7.2 - Problem 11: Sums of Continuous Random Variables

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?

lightbulbs_total <- 100
exp_lifetime <- 1000
cat(sprintf("%s = %f \n", c("Expected time (in hours) for the first of these bulbs to burn out: E(T)"), c(exp_lifetime/lightbulbs_total)))
## Expected time (in hours) for the first of these bulbs to burn out: E(T) = 10.000000

7.2 - Problem 14: Sums of Continuous Random Variables

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

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

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

The probability density function (pdf) of an exponential distribution is

\[ f_X(x) = f_Y(x) = \begin{cases}{\lambda e^{-\lambda x},} & \text{if } x\geq 0, \\ {0} & \text{otherwise;}\end{cases} \]

\(f_Z(z) = \int_{-\infty}^{+\infty} f_X(z + y) f_Y(y) \text{dy}\)

(from pdf) => \(f_Z(z) = \int_{0}^{z} \lambda e^{-\lambda (z + y)} \lambda e^{-\lambda y} dy\)

=> \(f_Z(z) = \int_{0}^{z} \lambda^2 e^{-\lambda z} dy\)

=> \(f_Z(z) = \frac{\lambda}{2} e^{-\lambda |z|}\)

Finally, \[ f_Z(z) = f_X(z) = f_Y(z) = \begin{cases}{(1/2)\lambda e^{-\lambda|z|},} & \text{if } z\geq 0, \\ {0} & \text{otherwise.}\end{cases} \]

8.2 - Problem 1: Continous Random Variables

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.

  1. P(|X - 10| >= 2).
  2. P(|X - 10| >= 5).
  3. P(|X - 10| >= 9).
  4. P(|X - 10| >= 20).

Given: \(\mu = 10\), \(\sigma ^2 = 100/3\)

sigma_square <- 100/3
sigma <- sqrt(sigma_square)
cat(sprintf("%s = %f \n", c(" Sigma"), c(sigma)))
##  Sigma = 5.773503

\[P(|X-\mu| \geq k\sigma) \leq 1/k^2 \] \[P(|X-10|) \geq k*5.773503 \leq 1/k^2 \] Let’s find and compare the found value to the proposed one originally.

  1. P(|X - 10|>=2).
k = 2/sigma
cat(sprintf("%s = %f \n", c(" 1/k^2"), c(1/k^2)))
##  1/k^2 = 8.333333

Upper bound: \(P(|X - 10|>=2) = 8.333333\)

  1. P(|X - 10| >= 5).
k = 5/sigma
cat(sprintf("%s = %f \n", c(" 1/k^2"), c(1/k^2)))
##  1/k^2 = 1.333333

Upper bound: \(P(|X - 10|>=5) = 1.333333\)

  1. P(|X - 10| >= 9).
k = 9/sigma
cat(sprintf("%s = %f \n", c(" 1/k^2"), c(1/k^2)))
##  1/k^2 = 0.411523

Upper bound: \(P(|X - 10|>=9) = 0.411523\)

  1. P(|X - 10| >= 20).
k = 20/sigma
cat(sprintf("%s = %f \n", c(" 1/k^2"), c(1/k^2)))
##  1/k^2 = 0.083333

Upper bound: \(P(|X - 10|>=20) = 0.083333\)