DATA605_HW8_SUMS OF RANDOM VARIABLES

Pg 303, Exercise 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.)

EX. 11 Solution

Expected time for bulbs to burn out.

n=100
lifetime=1000
lifetime/n
## [1] 10

Pg 303, Exercise 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 \(f_Z(z) = (1/2)e^{-\lambda|z|}\).

EX. 14 Solution

\(f_Z(z) = (1/2)e^{-\lambda|z|}\) can be re-written 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 is

\(f_{X_1}(x)=f_{X_2}(x)=\begin{cases} \lambda e^{-\lambda x}, & \mbox{if } x\ge 0, \\ 0, & \mbox{otherwise. }\end{cases}\)

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

Consider \(z=x_1-x_2\), then \(x_2=x_1-z\).

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

Pg 320-321 Exercise 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

Chebyshev Inequality: \(P(|X-\mu|\ge\epsilon) \le \frac{\sigma^2}{\epsilon^2}\) or, per example 8.4, \(P(|X-\mu|\ge k\sigma) \le \frac{1}{k^2}\).

Per problem, \(\mu=10\) 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 \(u = \frac{1}{k^2} = \frac{1}{(\epsilon\sqrt{3}/10)^2} = \frac{100}{3\epsilon^2}\).

  1. P(|X−10|≥2)
std = sqrt(100/3)
k = 2/std
1/k^2
## [1] 8.333333

Since probability cannot be greater than 1, the upper bound is 1

  1. P(|X − 10| ≥ 5)
std = sqrt(100/3)
k = 5/std
1/k^2
## [1] 1.333333

Since probability cannot be greater than 1, the upper bound is 1

  1. P(|X − 10| ≥ 9).
k = 9/std
1/k^2
## [1] 0.4115226
  1. P(|X − 10| ≥ 20).
k = 20/std
1/k^2
## [1] 0.08333333