p303 #11

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 burn out?

If \(X_i\) is an independent random variable for light bulb \(i\) with an expected lifetime of \(1000\) hours, then we know that the expected value of \(X_i\) is

\[ \begin{align} E[X_i] &= \mu = \frac{1}{\lambda_i} = \frac{1}{1000} \end{align} \].

Then, we can find the minimum value of \(X_i\), which will be the time it takes for the first light bulb in the series to burn out. This can be represented by the exponential probability distribution.

\[ min\{X_1, \ X_2, \ ... \ X_{100} \} \sim \text{exponential} \ (\sum\limits_{i=1}^{100} \lambda_i) \]

Calculating this out, we see

\[ (\sum\limits_{i=1}^{100} \lambda_i) = 100 \times \frac{1}{1000} = \frac{100}{1000} = \frac{1}{10}. \]

Thus, we can say that the expected minimum value of \(X_i\) is

\[ E[\text{min} \ X_i] = \frac{1}{1/10} = 10 \text{ hours}. \]

The first light bulb in the series will burn out in 10 hours.

p303 #14

Assume that \(X_1\) and \(X_2\) are independent random variables, each having an exponential density with paramater \(\lambda\). Show that \(Z = X_1 - X_2\) has density

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

The densities of the three random variables \(X_1\), \(X_2\), and \(Z\) can be represented by \(f_X{_1}\), \(f_X{_2}\), and \(f_Z\), respectively.

\[ f_X{_1} = \lambda e^{-\lambda x_1} \quad \text{for } X_1\ge0, \text{ and } 0 \text{ elsewhere} \\ f_X{_2} = \lambda e^{-\lambda x_2} \quad \text{for } X_2\ge0, \text{ and } 0 \text{ elsewhere} \\ \]

Then, the convolution of \(X_1\) and \(X_2\) is

\[ \begin{align} f_Z(z) &= \int_{-\infty}^{+\infty} f_X{_1}(z-x_1) \ f_{X_2}(x_1) \ dx_2 \\ \\ &= \int_{0}^{\infty} \lambda e^{-\lambda (z + 2x_2)} \lambda e^{-\lambda (z + 2x_2)} dx_1 \\ \\ &= \int_{0}^{\infty} \lambda^2 e^{-\lambda (z + 2x_2)} e^{\lambda z} dx_1 = \frac{\lambda}{2}e^{\lambda z} \qquad \text{When } z \text{ is negative} \\ \\ &= \int_{-z}^{\infty} \lambda^2 e^{-\lambda (z + 2x_2)} e^{\lambda z} dx_1 = \frac{\lambda}{2}e^{-\lambda z} \qquad \text{When } z \text{ is positive} \\ \\ &= \frac{1}{2}\lambda^{\lambda |z|} \end{align} \]

p.320-321 #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. \(P(|X-10|) \ge 2 \quad \le \frac{\sigma^2}{k^2} \approx \frac{33}{4} \approx 8.33 = 1\)

variance <- 100/3
k <- 2
k^2

## [1] 4

variance / k^2

## [1] 8.333333

2. \(P(|X-10|) \ge 5 \quad \le \frac{\sigma^2}{k^2} \approx \frac{33}{25} \approx 1.33 = 1\)

variance <- 100/3
k <- 5
k^2

## [1] 25

variance / k^2

## [1] 1.333333

3. \(P(|X-10|) \ge 9 \quad \le \frac{\sigma^2}{k^2} \approx \frac{33}{81} \approx 0.41\)

variance <- 100/3
k <- 9
k^2

## [1] 81

variance / k^2

## [1] 0.4115226

4. \(P(|X-10|) \ge 20 \quad \le \frac{\sigma^2}{k^2} \approx \frac{33}{400} \approx 0.08\)

variance <- 100/3
k <- 20
k^2

## [1] 400

variance / k^2

## [1] 0.08333333