The duration of Each lightbulb represent an indepeent random variable Xi. By exercise 10, the minimum value has the mean \(\mu/n\). The expected time for the first of the bulbs to burn out is \(1000h/100=10h\).
Assume that X1 and X2 are independent random variables, each having an exponential density with parameter \(\lambda\). Show that Z = X1 − X2 has density \(f_Z(z) = (1/2)e^{−|z|}\) .Z
Sincs Z is the difference of two positive numbers, it can be positive or negative. If \(Z\le0\), then \(X1\le X2\). The probability is defined from \(-\infty\) to 0. \(f_Z(z)=\int_{-\infty}^{0}{f_{X_1}(z+x_2)f(x_2)dx_2}\\\). \(=\int_{-\infty}^{0}{\lambda e^{-\lambda(z+x_2)}\lambda e^{-\lambda x_2}dx_2}\\=\lambda^2 e^{-\lambda z }\int_{-\infty}^{0}{e^{-2\lambda x_2}dx_2}\\=\lambda^2 e^{-\lambda z }(\frac{-1}{2\lambda})\\=\frac{-1}{2}\lambda e^{-\lambda z }\).
If \(Z\ge0\), then \(X1\ge X2\). The probability is defined from 0 to \(\infty\). \(f_Z(z)=\int_{0}^{\infty}{f_{X_1}(z+x_2)f(x_2)dx_2}\\\). \(=\int_{0}^{\infty}{\lambda e^{-\lambda(z+x_2)}\lambda e^{-\lambda x_2}dx_2}\\=\lambda^2 e^{-\lambda z }\int_{0}^{\infty}{e^{-2\lambda x_2}dx_2}\\=\lambda^2 e^{-\lambda z }(\frac{1}{2\lambda})\\=\frac{1}{2}\lambda e^{-\lambda z }\\\).
putting together, \(f_Z(z)=\begin{cases} \frac{-1}{2}\lambda e^{-\lambda z} \quad z\le0 \\ (\frac{1}{2}\lambda e^{-\lambda z } \quad z\ge 0\end{cases}\)
\(f(z)=\frac{\lambda}{2}e^{-\lambda|z|}\)
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 (a) \(P(|X − 10| \ge 2).\) (b) \(P(|X − 10| \ge 5).\) (c) \(P(|X − 10| \ge 9).\) (d) \(P(|X − 10| \ge 20).\)
Theorem 8.1 (Chebyshev Inequality) Let X be a discrete random variable with expected value μ = E(X), and let \(\epsilon > 0\) be any positive real number. Then \(P(|X − μ| \ge \epsilon )\le \frac{\sigma^2}{\epsilon^2}\).
We have \(P(|X − 10| \ge \epsilon )\le \frac{100}{3\epsilon^2}\).
\(\epsilon=2\) \(P(|X − 10| \ge 2 )\le \frac{100}{12}\). Thus \(P(12\ge X \ge 8 )\le \frac{25}{3}\)
\(P(|X − 10| \ge 5).\) Thus \(P(15\ge X \ge 5 )\le \frac{4}{3}\)
\(P(|X − 10| \ge 9).\) Thus \(P(19\ge X \ge 1 )\le \frac{100}{243}\)
\(P(|X − 10| \ge 20).\) Thus \(P(30\ge X \ge -10 )\le \frac{1}{27}\)