 2; k=2/sqrt((100/3)) = 0.3464102 Note: as K<1, the inequality is trivial as all the probabilities should be <=1

P(|X-10|\(\geq\) 2) = 1/(0.3464)^{2} = 8.3338223

(b) P(|X-10|\(\geq\) 5)
k*SD = 5; k=5/sqrt((100/3)) = 0.8660254 Note: as K<1, the inequality is trivial as all the probabilities should be <=1

P(|X-10|\(\geq\) 5) = 1/(0.8660)^{2} = 1.3333333

(c) P(|X-10|\(\geq\) 9)
k*SD = 9; k=9/sqrt((100/3)) = 1.5588457

P(|X-10|\(\geq\) 9) = 1/(1.5588)^{2} = 0.4115226

No more than 41% of the distribution’s values can be more than 1.55 standard deviations away from the mean
(d) P(|X-10|\(\geq\) 20)
k*SD = 20; k=20/sqrt((100/3)) = 3.4641016

P(|X-10|\(\geq\) 20) = 1/(3.4641)^{2} = 0.0833333

No more than 8% of the distribution’s values can be more than 3.46 standard deviations away from the mean