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?

n  <- 1000
lbulbs <- 100

# The distribution of the minimum value of n independent expontially distributed variables with mean ?? would also be exponential with mean ??(mean)/n(# of items)

b <- n/lbulbs 
b 
## [1] 10

Assume that X1 and X2 are independent random variables, each having an exponential density with parameter lambda. Show that Z = X1-X2 has density .

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

Ans :-

Both X1 and X2 are evaluated on the interval \[0\le x<\infty.\]

\[ 1) When\quad X2 \ge X1\]

\[\int _{ 0 }^{ \infty }{ { fX }^{ (x) }{ fY }^{ x-z }dx }\] \[\int _{ 0 }^{ \infty }{ { \lambda e }^{ -\lambda x }{ \lambda e }^{ -\lambda (x???z) } } dx\] \[{\lambda e}^{\lambda z}\int _{ 0 }^{ \infty }{ { \lambda e }^{ -2\lambda x } } dx\] \[{ \lambda e }^{ \lambda z }(\frac { -1 }{ 2 } { e }^{ -2\lambda x })\] \[fZ(z)=\frac { \lambda }{ 2 } { e }^{ \lambda z }\] \[2) When\quad X1 \ge X2\]

\[fZ(z)=\frac { \lambda }{ 2 } { e }^{ -\lambda z }\] \[fZ(z)=\frac { \lambda }{ 2 } { e }^{ -\lambda |z| }\] ## Let X be a continuous random variable with mean ?? = 10 and variance ??2 = 100/3. Using Chebyshev’s Inequality, find an upper bound for the following probabilities.

pX = var / e^2

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