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?
If each lightbulb’s lifespan is modeled as an independent random variable \(X_1, X_2, ...,X_{100}\) then each of them has an exponential density with mean \(\mu=1000\). Therefore, if M is the minimum lifespan of these light bulbs (ie, the first one to burn out) then the density of M is exponential with mean \(\mu=\frac{1000}{100}=10\) so the expected time for the first of these bulbs to burn out is 10 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 \(f_z(z)=(1/2)\lambda e^{-\lambda |z|}\)
We want to find the convolution of \(X_1\) and \(-X_2\), and we know that \(P(-X_2=x)=P(X_2=-x)\), therefore \(f_{-X_2}(x)=f_{X_2}(-x)\)
Then:
\(f_Z(z)=f_{X_1}(z+x)f_{X_2}(-x)dx\\=\int\lambda e^{-\lambda (z+x)} \lambda e^{-\lambda(-x)}\\=\int\lambda^2e^{-\lambda z-\lambda x+\lambda x} \\=\int\lambda^2e^{-\lambda z}\)
Which in turn evaluates to \((1/2)\lambda e^{-\lambda |z|}\)
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:
\(P(|X-10|\geq2)\leq\frac{1x}{2^2}\\\frac{\frac{100}{3}}{2^2}=\frac{\frac{100}{3}}{4}=\frac{100}{12}=\frac{25}{3}\\P(|X-10|\geq2)\leq\frac{25}{3}\)
\(P(|X-10|\geq5)\leq\frac{\frac{100}{3}}{5^2}\\\frac{\frac{100}{3}}{5^2}=\frac{\frac{100}{3}}{25}=\frac{100}{75}=\frac{4}{3}\\P(|X-10|\geq5)\leq\frac{4}{3}\)
\(P(|X-10|\geq9)\leq\frac{\frac{100}{3}}{9^2}\\\frac{\frac{100}{3}}{9^2}=\frac{\frac{100}{3}}{81}=\frac{100}{243}\\P(|X-10|\geq9)\leq\frac{100}{243}\)
\(P(|X-10|\geq20)\leq\frac{\frac{100}{3}}{20^2}\\\frac{\frac{100}{3}}{20^2}=\frac{\frac{100}{3}}{400}=\frac{100}{1200}=\frac{1}{12}\\P(|X-10|\geq20)\leq\frac{1}{12}\)