1000 hours. What is the expected time for the first of these bulbs to burn out?
\[99 = 100e^\frac{-t}{1000}\] \[e^\frac{-t}{1000} = \frac{99}{100}\] \[t = 1000*ln(0.99)\] \[t = 10\]
The expected time for the first bulb to burn out is 10 hours.
exponential density with parameter . Show that \(Z = X1 - X2\) has density \(f_Z(z) = (1/2)\lambda e^{-\lambda|z|}\)
Let \(X_1\) and \(X_2\) be two independent exponential random variables with parameter \(\lambda\).
The PDF of \(X_1\) is \(f(x_1) = \lambda e^{-\lambda x_1}, x_1 \geq 0\) The PDF of \(X_2\) is \(f(x_2) = \lambda e^{-\lambda x_2}, x_2 \geq 0\)
Since \(X_1\) and \(X_2\) are independent, the joint density function of \(X_1\) and \(X_2\) is:
\[f(x_1, x_2) = f(x_1)f(x_2)\] \[=\lambda e^{-\lambda x_1} * \lambda e^{-\lambda x_2}\] \[\lambda^2e^{-\lambda (x_1 + x_2)}\] Let \(Z = X_1 - X_2\) \(V = X_2\) -> \(x_1 = z + v\) \(x_2 = v\)
Then, \(J = \begin{vmatrix} 1 & 1 \\ 0 & 1\end{vmatrix}\) \[=1\] Thus, the joint PDF of Z and V becomes: \[g(z, v) = \lambda^2e^{-\lambda (z+2v)}, v>0, -\infty<z<\infty\] \[z = x_1 - v \Rightarrow v = x_1 - z\] Thus, \(v > -z,\) if \(-\infty<z<\infty\) \(v>0\) if \(z>0\)
For \(-\infty<z<\infty\), \[f(z) = \int_{-z}^\infty g(z,v)dv\] \[= \int_{-z}^\infty \lambda^2e^{-\lambda (z+2v)} dv\] \[\frac{\lambda}{2} e^{\lambda z}\]
For \(z>0\), \[f(z) = \int_{0}^\infty g(z,v)dv\] \[= \int_{0}^\infty \lambda^2e^{-\lambda (z+2v)} dv\] \[\frac{\lambda}{2} e^{-\lambda z}\]
Combining the two results, we have:
\[g(z) = \frac{\lambda}{2}e^{-\lambda |x|}, -\infty < z < \infty\] \[= \frac{\lambda}{2}e^{-\lambda |x_1-x_2|}\]
\(\sigma^2 = \frac{100}{3}\). Using Chebyshev’s Inequality, find an upper bound for the following probabilities:
For any real number \(k > 0,\) \(Pr(|X-\mu| \geq k\sigma) \leq \frac{1}{k^2}\)
\[\mu = 10\] \[\sigma^2 = \frac{100}{3}\] \[s = \frac{10}{\sqrt3} = 5.773\]
All probabilities are between 0 and 1, therefore this solution is trivial.
All probabilities are between 0 and 1, therefore this solution is trivial.
\(P(|X - 10| \geq 9)\) \[k(5.773) = 9\] \[k=1.5589\] \[\frac{1}{1.5589^2}\] The upper bound for this probability is 0.411494.
\(P({|X - 10|} \geq 20)\) \[k(5.773) = 10\] \[k=1.7322\] \[\frac{1}{1.7322^2}\] The upper bound for this probability is 0.3333.
a <- 1/(0.3464)^2
a## [1] 8.333822b <- 1/(0.8661)^2
b## [1] 1.333104c <- 1/(1.5589)^2
c## [1] 0.411494d <- 1/(1.7322)^2
d## [1] 0.3332759From this set of questions and solutions, we proved that k values less than 1 produce probabilities greater than 1. Therefore, we can only use Chebychev’s inequality when k is greater than 1.