7.2.11
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?
For this problem, we are interested in finding the expected value of hte distribution function that describes the minimum of N=100 exponential random variables. There is a great video on youtube by Marcus Emsermann that walks through the solution using volcanic eruptions as an example.
\[E(min) = \frac{\mu }{n}\]
num_lights <- 100
mu <- 1000 #exponential lifetime
E_min = mu / num_lights
paste( 'The Expected time for the first bulb to burn out is:', E_min )## [1] "The Expected time for the first bulb to burn out is: 10"7.2.14
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 \(f_Z (z) = \frac{1}{2}\lambda e^ {−\lambda |z|}\).
From example 7.4 we have:
\[f_X(x)= f_Y(x) = \left\{ \begin{array}{c} \lambda e^{-\lambda x}, & \mbox{if } x \geq 0,
\\ 0, & \mbox{otherwise};\end{array} \right. \]
we can adapt the convolutional formula for the sum of two independent exponential random variables to write \[f_Z(z) = \int _{- \infty}^{+\infty} f_X(z-y)f_{-Y}(z-x)dx\] now we use the fact that \(f_{-Y}(z-x) = f_Y(x-z)\) to rewrite the equation as \[f_Z(z) = \int _{- \infty}^{+\infty} f_X(z-y)f_{Y}(x-z)dx\] Now evaluate the case where \(z\lt 0\) \[= \int_{0}^{\infty } \lambda e^{-\lambda x} \lambda e^{-\lambda (x-z)} dx\] \[ = \lambda e^{\lambda z} \int_{0}^{\infty} \lambda e^{-2\lambda x} dx\] \[ = \lambda e^{\lambda z} \left( -\frac{1}{2} e^{-2\lambda x} \vert_0 ^{\infty}\right) = \frac{\lambda}{2}e^{\lambda z}\]
Now to evaluate at \(z \ge 0\) there is a shortcut because \(Z\) and \(-Z\) are symmetric, we can assume that \(f_Z(z) = f_Z(-z)\) \[\therefore f_Z(z) = \left\{ \begin{array}{c} \frac{\lambda}{2} e^{\lambda x}, & z \lt 0, \\ \frac{\lambda}{2} e^{-\lambda x}, & z \geq 0;\end{array} \right. = \frac{\lambda}{2}e^{-\lambda |z|}\]
This is explained further & with better detail in the MIT Open courseware video, The Difference of Two Independent Exponential Random Variables
8.2.1
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.
The Chebyshev Inequality states \(P(|X-\mu| \ge \in) \le \frac{\sigma^2}{\in^2}\)
(a)
\[P (|X − 10| ≥ 2)\]
posnum <- 2
P_Xmin10_2 <- sigmasigma / posnum^2
if (P_Xmin10_2 >= 1){
paste( 'The upper bound for the probability is: 1' )
} else {
paste( 'The upper bound for the probability is:', round( P_Xmin10_2, 4 ) )
}## [1] "The upper bound for the probability is: 1"(b)
\[P (|X − 10| ≥ 5)\]
posnum <- 5
P_Xmin10_5 <- sigmasigma / posnum^2
if (P_Xmin10_5 >= 1){
paste( 'The upper bound for the probability is: 1' )
} else {
paste( 'The upper bound for the probability is:', round( P_Xmin10_5, 4 ) )
}## [1] "The upper bound for the probability is: 1"(c)
\[P (|X − 10| ≥ 9)\]
posnum <- 9
P_Xmin10_9 <- sigmasigma / posnum^2
if (P_Xmin10_9 >= 1){
paste( 'The upper bound for the probability is: 1' )
} else {
paste( 'The upper bound for the probability is:', round( P_Xmin10_9, 4 ) )
}## [1] "The upper bound for the probability is: 0.4115"(d)
\[P (|X − 10| ≥ 20)\]
posnum <- 20
P_Xmin10_20 <- sigmasigma / posnum^2
if (P_Xmin10_20 >= 1){
paste( 'The upper bound for the probability is: 1' )
} else {
paste( 'The upper bound for the probability is:', round( P_Xmin10_20, 4 ) )
}## [1] "The upper bound for the probability is: 0.0833"