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?

\[E[X_i] = \frac{1}{\lambda_i} = 1000\]

Expected lifetime of a bulb is 1000 hours.

\[\lambda_i = \frac{1}{1000}\]

\(X_i\) is exponential so

\[min\{X_1,X_2,...,X_{100}\} \sim exponential(\sum\limits_{i=1}^{100} \lambda_i)\]

\[E[min X_i] = \frac{1}{\frac{1}{10}} = 10\]

Expected time for the first of these bulbs to burn out: 10 hours

14. Assume that X1 and X2 are independent random variables, each having an exponential density with parameter \(\lambda\). Show that \(Z = X_1 − X_2\) has density

\[f_Z(z)=(1/2)\lambda e^{−\lambda|z|}\]

Probability Density Function

\[f(x_1) = \lambda e^{-\lambda x_1}\]

\[f(x_2) = \lambda e^{-\lambda x_2}\]

Joint density

\[\lambda^2 e^{-\lambda(x_1 + x_2)}\]

Substitute to get joint density \(Z\) and \(X_2\) \[\lambda^2 e^{-\lambda(z + 2x_2)}\]

When z is positive:

\[\int_{0}^{\infty} \lambda^2 e^{-\lambda(z + 2x_2)} dx = \frac{\lambda}{2} e^{-\lambda z}\]

When z is negative:

\[\int_{-z}^{\infty} \lambda^2 e^{-\lambda(z + 2x_2)} dx = \frac{\lambda}{2} e^{\lambda z}\]

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

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 <- function(e, var){
    pX = var / e^2
    return(pX)
}
  1. \(P(|X−10| \geq 2)\)
var <- 100/3
e <- 2
p <- chebyshev(e, var)
p
## [1] 8.333333
  1. \(P(|X−10| \geq 5)\)
e <- 5
p <- chebyshev(e, var)
p
## [1] 1.333333
  1. \(P(|X−10| \geq 9)\)
e <- 9
p <- chebyshev(e, var)
p
## [1] 0.4115226
  1. \(P(|X − 10| \geq 20)\)
e <- 20
p <- chebyshev(e, var)
p
## [1] 0.08333333