Introduction to Probability, Grinstead, C. Snell, J., 1997
Page 320
Question 1

Question
1) Let X be a continuous random variable with mean µ = 10 and variance σ2 = 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).

Chebyshev’s inequality provides an estimate of how a distribution varies from it’s mean.
The formula results in an upper bound of probability that a random variable varies from the mean by a certain number of standard deviations.

Below is Chebyshev’s inequality formula:
\(P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2}\)

Because we are provided variance we can calculate standard deviation, which is equal to the square root of the variance. We can then plug the standard deviations into the formular.

#Given
mean <- 10
variance <- 100/3

#SD
sd <- sqrt(variance)

#k the number of standard deviations away from the mean
k_1 <- 2 / sd
k_2 <- 5 / sd
k_3 <- 9 / sd
k_4 <- 20 / sd

#upper bounds 
upper_bound_1 <- 1 / (k_1^2)
upper_bound_2 <- 1 / (k_2^2)
upper_bound_3 <- 1 / (k_3^2)
upper_bound_4 <- 1 / (k_4^2)

#results
u_b_1<-bound<-cat("Upper bound for P(|X - 10| >= 2):", upper_bound_1, "\n")
## Upper bound for P(|X - 10| >= 2): 8.333333
u_b_2<-cat("Upper bound for P(|X - 10| >= 5):", upper_bound_2, "\n")
## Upper bound for P(|X - 10| >= 5): 1.333333
u_b_3<-cat("Upper bound for P(|X - 10| >= 9):", upper_bound_3, "\n")
## Upper bound for P(|X - 10| >= 9): 0.4115226
u_b_4<-cat("Upper bound for P(|X - 10| >= 20):", upper_bound_4, "\n")
## Upper bound for P(|X - 10| >= 20): 0.08333333

Page 303 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.)

Rate for 1 Light Bulb
\(\lambda= 1/1000\)

Rate for 100 Light Bulb First Failure
\(nλ\) \(100*1/1000=0.1\)

Expected Time for 1 Light Bulb to fail is inverse of rate
\(E|T| = \frac{1}{n*\lambda}\)
\(E|T| = \frac{1}{1/1000}\)
\(E|T| = \frac{1}{0.1}\)
\(E|T| = 10\)

#*Rate for 100 Light Bulb First Failure*   
p_fails <- 100/1000 
#expected time
evalue <- 1 / p_fails

print(paste("The expected time of first bulb failure is", evalue, "hours"))
## [1] "The expected time of first bulb failure is 10 hours"

Page 303 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 \(fZ(z) = (1/2)λe^{−λ|z|}\)
\(f_Z(z) = \int_{-\infty}^{\infty} f_{X_1}(x) \, f_{X_2}(x - z) \, dx\)

Since x2 has the same distribution as x1, we substitute \((x - z)\) with \(\lambda e^{-\lambda(x-z)}\) and solve.

\(= \int_{0}^{\infty} \lambda e^{-\lambda x} \cdot \lambda e^{-\lambda (x - z)} \, dx\)
\(= \int_{0}^{\infty}\lambda^2 *e^{-2\lambda x1} e^{\lambda z} \, dx1\)
\(= \int_{0}^{\infty}\lambda^2 *e^{\lambda(z-2*x1)}\, dx1\)
\(= (1/2)λe^{−λ|z|}\)