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.)

Going through exercise 10, we get that in the exponential case, we have a minimum value as E(M) = μ/n, where μ is the mean i.e., 1000, and n is the number of bulbs i.e., 100. Therefore, E(M) = 1000/100 = 10

14 Assume that X1 and X2 are independent random variables, each having an exponential density with parameter . Show that Z = X1 − X2 has density fZ(z) = (1/2) e− |z| .

\[\begin{align} z &= {x_1} - {x_2} \\ {x_2} &= {x_1} - z \\ f_Z(z) &= \int_{-\infty}^{\infty} f_{x_1}(x_1)\; f_{x_2}(x_1-z)\; dx_1 \\ Going\;by\;the\;exponential\;distribution \\ f_X(x) = \lambda e^{-\lambda x} \end{align}\]

Assuming z>0,

\[\begin{align} f_Z(z) &= \int_{0}^{\infty} \lambda e^{-\lambda x_1} \lambda e^{-\lambda(x_1-z)} dx_1 \\ &= \int_{0}^{\infty} \lambda^2 e^{-2\lambda x_1+\lambda z}dx_1 \\ &= \int_{0}^{\infty} \lambda^2 e^{\lambda(-2 x_1+ z)}dx_1 \\ &= \int_{0}^{\infty} \lambda^2 e^{\lambda(z-2 x_1)}dx_1 \\ &= (1/2)\lambda e^{-\lambda(z)} \\ \end{align}\]

Let X be a continuous random variable with mean μ = 10 and variance 2 = 100/3. Using Chebyshev’s Inequality, find an upper bound for the following probabilities.

According to Chebyshev’s inequality,

\[\begin{align} P(|X-μ|≥E)≤ \frac{σ^2}{E^2} \end{align}\]

  1. P(|X − 10| >= 2)
var <- 100/3
E <- 2
var/E^2
## [1] 8.333333
  1. P(|X − 10| >= 5)
var <- 100/3
E <- 5
var/(E^2)
## [1] 1.333333
  1. P(|X − 10| >= 9)
var <- 100/3
E <- 9
var/(E^2)
## [1] 0.4115226
  1. P(|X − 10| >= 20)
var <- 100/3
E <- 20
var/E^2
## [1] 0.08333333