Chapter 7, Question 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?

Let \(X_i\) be the random variable dictating if the \(i\)th lightbulb has burned out. \(P(X_i)=\lambda e^{-\lambda x_i}\). The odds of two lightbulbs not burning out is \(P(X_i \cap X_j)=P(X_i)\times P(X_j)\). Because all lightbulbs have the same probability, this is equal to \(P(X_1)^2\). The probability of no lightbulb burning out is thus

\[ \bigcap_{i=1}^{100}P(X_i)=P(X_1)^{100}=\bigg( \lambda e^{-\lambda x_1}\bigg)^{100}=\lambda^{100}e^{-100\lambda x_1} \]

So

\[ E(X_1)=\frac{1}{\lambda}=1000\\ E(\bigcap_{i=1}^{100}X_i)=\frac{1}{100\lambda}=\frac{1000}{100}=10 \]

So the expected time for a bulb to burn out is 10 years.

Question 14.

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 \[ f_z(z)=\frac{1}{2}\lambda e^{-\lambda|z|} \]

Because the exponential distribution requires a negative exponent on \(e\), we can break this down into two cases, \(X<Y\) and \(X>Y\). For \(X>Y\), we have

\[ f_z(z)=\int_0^\infty f_x(z+y)f_y(y)dy\\ = \int_0^\infty e^{-\lambda(z+y)}e^{-\lambda y}dy\\ =e^{\lambda z}\int_0^\infty e^{-\lambda 2y}dy\\ =\frac{-\lambda^2}{2}e^{\lambda z}(0-1)=\frac{\lambda^2}{2}e^{\lambda z} \]

If \(X<Y\), we must work with a \(-z\) in the same equation set, in order to keep the exponent negative and the function a distribution. So, we have

\[ f_z(z)=\int_{-\infty}^0 f_x(-z+y)f_y(y)dy\\ = \int_{-\infty}^0 e^{-\lambda(-z+y)}e^{-\lambda y}dy\\ =e^{-\lambda z}\int_{-\infty}^0 e^{-\lambda 2y}dy\\ =\frac{\lambda^2}{2}e^{-\lambda z}(1-0)=\frac{\lambda^2}{2}e^{-\lambda z} \]

Combining the two, this gives us the original:

\[ f_z(z)=\frac{1}{2}\lambda e^{-\lambda|z|} \]

Problem 1.

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

  1. \(P(|X - 10| \geq 2).\)

\(P(|X-\mu|\geq \varepsilon)\leq \frac{\sigma^2}{\varepsilon}\)

\(\frac{\sigma^2}{\varepsilon}=\frac{100}{3\varepsilon}\)

\(\frac{50}{3}\)

  1. \(P(|X - 10| \geq 5).\)

\(\frac{20}{3}\)

  1. \(P(|X - 10| \geq 9).\)

\(\frac{100}{27}\)

  1. \(P(|X - 10| \geq 20).\)

\(\frac{5}{3}\)