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?

The density of minimum value among n independent random variables with an exponential density has mean \(\mu\)/n.

mu <- 1000
n <- 100

exp_time <- mu/n 
exp_time
## [1] 10

14. 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) \(\lambda\) \(e^{-\lambda |z|}\)

\(f_{z}\)(z) = (1/2) \(\lambda\) \(e^{-\lambda |z|}\) can be written as

\[\begin{equation} f_{z}(z)=\begin{cases} (1/2) \lambda e^{-\lambda z}, & \text{if $z \geq 0$}.\\ (1/2) \lambda e^{\lambda z}, & \text{otherwise}. \end{cases} \end{equation}\]

exponential density function is defined as

\[\begin{equation} f(x)=\begin{cases} (1/2) \lambda e^{-\lambda x}, & \text{if $x \geq 0$}.\\ 0, & \text{otherwise}. \end{cases} \end{equation}\]

In this case Z = X1-X2

\(f_{z}\)(z)= \(f_{x1+(-x2)}\)(z) = \(\int_{-\infty}^\infty\) \(f_{x1}(x1)\) \(f_{-x2}(z-x1)\) dx1 = \(\int_{-\infty}^\infty\) \(\lambda e^{-\lambda x1}\) \(\lambda e^{-\lambda (x1-z)}\) dx1 = \(\int_{-\infty}^\infty\) \(\lambda^{2}\) \(e^{\lambda (z-2x1)}\) dx1

for \(z \geq 0\)

\(f_{z}\)(z)= = \(\int_{z}^\infty\) \(\lambda^{2}\) \(e^{\lambda (z-2x1)}\) dx1 = (1/2) \(\lambda e^{-\lambda z}\)

for z < 0

\(f_{z}\)(z)= = \(\int_{0}^\infty\) \(\lambda^{2}\) \(e^{\lambda (z-2x1)}\) dx1 = (1/2) \(\lambda e^{\lambda z}\)

After merging these two we will have

\[\begin{equation} f_{z}(z)=\begin{cases} (1/2) \lambda e^{-\lambda z}, & \text{if $z \geq 0$}.\\ (1/2) \lambda e^{\lambda z}, & \text{otherwise}. \end{cases} \end{equation}\]

Reference https://www.youtube.com/watch?v=f8Nli1AfygM

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 Inequality

P(|x-\(\mu\)|\(\geq e\)) \(\leq\) \(\frac {\sigma ^2}{e^2}\)

  1. P(|X-10| \(\geqslant\) 2)
var <- 100/3
e <- 2
round(var/e^2, 4)
## [1] 8.3333

In this case upper bound would be 1 as probability can’t be greater than 1.

  1. P(|X-10| \(\geqslant\) 5)
e <- 5
round(var/e^2, 4)
## [1] 1.3333

In this case upper bound would be 1 as probability can’t be greater than 1.

  1. P(|X-10| \(\geqslant\) 9)
e <- 9
round(var/e^2, 4)
## [1] 0.4115
  1. P(|X-10| \(\geqslant\) 20)
e <- 20
round(var/e^2, 4)
## [1] 0.0833