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?
According problem set 10, the exponential density has mean = \(\frac {\mu}{n}\),
If \(X_{i}\) is an independent random variable, \(E[X_{i}] = \frac {1}{\lambda_{i}} = 1000\).
Since, \(\frac{1}{\lambda} = 1000\) , then \(E[min X_{i}] = \frac {1000}{100} = 10\)
Therefore, 10 hours is expected to pass before the first bulb burn out.
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 \({f}_{Z}(z)=\frac {1}{2}e^{-\lambda{z}}\).
\(f(x_1) = {\lambda}{e}^{-\lambda{x1}}\)
\(f(x_2) = {\lambda}{e}^{-\lambda{x2}}\)
As given, \(Z = X_1 - X_2\) , so \(X_1 = Z + X_2\) and to get PDF, \(\lambda{e}^{-\lambda{x_1}} * \lambda{e}^{-\lambda{x_2}} = \lambda^2e^{-\lambda(X_1+X_2)}\) . With substitution \(Z\) and \(X_2\) to \(X_1\), \(\lambda^2e^{-\lambda((z+x_2)+x_2)}= \lambda^2e^{-\lambda(z+2x_2)}\)
So, when \(z\geq0\),
\(\int _{0}^{\infty}{\frac{\lambda}{2}{e}}^{-\lambda{(z+2x_2)}}dx = {\frac{\lambda}{2}{e}}^{-\lambda{z}}\)
and, when z<0,
\(\int _{-}^{\infty}z{{\lambda}^{2}{e}}^{-\lambda{(z+2x_2)}}dx = {\frac{\lambda}{2}{e}}^{\lambda{z}}\)
Let \(X\) be a continuous random variable with mean \(\mu\) = 10 and variance \(\sigma^{2}\) = \(\frac {100}{3}\). Using Chebyshev’s Inequality, find an upper bound for the following probabilities.
Chebyshev’s Inequality is given by \(P(|X−\mu|≥kσ)\geq\frac{1}{k^2}\)
k <- 2/sqrt(100/3)
up<- 1/k^2
up## [1] 8.333333The upper bound is 1 since the highest value of probability is 1.
k <- 5/sqrt(100/3)
up<-1/k^2
up## [1] 1.333333The upper bound is 1 since the highest value of probability is 1.
k <- 9/sqrt(100/3)
up<-1/k^2
up## [1] 0.4115226The upper bound is 0.412.
k <- 20/sqrt(100/3)
up<-1/k^2
up## [1] 0.08333333The upper bound is 0.083.