DATA605 - Homework 08

SUMS OF CONTINUOUS RANDOM VARIABLES

Exercise 11 (Page 303)

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.)

Exercise 10: Let \(X_1, X_2, . . . , X_n\) be n independent random variables each of which has an exponential density with mean \(\mu\). Let \(M\) be the minimum value of the \(X_j\). Show that the density for \(M\) is exponential with mean \(\mu/n\). Hint: Use cumulative distribution functions.

SOLUTION:

Based on Exercise 10, we know that the density for \(M\) is exponential with mean \(\mu/n\),

We also know that

• \(n = 100 lightbulbs\)

Therefore, the expected time for the first of these bulb to burn out is:

• \(E(M) = \mu/n = \frac{1000 hours}{100} = 10 hours\)

Exercise 14 (Page 303)

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

\(f_Z(z) = (1/2)\lambda e^{-\lambda |z|}\)

SOLUTION:

Since \(f_Z(z)\) is an exponential function, \(z\) can only take positive values. This it can be written as:

\[\begin{equation} f_Z(z) = \begin{cases} \frac{1}{2}\lambda e^{-\lambda z} & \text{if } z \geq 0\\ \frac{1}{2}\lambda e^{\lambda z} & \text{if } z < 0 \end{cases} \end{equation}\]

The PDF for \(X1\) and \(X2\) is:

\[\begin{equation} fX_1(x) = fX_2(x) = \begin{cases} \lambda e^{-\lambda x} & \text{if } x \geq 0\\ 0 & \text{if } z < 0 \end{cases} \end{equation}\]

\[f_Z(z) = \int_{-\infty}^{\infty} f(x_1)(x)f(x_2)(x-z) \,dx \]

\[ = \int_{0}^{\infty} \lambda e^{-\lambda z} \lambda e^{-\lambda (x-z)} \,dx \]

\[ = \int_{0}^{\infty} \lambda ^2 e^{-2\lambda x + \lambda z} \,dx \]

\[ = \lambda e^{\lambda z} \int_{0}^{\infty} e^{-2\lambda x} \,dx \]

\[ = \frac{1}{2} \lambda e^{-\lambda|z|} \]

CONTINUOUS RANDOM VARIABLES

Exercise 1 (Pages 320-321)

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. \(P(|X - 10| >= 2)\).
2. \(P(|X - 10| >= 5)\).
3. \(P(|X - 10| >= 9)\).
4. \(P(|X - 10| >= 20)\).

SOLUTION:

The theorem of Chebyshev’s Inequality is defined as:

Let \(X\) be a discrete random variable with expected value \(\mu = E(X)\), and let \(\epsilon > 0\) be any positive real number. Then

\(P(|X-\mu | \geq \epsilon) \leq \frac{V(X)}{\epsilon ^2}\)

We also know that

• The mean \(E(X) = \mu = 10\)

• The variance \(V(X) = \sigma^2 = 100/3\)

• The standard deviation \(\sigma = \sqrt{\frac{100}{3}}\)

Then, if \(\epsilon = k \sigma\), Chebyshev’s Inequality states that

\(P(|X-\mu | \geq k \sigma) \leq \frac{\sigma ^2}{k^2 \sigma ^2} = \frac{1}{k^2}\) *** (CI Ineq).

Exercise 1.a : \(P(|X - 10| \geq 2)\).

If \(k \sigma = 2 => k = \frac{2}{\sqrt{\frac{100}{3}}}\)

Using inequality (CI Ineq) we get:

\(P(|X-10| \geq 2) = \frac{1}{k^2} = 8.333333\)

1/ (2/sqrt(100/3))^2
#> [1] 8.333333

Exercise 1.b : \(P(|X - 10| \geq 5)\).

If \(k \sigma = 5 => k = \frac{5}{\sqrt{\frac{100}{3}}}\)

Using inequality (CI Ineq) we get:

\(P(|X-10| \geq 5) = \frac{1}{k^2} = 1.333333\)

1/ (5/sqrt(100/3))^2
#> [1] 1.333333

Exercise 1.c : \(P(|X - 10| \geq 9)\).

If \(k \sigma = 9 => k = \frac{9}{\sqrt{\frac{100}{3}}}\)

Using inequality (CI Ineq) we get:

\(P(|X-10| \geq 9) = \frac{1}{k^2} = 0.4115226\)

1/ (9/sqrt(100/3))^2
#> [1] 0.4115226

Exercise 1.d : \(P(|X - 10| \geq 20)\).

If \(k \sigma = 20 => k = \frac{20}{\sqrt{\frac{100}{3}}}\)

Using inequality (CI Ineq) we get:

\(P(|X-10| \geq 20) = \frac{1}{k^2} = 0.08333333\)

1/ (20/sqrt(100/3))^2
#> [1] 0.08333333