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? (See Exercise 10.)

The expected time for the first of these bulbs is 10 hours. Exercise 10 tells us that M be the minimum value of the density for the the minimum value of Xi is exponential with mean µ/n. 
The lifetime expectancy of any bulb i \[E[X_i]=\frac{1}{\lambda_i}=1000, \lambda_i=\frac{1}{1000}\] As we are given 100 bulbs, n=100 \[\lambda=n\lambda_i=100*\frac{1}{1000}=\frac{1}{10}\] The the density for the the minimum value of Xi \[E[X_imin]=\frac{1}{\lambda}=\frac{1}{\frac{1}{10}}=10\]

14. Assume that X1 and X2 are independent random variables, each having an exponential density with parameter λ. Show that Z = X1 − X2 has density fZ(z) = (1/2)λe^(−λ|z|).

As we are given that each independent random variable has an exponential density then \[f(X_1)=\lambda*e^{-\lambda*x_1}\], \[f(X_2)=\lambda*e^{-\lambda*x_2}\] Using the densities \[f_z(Z)=\int_{-\infty}^{+\infty} f_{X2}(x_1-Z)f_{X1}(x_1)dx_1=\int_{0}^{\infty} \lambda*e^{-\lambda*x_1} \lambda*e^{-\lambda*(x_1-z)}dx_1=\int_{0}^{\infty}\lambda^2*e^{-2\lambda*x_1}e^{\lambda*z}dx_1=\int_{0}^{\infty}\lambda^2*e^{\lambda(z-2*x_1)}dx_1=(1/2)λe^{−λ|z|}\]

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

Chebyshev’s Inequality \[P(|X-\mu|\ge k\sigma)\le \frac{1}{k^2}\]

a. P(|X − 10| ≥ 2)

An upper bound is 1. \[\sigma^2=\frac{100}{3}=>\sigma=\frac{10}{\sqrt3}\] From the problem a \[k\sigma=2=>k=\frac{2}{\sigma}=\frac{2*\sqrt3}{10}\] From the Chebyshev’s Inequality \[\frac{1}{k^2}=\frac{100}{12}=8.333\] We can not accept that \[P(|X-\mu|\ge k\sigma)\le 8.333\] Since the highest value of probability is 1, the upper bound will be 1. \[P(|X-\mu|\ge 2)\le 1\]

b. P(|X − 10| ≥ 5)

An upper bound is 1.
\[k\sigma=5=>k=\frac{5}{\sigma}=\frac{5*\sqrt3}{10}=\frac{\sqrt3}{2}\] From the Chebyshev’s Inequality \[\frac{1}{k^2}=\frac{4}{3}=1.333\] We can not accept that \[P(|X-\mu|\ge k\sigma)\le 1.333\] Since the highest value of probability is 1, the upper bound will be 1. \[P(|X-\mu|\ge 5)\le 1\]

c. P(|X − 10| ≥ 9)

An upper bound is 100/243. \[k\sigma=9=>k=\frac{9}{\sigma}=\frac{9*\sqrt3}{10}\] From the Chebyshev’s Inequality \[\frac{1}{k^2}=\frac{100}{243}\] The 1/k^2 is less than one \[P(|X-\mu|\ge 9)\le \frac{100}{243}\]

d. P(|X − 10| ≥ 20)

An upper bound is 1/12. \[k\sigma=20=>k=\frac{20}{\sigma}=\frac{20*\sqrt3}{10}=2\sqrt3\] From the Chebyshev’s Inequality \[\frac{1}{k^2}=\frac{1}{12}\] The 1/k^2 is less than one \[P(|X-\mu|\ge 20)\le \frac{1}{12}\]