Question 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.)

Given, μ = 1000 hours n = 100

expected_time = μ/n = 1000/100 = 10 hours

R Code:

mu <- 1000   
n <- 100   

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

Question 14:

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

Let’s use convolution formula for the probability density function to density function of \[Z = X_1 - X_2\]

\[f_Z(z) = \int_{-\infty}^{\infty} f_{X_1}(x) f_{X_2}(z-x)dx\] Density functions:

\[f_{X_1}(x) = \begin{cases} \lambda e^{-\lambda x} & x \geq 0 \ 0 & x < 0 \end{cases}\]

\[f_{X_2}(x) = \begin{cases} \lambda e^{-\lambda x} & x \geq 0 \ 0 & x < 0 \end{cases}\]

Substituting these expressions into formula:

\[f_Z(z) = \int_{-\infty}^{\infty} f_{X_1}(x) f_{X_2}(z-x)dx = \int_{0}^{\infty} \lambda e^{-\lambda x} \lambda e^{-\lambda(z-x)}dx\]

\[= \lambda^2 \int_{0}^{\infty} e^{-\lambda z}dx = \lambda^2 e^{-\lambda z} \int_{0}^{\infty} dx = \lambda^2 ze^{-\lambda z}\] We have shown that the density function of \[Z = X_1 - X_2\] \[f_Z(z) = \frac{1}{2}\lambda e^{-\lambda |z|}\]

Question 1:

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

Chebyshev’s Inequality formula: \[P(|X - µ| ≥ kσ) ≤ 1/k^2\] (a) P(|X − 10| ≥ 2).

\[\sigma^2 = \frac{100}{3}\]

\[k\sigma=2\]

\[k=\frac{2}{\sqrt{\frac{100}{3}}}\]

\[\frac{1}{k^2}=\frac{1}{\left(\frac{2}{\sqrt{\frac{100}{3}}}\right)^2}\] (b) P(|X - 10| ≥ 5)

\[k\sigma=5\] \[k=\frac{5}{\sqrt{\frac{100}{3}}}\]

\[\frac{1}{k^2}=\frac{1}{\left(\frac{5}{\sqrt{\frac{100}{3}}}\right)^2}\] (c) P(|X - 10| ≥ 9)

\[k\sigma=9 \] \[k=\frac{9}{\sqrt{\frac{100}{3}}}\]

\[\frac{1}{k^2}=\frac{1}{\left(\frac{9}{\sqrt{\frac{100}{3}}}\right)^2}\] (d) P(|X - 10| ≥ 20)

\[k\sigma=20 \] \[k=\frac{20}{\sqrt{\frac{100}{3}}}\]

\[\frac{1}{k^2}=\frac{1}{\left(\frac{20}{\sqrt{\frac{100}{3}}}\right)^2}\]

R Code:

## mean and variance
mean <- 10
sigma2 <- 100/3

## Values of k for each probability
k1 <- 2/sqrt(sigma2)
k2 <- 5/sqrt(sigma2)
k3 <- 9/sqrt(sigma2)
k4 <- 20/sqrt(sigma2)

## Upper bounds for each probability 
(p1 <- 1/k1^2)
## [1] 8.333333
(p2 <- 1/k2^2)
## [1] 1.333333
(p3 <- 1/k3^2)
## [1] 0.4115226
(p4 <- 1/k4^2)
## [1] 0.08333333