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?
Based on Ex.10, let \(X_1\), \(X_2\),…,\(X_n\) be \(n\) independent random variables each of which has an exponential density with mean \(\mu\) and \(M\) be the minimum value of the \(X_j\), then the density for \(M\) is exponential with mean \(\frac{\mu}{n}\).
In another word, if \(X_i\)~\(Exp(\lambda)\), then \(min\){\(X_1\), \(X_2\),…,\(X_n\)}~\(Exp(n\lambda)\)
For an exponential distribution:
\(E[X]\)= 1000 = \(\frac{1}{\lambda}\) \(\Rightarrow\) \(\lambda\) = \(\frac{1}{1000}\)
\(n\) = 100 bulbs, \(n\lambda\) = 100\(\times \frac{1}{1000}\) = \(\frac{1}{10}\)
Cumulative distribution function for exponential distribution: 1 - \(e^{-\lambda x}\)
First to burn out = \(min({X_1, X_2,...,X_n})\)
\(P(min({X_1, X_2,...,X_n}))\) = 1 - \(e^{-\frac{1}{10} x}\)
\(E[X] = \frac{1}{\frac{1}{10}}\) = 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
\(f_Z(z)=(\frac{1}{2}) \lambda e^{-\lambda|z|}\)
\(f_{X_1}(x)\) = \(f_{X_2}(x)\) = \(\begin{cases}\lambda e^{-\lambda x} & x\ge0\\0 & otherwise\end{cases}\)
\(f_Z(z) = \int_{-\infty}^{+\infty}f_{X_1}(x)f_{X_2}(x-z) dx\)
= \(\int_{0}^{+\infty}\lambda e^{-\lambda x} \lambda e^{-\lambda(x-z)} dx\)
= \(\int_{0}^{+\infty}\lambda^2 e^{-2\lambda x +\lambda z} dx\)
= \(\lambda e^{\lambda z}\int_{0}^{+\infty}e^{-2\lambda x} dx\)
= \(\lambda e^{\lambda z}(-\frac{1}{2}e^{-2\lambda x}]_0^{\infty}\)
= \((\frac{1}{2}) \lambda e^{\lambda z}\)
Since \(X_1\) and \(X_2\) are independent random variable having an exponential density with the same parameter \(\lambda\), \(Z=X_1-X_2=-(X_2-X_1)=-Z\), therefore
\(f_Z(z)=\begin{cases}\frac{1}{2} \lambda e^{\lambda z} & z < 0\\\frac{1}{2}\lambda e^{-\lambda z} & otherwise\end{cases}\) \(\Rightarrow\) \(f_Z(z)=(\frac{1}{2}) \lambda e^{-\lambda|z|}\).
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.
Chebyshev’s Inequality states that
Let \(X\) be any random variable with \(E[X]=\mu\) and \(V(X)=\sigma^2\).
\(P(|X - \mu|\ge k\sigma) \le \frac{\sigma ^2}{k^2\sigma ^2} = \frac{1}{k^2}\)
Thus, for any random variable, the probability of a deviaion from the mean of more than \(k\) standard deviation is \(\le \frac{1}{k^2}\).
For any of the following inequality, \(\mu\) = 10 and \(\sigma ^2 = \frac{100}{3}\).
sigma_sqare <- 100/3
sigma <- sqrt(sigma_sqare)\(k = \frac{k\sigma}{\sigma}\).
k <- 2/sigma
1/(k^2)## [1] 8.333333k <- 5/sigma
1/(k^2)## [1] 1.333333k <- 9/sigma
1/(k^2)## [1] 0.4115226k <- 20/sigma
1/(k^2)## [1] 0.08333333