Let \(X_{1}, X_{2},...,X_{100}\) be 100 lightbulbs with exponential lifetime \(\mu=1000\). The first lightbulb will burn out at minimum value of \(X_{j}\). Assume it is F. The density of F is then exponential with mean \(\mu/100\)
Answer:
mu <- 1000
num_Bulbs <- 100
mu/num_Bulbs## [1] 10The expected time for the first bulb to burn out is 10 hours.
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 \(f_Z(z) = (1/2)e^{-\lambda|z|}\).
Answer:
\(f_Z(z) = (1/2)e^{-\lambda|z|}\) can be re-written as \(f_Z(z) = \begin{cases} (1/2)e^{-\lambda z}, & \mbox{if } z \ge 0, \\ (1/2)e^{\lambda z}, & \mbox{if }z <0. \end{cases}\)
Since \(X_1\) and \(X_2\) have exponential density:
\[ \begin{split} f_Z(z) &= f_{X_1+(-X_2)}(z) \\ &= \int_{-\infty}^{\infty} f_{-X_2}(z-x_1) f_{X_1}(x_1) dx_1 \\ &= \int_{-\infty}^{\infty} f_{X_2}(x_1-z) f_{X_1}(x_1) dx_1 \\ &= \int_{-\infty}^{\infty} \lambda e^{-\lambda(x_1-z)} \lambda e^{-\lambda x_1} dx_1 \\ &= \int_{-\infty}^{\infty} \lambda^2 e^{-\lambda x_1 + \lambda z} e^{-\lambda x_1} dx_1 \\ &= \int_{-\infty}^{\infty} \lambda^2 e^{\lambda z - \lambda x_1 - \lambda x_1} dx_1 \\ &= \int_{-\infty}^{\infty} \lambda^2 e^{\lambda(z-2x_1)} dx_1 \end{split} \]
Consider \(z=x_1-x_2\), then \(x_2=x_1-z\).
If \(z \ge 0\), then \(x_2=(x_1-z) \ge 0\), and \(x_1 \ge z\), and, using WolframAlpha (Mathematica), \(f_Z(z) = \int_{z}^{\infty} \lambda^2 e^{\lambda(z-2x_1)} dx_1 = \frac{1}{2} \lambda e^{-\lambda z}\).
If \(z < 0\), then \(x_2=(x_1-z) \ge 0\), and \(x_1 \ge 0\), and \(f_Z(z) = \int_{0}^{\infty} \lambda^2 e^{\lambda(z-2x_1)} dx_1 =\frac{1}{2} \lambda e^{\lambda z}\).
through merging the two we will have \(f_Z(z) = \begin{cases} (1/2)e^{-\lambda z}, & \mbox{if } z \ge 0, \\ (1/2)e^{\lambda z}, & \mbox{if }z <0. \end{cases}\)