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.)
Let \(X_i\) be the independent random variable for light bulb \(i\). Then \(E[X_i] = \frac{1}{\lambda_i} = 1000\), because the expected lifetime of a bulb is 1000 hours. This means \(\lambda_i = \frac{1}{1000}\).
Since \(X_i\) is given as exponential, we know that:
\(min\{X_1,X_2,...,X_{100}\} \sim exponential(\sum\limits_{i=1}^{100} \lambda_i)\)
And \(\sum\limits_{i=1}^{100} \lambda_i = 100 \times \frac{1}{1000} = \frac{1}{10}\)
Therefore, the expected value \(E[min X_i] = \frac{1}{\frac{1}{10}} = 10\)
So the expected time for the first of these bulbs to burn out is 10 hours.
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
\(\boldsymbol{f_Z(z) = \frac{1}{2} \lambda e^{\lambda|z|}}\)
I’ll start by finding the probability density function for \(X_1\) and \(X_2\) using the exponential distribution function:
\(f(x_1) = \lambda e^{-\lambda x_1}\)
\(f(x_2) = \lambda e^{-\lambda x_2}\)
Using convolutions, the joint density of \(X_1\) and \(X_2\) is \(\lambda e^{-\lambda x_1} \times \lambda e^{-\lambda x_2}\), or \(\lambda^2 e^{-\lambda(x_1 + x_2)}\).
Because \(Z = X_1 − X_2\), we know that \(x_1 = z + x_2\). Then, we can use substition to get the joint density of \(Z\) and \(X_2\), which is \(\lambda^2 e^{-\lambda((z + x_2) + x_2)}\) or \(\lambda^2 e^{-\lambda(z + 2x_2)}\).
We know that \(x_2 = x_1 - z\). Therefore, if z is negative, \(x_2\) is greater than \(-z\), and if z is positive, then \(x_2\) is positive. So:
When z is negative, \(\int_{-z}^{\infty} \lambda^2 e^{-\lambda(z + 2x_2)} dx = \frac{\lambda}{2} e^{\lambda z}\).
When z is positive, \(\int_{0}^{\infty} \lambda^2 e^{-\lambda(z + 2x_2)} dx = \frac{\lambda}{2} e^{-\lambda z}\).
Or, \(f_Z(z) = \frac{1}{2} \lambda e^{\lambda|z|}\)
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’s Inequality: \(P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2}\)
\(\sigma = \sqrt{\frac{100}{3}} = \frac{10}{\sqrt{3}}\)
(a) \(\boldsymbol{P(|X − 10| \geq 2)}\).
\(k\sigma = 2\), so \(k\frac{10}{\sqrt{3}} = 2\) or \(k = \frac{2\sqrt{3}}{10}\). Therefore, \(\frac{1}{k^2} = \frac{1}{(\frac{2\sqrt{3}}{10})^2}\) or \(\frac{1}{0.12} = 8.333\). But the probability can’t be more than 1, so the bound is 1.
(b) \(\boldsymbol{P(|X − 10| ≥ 5)}\).
\(k\sigma = 5\), so \(k\frac{10}{\sqrt{3}} = 5\) or \(k = \frac{\sqrt{3}}{2}\). Therefore, \(\frac{1}{k^2} = \frac{1}{(\frac{\sqrt{3}}{2})^2}\) or \(\frac{1}{\frac{3}{4}} = 1.333\). But the probability can’t be more than 1, so the bound is 1.
(c) \(\boldsymbol{P(|X − 10| ≥ 9)}\).
\(k\sigma = 9\), so \(k\frac{10}{\sqrt{3}} = 9\) or \(k = \frac{9\sqrt{3}}{10}\). Therefore, \(\frac{1}{k^2} = \frac{1}{(\frac{9\sqrt{3}}{10})^2}\) or \(\frac{1}{2.43} = 0.412\).
(d) \(\boldsymbol{P(|X − 10| ≥ 20)}\).
\(k\sigma = 20\), so \(k\frac{10}{\sqrt{3}} = 20\) or \(k = 2\sqrt{3}\). Therefore, \(\frac{1}{k^2} = \frac{1}{(2\sqrt{3})^2}\) or \(\frac{1}{12} = 0.083\).