ex11 pg303
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?
The individual mean is:
\(\lambda_i = \frac{1}{1000}\)
while the company buys 100 lightbulbs, the accumulated mean is:
\(\lambda = \sum \lambda_i = 100 \times \frac{1}{1000} = \frac{1}{10}\)
the expected value is:
\(E(X) = \frac{1}{\lambda} = 10\)
ex14 pg303
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 \(f_Z(z)=(1/2)\lambda e^{-\lambda |z|}\)
exponential function formula:
\[f(x) = \lambda e^{-\lambda x}\] where \(x \geq 0\)
\[f_{X_{1}}(x_1) = \lambda e^{-\lambda x_1} \\ f_{X_{2}}(x_2) = \lambda e^{-\lambda x_2}\]
\[f_Z(z) = \int f_{X_1}f_{X_2} \\=\int \lambda e^{-\lambda x_1} * \lambda e^{-\lambda*x_2} \\ =\int \lambda^2 e^{-\lambda(x_1+x_2)}\]
\[\because Z = X_1 - X_2 \\ \therefore X_1 = Z + X_2\]
\[f_Z(z) = \int \lambda^2 e^{-\lambda(z+2x_2)} \\ =\frac{1}{2} \lambda e^{-\lambda z}\]
while z > 0:
\[f_Z(z) = \frac{1}{2} \lambda e^{-\lambda z}\]
while z < 0:
\[f_Z(z) = \frac{1}{2} \lambda e^{-\lambda|z|}\]
ex1 pg320-321
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.
\(P(|X-10| \geq 2).\)
\(P(|X-10| \geq 5).\)
\(P(|X-10| \geq 9).\)
\(P(|X-10| \geq 20).\)
according to Chebyshev Inequality theorem for continuous random variable for any positive number \(\epsilon \gt 0\), we have:
\(P(|X-\mu| \geq \epsilon) \leq \frac{\sigma^2}{\epsilon^2}\)
mu = 10
sigma_sq = 100/3
# >=2
epsilon2 = 2
sigma_sq / epsilon2^2## [1] 8.333333# >= 5
epsilon5 = 5
sigma_sq / epsilon5^2## [1] 1.333333# >= 9
epsilon9 = 9
sigma_sq / epsilon9^2## [1] 0.4115226# >= 20
epsilon20 = 20
sigma_sq / epsilon20^2## [1] 0.08333333