Problem 1 - Pg 303 Ex. 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.)

Solution 1
Expected time for the first of these bulbs to burn out:
The expected value of an exponential distribution is given by \(E(X) = 1/\lambda\)

n = 100
lifetime = 1000 # This means that 1 bulb will bulb in 1000 hours
problem1_soln = lifetime / n
problem1_soln
## [1] 10

Problem 2 - Pg 303 Ex. 14
Assume that \(X_1\) and \(X_2\) are independent random variables, each having an exponential density with parameter \(\lambda\). Show that \(Z = X_1-X_2\) has density:
\(f_Z(z) = (1/2)e^{-\lambda|z|}\).


Solution 2
\(f_Z(z) = (1/2)e^{-\lambda|z|}\) can be re-written as a piecewise function as: \(f_Z(z) = \begin{cases} (1/2)e^{-\lambda z}, & \mbox{if } z \ge 0, \\ (1/2)e^{\lambda z}, & \mbox{if }z <0. \end{cases}\)

Since \(X_1\) and \(X_2\) have exponential density, their PDF (Probability Density Function) is given by:
\(f_{X_1}(x)=f_{X_2}(x)=\begin{cases} \lambda e^{-\lambda x}, & \mbox{if } x\ge 0, \\ 0, & \mbox{otherwise. }\end{cases}\)

\[ \begin{split} f_Z(z) &= f_{X_1+(-X_2)}(z) \\ &= \int_{-\infty}^{\infty} f_{-X_2}(z-x_1) f_{X_1}(x_1) dx_1 \\ &= \int_{-\infty}^{\infty} f_{X_2}(x_1-z) f_{X_1}(x_1) dx_1 \\ &= \int_{-\infty}^{\infty} \lambda e^{-\lambda(x_1-z)} \lambda e^{-\lambda x_1} dx_1 \\ &= \int_{-\infty}^{\infty} \lambda^2 e^{-\lambda x_1 + \lambda z} e^{-\lambda x_1} dx_1 \\ &= \int_{-\infty}^{\infty} \lambda^2 e^{\lambda z - \lambda x_1 - \lambda x_1} dx_1 \\ &= \int_{-\infty}^{\infty} \lambda^2 e^{\lambda(z-2x_1)} dx_1 \end{split} \]

recall that \(z=x_1-x_2\), therefore, \(x_2=x_1-z\).

If \(z \ge 0\), then \(x_2=(x_1-z) \ge 0\), and \(x_1 \ge z\), and, using WolframAlpha, \(f_Z(z) = \int_{z}^{\infty} \lambda^2 e^{\lambda(z-2x_1)} dx_1 = \frac{1}{2} \lambda e^{-\lambda z}\).

If \(z < 0\), then \(x_2=(x_1-z) \ge 0\), and \(x_1 \ge 0\), and \(f_Z(z) = \int_{0}^{\infty} \lambda^2 e^{\lambda(z-2x_1)} dx_1 =\frac{1}{2} \lambda e^{\lambda z}\).

Combining two sides we get \(f_Z(z) = \begin{cases} (1/2)e^{-\lambda z}, & \mbox{if } z \ge 0, \\ (1/2)e^{\lambda z}, & \mbox{if }z <0. \end{cases}\)


Problem 3 - Pg 320-321 Ex. 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.

  1. \(P(|X - 10| \ge 2)\)
  2. \(P(|X - 10| \ge 5)\)
  3. \(P(|X - 10| \ge 9)\)
  4. \(P(|X - 10| \ge 20)\)
Solution 3: Chebyshev Inequality is defined by: Let X be a continous random variable with density \(f(x)\). Suppose X has a finite expected value \(\mu = E(X)\) and finite variance \(\sigma^2 = V(X)\). Then for any positive number \(\epsilon>0\), we have:
\(P(|X-\mu|\ge\epsilon) \le \frac{\sigma^2}{\epsilon^2}\).
.
Also, we know that, if \(\epsilon = k\sigma\), then:
\(P(|X-\mu|\ge k\sigma) \le \frac{1}{k^2}\).

In this case, we have, \(\mu=10\) and \(\sigma = \sqrt{\frac{100}{3}} = \frac{10}{\sqrt{3}}\).
If \(\epsilon = k\sigma\), then \(k=\frac{\epsilon}{\sigma} = \frac{\epsilon\sqrt{3}}{10}\).

Let \(u\) be upper bound in Chebyshev’s Inequality, then \(u = \frac{1}{k^2} = \frac{1}{(\epsilon\sqrt{3}/10)^2} = \frac{100}{3\epsilon^2}\).

std = sqrt(100/3)


Solution 3a: P(|X−10| ≥ 2)

k = 2/std
p_upper_3a = round((1 / (k^2)), 4)
p_upper_3a
## [1] 8.3333

Since probability cannot be greater than 1, therefore, the upper bound is 1 for this case

Solution 3b: P(|X−10| ≥ 5)

k = 5/std
p_upper_3b = round((1 / (k^2)), 4)
p_upper_3b
## [1] 1.3333

Also, since probability cannot be greater than 1, the upper bound is 1 for this case as well

Solution 3c: P(|X−10| ≥ 9)

k = 9/std
p_upper_3c = round((1 / (k^2)), 4)
p_upper_3c
## [1] 0.4115

The probability is less than 1 here, hence the upper bound is 0.4115

Solution 3c: P(|X−10| ≥ 20)

k = 20/std
p_upper_3d = round((1 / (k^2)), 4)
p_upper_3d
## [1] 0.0833

Also, the probability is less than 1 here, hence the upper bound is 0.0833