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.)
From,
\[ min\{X_1, X_2, .. X_n\} \sim exponential(\sum_{i=1}^{n} \lambda_i) \]
We get,
\[ \lambda_1 + \lambda_2 + ... + \lambda_{100}= 100 \lambda = \frac{100}{1000} = \frac{1}{10} \\ E[min(X_i)] = \frac{1}{\lambda} = \frac{1}{\frac{1}{10}} = 10 \]
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: \[f_Z(z) = (1/2) \lambda e^{-\lambda |z|}\]
We know, \(X, Y \sim exp(\lambda)\), \[ f_X(x) = f_Y(y) = \begin{cases} \lambda e^{-\lambda |\cdot|} &,\text{for x,y} \geq 0, \\ 0 &,\text{Otherwise} \end{cases} \]
Following the convolution formula for continuous random variables, we have,
\[ \begin{align} f_{Z}(z) &= \int_{-\infty}^{\infty} f_X(z+y) f_Y(y) \ dy \\ &= \int_{0}^{\infty} \lambda e^{-\lambda(z+y)} \lambda e^{-\lambda y} \ dy & \text{pull-out the constants} \\ &= \lambda^2 e^{-\lambda z} \int_{0}^{\infty} e^{-2 \lambda y} \ dy & \text{apply u-sub} \\ &= \frac{\lambda ^2}{2\lambda}e^{-\lambda z } \Leftrightarrow \frac{\lambda}{2} e^{-\lambda z},\ \text{for} \ z \geq 0 \end{align} \] This is the pdf of z for \(z \in \mathbb{R^+}\).
For \(z < 0\), we can obtain the density function by the fact that the density of z is symmetric around zero (i.e. \(f_Z(1) = f_Z(-1)\))
Thus, the difference of the two independent exponential random variable has the following piecewise density function; \[ f_Z(z) = \begin{cases} \frac{\lambda}{2} e^{-\lambda z},\ \text{for} \ z \geq 0 \\ \frac{\lambda}{2} e^{\lambda z},\ \text{for} \ z < 0 \end{cases} \]
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 \[ P(|X- \mu| \geq k \sigma) \leq \frac{1}{k^2},\ k \in \mathbb{R^+} \\ \sigma = \frac{10}{\sqrt{3}} \]
\[ \begin{align} k \sigma = 2 \Rightarrow k \frac{10}{\sqrt{3}} = 2 \Rightarrow k = \frac{\sqrt{3}}{5} \\ P(|X-10| \geq 2) \leq \frac{1}{k^2} = \frac{25}{3} \\ P(|X-10| \geq 2) \leq 1 \end{align} \]
\[ \begin{align} k \sigma = 5 \Rightarrow k \frac{10}{\sqrt{3}} = 5 \Rightarrow k = \frac{\sqrt{3}}{2} \\ P(|X-10| \geq 5) \leq \frac{1}{k^2} = \frac{4}{3} \\ P(|X-10| \geq 5) \leq 1 \end{align} \]
\[ \begin{align} k \sigma = 9 \Rightarrow k \frac{10}{\sqrt{3}} = 9 \Rightarrow k = 9 \frac{\sqrt{3}}{10} \\ P(|X-10| \geq 9) \leq \frac{1}{k^2} = \frac{100}{243} \\ P(|X-10| \geq 9) \leq 0.4115 \end{align} \]
\[ \begin{align} k \sigma = 20 \Rightarrow k \frac{10}{\sqrt{3}} = 20 \Rightarrow k = 2 \sqrt{3} \\ P(|X-10| \geq 20) \leq \frac{1}{k^2} = \frac{1}{12} \\ P(|X-10| \geq 20) \leq 0.0833 \end{align} \]