A company buys 100 lightbulbs, each of which has an exponential lifetime of 1000 hours. What is the expected time for the flrst of these bulbs to burn out?
Answer
From Exercise 10, for n independent random variables that have an exponential density and mean μ, the minimum value M is exponential density with a mean as \(\frac{\mu}{n}\) For any of the bulbs i, let its independent random variable \(X_i\) be: \[ E[X_i] = \frac{1}{\lambda_i} = 1000 \\ Since X_i \ is \ given\ as\ exponential,\ we\ have: \\ minX_1,X_2,\dots, X_100\sim e^{\sum_{i = 1}^{100}{\lambda_i}} \\ \sum_{i = 1}^{100}{\lambda_i} = 100 \times\ \frac{1}{1000} = \frac{1}{10} \\ E[minX_i] = \frac{1}{\frac{1}{10}} = 10 \ years \]
Assume that X1 and X2 are independent random variables, each having an exponential density with parameter ‚. Show that Z = X1 - X2 has density: \(f_Z(z) = (\frac{\lambda}{2}) e^{-\lambda |z|}\)
Answer
Both X1 and X2 are evaluated on the interval \(0\le\infty\). Since X2 is subtracted we can evaluate the convolution of X1 and X2 into two parts, \(-\infty\to 0\) when X2 ≥ X1 and \(0\to\infty\) when X1 ≥ X2
For X1 ≥ X2: \[ f_Z(z) = \int_{-\infty}^\infty {f_{X_1}(z+x_2)f_{X_2}(x_2)} \,dx_2 \\ f_Z(z) = \int_{0}^\infty {\lambda e^{-\lambda(z+x_2)} \lambda e^{-\lambda x_2}} \,dx_2 \\ f_Z(z) = \int_{0}^\infty {\lambda e^{-\lambda(z)} \lambda e^{-2\lambda x_2}} \,dx_2 \\ f_Z(z) = \lambda^2 e^{-\lambda z} (\int_{0}^\infty {\lambda e^{-\lambda x_2}} \,dx_2) \\ f_Z(z) = \lambda^2 e^{-\lambda z} (\frac{1}{2\lambda}) \\ f_Z(z) = \frac{\lambda}{2} (e^{-\lambda z}) \]
For X2 ≥ X1: \[ f_Z(z) = \int_{-\infty}^\infty {f_{X_1}(z+x_2)f_{X_2}(x_2)} \,dx_2 \\ f_Z(z) = \int_{-\infty}^0 {\lambda e^{-\lambda(z+x_2)} \lambda e^{-\lambda x_2}} \,dx_2 \\ f_Z(z) = \int_{-\infty}^0 {\lambda e^{-\lambda(z)} \lambda e^{-2\lambda x_2}} \,dx_2 \\ f_Z(z) = \lambda^2 e^{-\lambda z} (\int_{-\infty}^0 {\lambda e^{-\lambda x_2}} \,dx_2) \\ f_Z(z) = \lambda^2 e^{-\lambda z} (\frac{-1}{2\lambda}) \\ f_Z(z) = \frac{-\lambda}{2} (e^{-\lambda z}) \] These can be combined as follows: \[ \begin{equation} f(z) = \begin{cases} \frac{\lambda}{2} (e^{-\lambda z}) & \text z \geq 0 \\ \frac{-\lambda}{2} (e^{-\lambda z}) & \text z \lt 0 \end{cases} \end{equation} \] This can be ssummarized as: \[ f_Z(z) = (\frac{\lambda}{2}) e^{-\lambda |z|} \]
Let X be a continuous random variable with mean \(\mu\)= 10 and variance \(\sigma^2\) = \(\frac{100}{3}\). Using Chebyshev’s Inequality, find an upper bound for the following probabilities:
Solution
Chebyshev inequality:
\(P(|X−\mu| \ge \epsilon) \le\frac{\sigma^2}{\epsilon^2}\)
In this case, \(u = 10\) and \(\sigma = \sqrt{\frac{100}{3}} = \frac{10}{3}\) \(\epsilon=k\sigma\), hence \(k=\frac{\epsilon}{\sigma} = \frac{\epsilon \sqrt 3}{10}\)
Let \(u\) be the upper bound in Chebyshev inequaity
Then, \(u=\frac{1}{k^2} = \frac{1}{(\frac{\epsilon \sqrt 3}{10})^2} = \frac{100}{3\epsilon^2}\)
Answer
\(\epsilon=2\), upper bound: \(u = \frac{100}{3\epsilon^2} = \frac{100}{3\times2^2}\)
epsilon1 <- 2
upper_bound1 = 100 / (3 * epsilon1 ^2)
upper_bound1## [1] 8.333333Answer
\(\epsilon=2\), upper bound: \(u = \frac{100}{3\epsilon^2} = \frac{100}{3\times5^2}\)
epsilon2 <- 5
upper_bound1 = 100 / (3 * epsilon2 ^2)
upper_bound1## [1] 1.333333Answer
\(\epsilon=2\), upper bound: \(u = \frac{100}{3\epsilon^2} = \frac{100}{3\times9^2}\)
epsilon3 <- 9
upper_bound1 = 100 / (3 * epsilon3 ^2)
upper_bound1## [1] 0.4115226Answer
\(\epsilon=2\), upper bound: \(u = \frac{100}{3\epsilon^2} = \frac{100}{3\times20^2}\)
epsilon4 <- 20
upper_bound1 = 100 / (3 * epsilon4 ^2)
upper_bound1## [1] 0.08333333