Exercise 11 pg. 303

  1. A company buys 100 light bulbs, each of which has an exponential lifetime of 1000 hours. What is the expected time for the first of these bulbs to burnout? (See Exercise 10.)

Note that \(X_i \sim exp(\lambda_i )\), for \(i = 1,2,3 ...n\), and \(X_1 , X_2 ... X_n\) are mutually independent random variables, then \[min \{ X_1, X_2 ... X_n \} \sim exp( \Sigma_{i=1}^{n} \lambda_i)\].

Therefore, the expected value of the above probability distribution \[ E[min\{X_1,X_2 ...X_{100}\}] = \frac{1}{ \Sigma_{i=1}^{100} \lambda_i} = \frac{1}{100\lambda} = \frac{1}{100\cdot\frac{1}{1000}}=10\].

ANS. The expected time for the first of these bulbs to burnout is 10.

Exercise 14 pg 303

  1. Assume that X1 and X2 are independent random variables, each having an exponential density with parameter λ. Show that \(Z = X1 - X2\) has density \[ f_Z (z) = \frac{1}{2} \lambda e^{-\lambda |z|} \].

\[ f_Z(z) = \int_{-\infty}^{\infty}f(z+t)f(t)dt \] \[ f_Z(z) = \int_{-\infty}^{\infty}\lambda^2 e^{-\lambda (z+t)}e^{-\lambda t}dt \] \[=\int_{-\infty}^{\infty}\lambda^2 e^{ -\lambda ( 2t+ z)}dt \] When \(z > 0\) \[ = - \frac{1}{2} \lambda e^{-\lambda(2t+z)} \bigr|^{\infty}_{0} =\frac{1}{2} \lambda e^{-\lambda z} \] When \(z< 0\) \[ = - \frac{1}{2} \lambda e^{-\lambda(2t+z)} \bigr|^{0}_{-\infty} =-\frac{1}{2} \lambda e^{-\lambda z} \]

Exercise 1 pg. 320

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. (a) P(|X −10| ≥ 2). (b) P(|X −10| ≥ 5). (c) P(|X −10| ≥ 9). (d) P(|X −10| ≥ 20).

Chebyshev’s Inequality Formula:

\(Pr(|X-\mu|\ge k\sigma) \le \tfrac{1}{k^2} \\ X = random \ variable \\ \mu = expected \ value \\ \sigma = standard \ deviation \\ k = number \ of \ standard \ deviations \ from \ the \ mean\)

Ans.

## Standard Deviation = sigma
mu = 10 
sigmasquare = 100/3
sigma= sqrt(sigmasquare)

\(A)\ P(|X-10| \ge 2)\)

k_A = 2/sigma
bound_A = min(1/k_A^2, 1)

bound_A
## [1] 1
fractions(bound_A)
## [1] 1

The upper bound is 1

\(B)\ P(|X-10| \ge 5)\)

k_B = 5/sigma
bound_B = min(1/k_B^2, 1)

bound_B
## [1] 1
fractions(bound_B)
## [1] 1

The upper bound is 1

\(C)\ P(|X-10|) \ge 9\)

k_C = 9/sigma
bound_C = min(1/k_C^2, 1)

bound_C 
## [1] 0.4115226
fractions(bound_C) 
## [1] 100/243

The upper bound is 0.4115226 or \(\tfrac{100}{243}\)

\(D)\ P(|X-10| \ge 20)\)

k_D = 20/sigma
bound_D = min(1/k_D^2, 1)

bound_D
## [1] 0.08333333
fractions(bound_D)
## [1] 1/12

The upper bound is 0.0833333 or \(\tfrac{1}{12}\)