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? (See Exercise 10.)
Exercise 10 states that for n independent random variables with exponential density and mean \(\mu\), the minimum value M is exponential density with mean \(\frac{\mu}{n}\).
In this case \(\mu = 1000\space hours\) and \(n = 100\) such that the expected value of M is \(\frac{1000}{100} =10\space hours\) since the expected value of a random variable is the mean of that random variable.
Note from last week’s assignment that the minimum density for \(X_1 ,X_2, ... X_n\) iid with pdf f(x) and cdf F(x) is given by:
\[ f_{1}(x) = nf(x)(1-F(x))^{n-1} \]
In this case \(\lambda = \frac{1}{\mu} = 0.001\space year^{-1}\) and \(n = 100\).
\[ f(x) = \lambda e^{-\lambda x} \\ F(x) = 1 - e^{-\lambda x} \\ f_{1}(x) = n \lambda e^{-\lambda x} (1-[1-e^{-\lambda x}])^{n-1} \\ f_{1}(x) = n \lambda e^{-\lambda x}(e^{-\lambda x})^{n-1} \\ f_{1}(x) = n \lambda (e^{-\lambda x})^n \\ f_{1}(x) = n \lambda e^{-n\lambda x} \] https://www2.stat.duke.edu/courses/Spring12/sta104.1/Lectures/Lec15.pdf
Note that the density for the minimum is itself an exponential density with \(\lambda^{\prime} = n\lambda = \frac{n}{\mu} = 1/\mu^\prime\) which is what Exercise 10 proved.
The expected value for an exponential density is \(\mu\). In this case, \(\mu = \frac{1000\space years}{100} = \bf{10\space years}\).
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 \(f(z) = (1/2)\lambda e^{-\lambda|z|}\).
Both \(X_1\) and \(X_2\) are evaluated on the interval \(0 \leq x < \infty\). Since \(X_2\) is subtracted we can evaluate the convolution of \(X_1\) and \(X_2\) into two parts, \(-\infty \rightarrow 0\) when \(X_2 \geq X_1\) and \(0 \rightarrow \infty\) when \(X_1 \geq X_2\).
For \(X_2 \geq X_1\) \[ f_Z(z) = \int_{-\infty}^{\infty}f_{X_1}(z+x_2)f_{X_2}(x_2)dx_2 \\ f_Z(z) = \int_{-\infty}^0\lambda e^{-\lambda (z+x_2)} \lambda e^{-\lambda x_2}dx_2 \\ f_Z(z) = \int_{-\infty}^0\lambda e^{-\lambda (z)} \lambda e^{-2\lambda x_2}dx_2 \\ f_Z(z) = \lambda^2 e^{-\lambda z}(\int_{-\infty}^0 e^{-2\lambda x_2}dx_2) \\ f_Z(z) = \lambda^2 e^{-\lambda z}(\frac{ -1}{2\lambda}) \\ f_Z(z) = \frac{-\lambda}{2} (e^{-\lambda z}) \]
For \(X_1 \geq X_2\) \[ f_Z(z) = \int_{-\infty}^{\infty}f_{X_1}(z+x_2)f_{X_2}(x_2)dx_2 \\ f_Z(z) = \int_0^{\infty}\lambda e^{-\lambda (z+x_2)} \lambda e^{-\lambda x_2}dx_2 \\ f_Z(z) = \int_0^{\infty}\lambda e^{-\lambda (z)} \lambda e^{-2\lambda x_2}dx_2 \\ f_Z(z) = \lambda^2 e^{-\lambda z}(\int_0^{\infty} e^{-2\lambda x_2}dx_2) \\ f_Z(z) = \lambda^2 e^{-\lambda z}(\frac{ -1}{2\lambda}) \\ f_Z(z) = \frac{\lambda}{2} (e^{-\lambda z}) \]
To combine:
\[ f(z) = \begin{cases} \frac{-\lambda}{2} (e^{-\lambda z}) & z < 0 \\ \frac{\lambda}{2} (e^{-\lambda z}) & z \geq 0 \end{cases} \]
This can be rewritten: \[ f(z) = \frac{\lambda}{2} e^{-\lambda |z|} \]
Because the negative sign will cancel out under the integral when integrating from \(-\infty \rightarrow 0\).
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.
\(P(|X - \mu| \geq \epsilon) \leq \frac{\sigma^2}{\epsilon^2}\)
Here \(\epsilon = 2\) such that:
\[ P(|X - 10| \geq 2) \leq \frac{\frac{100}{3}}{2^2}\\ P(|X - 10| \geq 2) \leq \frac{25}{3} \]
Here \(\epsilon = 5\) such that:
\[ P(|X - 10| \geq 5) \leq \frac{\frac{100}{3}}{5^2}\\ P(|X - 10| \geq 5) \leq \frac{4}{3} \]
Here \(\epsilon = 9\) such that:
\[ P(|X - 10| \geq 9) \leq \frac{\frac{100}{3}}{9^2}\\ P(|X - 10| \geq 9) \leq \frac{100}{243} \]
Here \(\epsilon = 20\) such that:
\[ P(|X - 10| \geq 20) \leq \frac{\frac{100}{3}}{20^2}\\ P(|X - 10| \geq 20) \leq \frac{100}{1200} \]