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.)

From exercise 10, the minimum value of an exponential density of \(n\) independent random variables is \(\mu\)/n.

Here \(\mu\) = 1000 hours. n=100

mu = 1000
n = 100

(excected_hours = mu/n)
## [1] 10

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

The PDF is convolution of two exponential is: \[ f_{X1}(x)= f_{X2}(x) = \begin{equation*} \left\{ \begin{array}{ll} \lambda e^{-\lambda x}, & \quad x > 0, \\ 0, & \quad otherwise. \end{array} \right. \end{equation*} \]

The convolution of two exponential densities if \(Z = X_1 - X_2\) is: \[ f_Z(z)=\int_{-\infty }^{\infty } f_{X_2}(x_1-z)f_{X_1}(x_1)dx_1 =\int_{0 }^{\infty } \lambda e^{-\lambda (x_1-z)}\lambda e^{-\lambda x_1} = \int_{0 }^{\infty } \lambda^2 e^{\lambda (z-2x_1)}\ \] Going to https://www.integral-calculator.com/ to calculate the integral gets:

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

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.

If for any positive number = \(k\sigma = k\) standard deviations for some integer \(k\), then

\[ P(|X-\mu|\ge k\sigma) \le\frac{1}{k^2} \]

(a) P(|X − 10| ≥ 2).

\[ k\sigma = 2 \] \[ k= \frac{2}{\sqrt{100/3}} \] \[ \frac{1}{k^2}= 1/(\frac{2}{\sqrt{100/3}})^2 \]

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

Here the upper bound is greater than 1 which is not possible therefore the answer is 1.

(b) P(|X − 10| ≥ 5).

\[ k\sigma = 5 \]

\[ k= \frac{5}{\sqrt{100/3}} \] \[ \frac{1}{k^2}= 1/(\frac{5}{\sqrt{100/3}})^2 \]

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

Here the upper bound is greater than 1 which is not possible therefore the answer is 1.

(c) P(|X − 10| ≥ 9).

\[ k\sigma = 9 \]

\[ k= \frac{9}{\sqrt{100/3}} \] \[ \frac{1}{k^2}= 1/(\frac{9}{\sqrt{100/3}})^2 \]

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

(d) P(|X − 10| ≥ 20).

\[ k\sigma = 20 \]

\[ k= \frac{20}{\sqrt{100/3}} \] \[ \frac{1}{k^2}= 1/(\frac{20}{\sqrt{100/3}})^2 \]

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