Data 605 Homework 8

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?

To solve this problem, we can use the concept of exponential distribution and the memorylessness property of exponential random variables.

Let \(X_i\) be the lifetime of the \(i^{th}\) lightbulb, which follows an exponential distribution with a rate parameter \(\lambda = \frac{1}{1000}\) (since the mean lifetime is 1000 hours).

The expected time for the first lightbulb to burn out (the expected value of the minimum of the lifetimes of all 100 bulbs) is given by the sum of the individual lifetimes:

\[ E(X_{\text{min}}) = E(X_1 + X_2 + \ldots + X_{100}) \]

Since the lightbulbs have independent lifetimes, the expected value of the sum is the sum of the expected values:

\[ E(X_{\text{min}}) = E(X_1) + E(X_2) + \ldots + E(X_{100}) \]

Each individual lightbulb has the same exponential distribution, so their expected lifetimes are all the same:

\[ E(X_{\text{min}}) = 100 \cdot E(X_1) \]

We know that the expected value of an exponential distribution with rate parameter \(\lambda\) is \(\frac{1}{\lambda}\). Therefore:

\[ E(X_1) = \frac{1}{\lambda} = \frac{1}{\frac{1}{1000}} = 1000 \]

So, the expected time for the first lightbulb to burn out is 1000 hours.

Exercise 14

Assume that X1 and X2 are independent random variables, each having an exponential density with parameter ‚. Show that Z = X1 - X2 has density: fZ(z)=(λ2)e−λ|z|

Assume that \(X_1\) and \(X_2\) are independent random variables, each having an exponential density with parameter \(\lambda\). We want to show that \(Z = X_1 - X_2\) has density given by:

\[ f_Z(z) = \frac{\lambda}{2} e^{-\lambda |z|} \]

To derive this, let’s first find the joint PDF of \(X_1\) and \(X_2\). Since \(X_1\) and \(X_2\) are independent, the joint PDF \(f_{X_1,X_2}(x_1,x_2)\) is the product of their individual PDFs:

\[ f_{X_1,X_2}(x_1,x_2) = f_{X_1}(x_1) \times f_{X_2}(x_2) = \lambda^2 e^{-\lambda (x_1 + x_2)} \]

Now, let \(Z = X_1 - X_2\). We want to find the PDF of \(Z\), denoted as \(f_Z(z)\). We’ll use the convolution formula to find \(f_Z(z)\):

\[ f_Z(z) = \int_{-\infty}^{\infty} f_{X_1,X_2}(x, x+z) \, dx \]

Substituting the joint PDF \(f_{X_1,X_2}(x_1,x_2) = \lambda^2 e^{-\lambda (x_1 + x_1 + z)}\), we get:

\[ f_Z(z) = \int_{-\infty}^{\infty} \lambda^2 e^{-\lambda (x_1 + x_1 + z)} \, dx_1 \]

\[ = \lambda^2 e^{-\lambda z} \int_{-\infty}^{\infty} e^{-2\lambda x_1} \, dx_1 \]

\[ = \lambda^2 e^{-\lambda z} \left[-\frac{1}{2\lambda} e^{-2\lambda x_1}\right]_{-\infty}^{\infty} \]

\[ = \frac{\lambda}{2} e^{-\lambda z} \left(0 - (-1)\right) \]

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

Therefore, the PDF of \(Z = X_1 - X_2\) is \(f_Z(z) = \frac{\lambda}{2} e^{-\lambda |z|}\). This matches the form \(f_Z(z) = \lambda^2 e^{-\lambda |z|}\), as expected.

Exercise 1

Let X be a continuous random variable with mean µ = 10 and variance σ 2 = 100/3. Using Chebyshev’s Inequality, find an upper bound for the following probabilities. 8.2. CONTINUOUS RANDOM VARIABLES 321 (a) P(|X − 10| ≥ 2). (b) P(|X − 10| ≥ 5). (c) P(|X − 10| ≥ 9). (d) P(|X − 10| ≥ 20).

1. \(P(|X-10| \geq 2) \leq \frac{1}{{\left(\frac{2\sqrt{3}}{10}\right)}^2} = \frac{1}{\left(\frac{2\sqrt{3}}{10}\right)^2} = \frac{1}{{\left(\frac{2\sqrt{3}}{10}\right)} \times {\left(\frac{2\sqrt{3}}{10}\right)}} = \frac{1}{\frac{4 \times 3}{100}} = \frac{100}{12} = \frac{25}{3}\)

2. \(P(|X-10| \geq 5) \leq \frac{1}{{\left(\frac{5\sqrt{3}}{10}\right)}^2} = \frac{1}{\left(\frac{5\sqrt{3}}{10}\right)^2} = \frac{1}{{\left(\frac{5\sqrt{3}}{10}\right)} \times {\left(\frac{5\sqrt{3}}{10}\right)}} = \frac{1}{\frac{25 \times 3}{100}} = \frac{100}{75} = \frac{4}{3}\)

3. \(P(|X-10| \geq 9) \leq \frac{1}{{\left(\frac{9\sqrt{3}}{10}\right)}^2} = \frac{1}{\left(\frac{9\sqrt{3}}{10}\right)^2} = \frac{1}{{\left(\frac{9\sqrt{3}}{10}\right)} \times {\left(\frac{9\sqrt{3}}{10}\right)}} = \frac{1}{\frac{81 \times 3}{100}} = \frac{100}{243}\)

4. \(P(|X-10| \geq 20) \leq \frac{1}{(2\sqrt{3})^2} = \frac{1}{4 \times 3} = \frac{1}{12}\)