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? (See Exercise 10.)
Answer
Let \(X_{1}, X_{2},...,X_{100}\) be 100 lightbulbs with exponential lifetime \(\mu=1000\). The first lightbulb will fail at minimum value of \(X_{j}\) suppose that is M. The denisity of M is exponential with mean \(\mu/100\)
\[ u=1000\\ n=100\\ E[x]=\frac{u}{n}=\frac{1000}{100}=10 \]
mu <- 1000
n <- 100
mu/n## [1] 10The expected time for the first bulb to fail out is 10 hours with out disregading other factors that may also influence like electric grid and spikes.
The distribution of the minimum value of n independent expontially distributed variables with mean \(\mu\) would also be exponential with mean \(\mu/n\). So in this case the expected value would be 1000 hrs/100 or 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 \(f_Z(z) = (1/2)e^{-\lambda|z|}\).
Answer
\[ f_z(Z)=(1/2)\lambda e^{-\lambda |z|} \]
\(f_Z(z) = (1/2)e^{-\lambda|z|}\) can be re-written as \(f_Z(z) = \begin{cases} (1/2)e^{-\lambda z}, & \mbox{if } z \ge 0, \\ (1/2)e^{\lambda z}, & \mbox{if }z <0. \end{cases}\)
Since \(X_1\) and \(X_2\) have exponential density, in pdf format, lets compute the PDF for our x1 and x2
\(f_{X_1}(x)=f_{X_2}(x)=\begin{cases} 0, & \mbox{if }x\lt 0\,\\\lambda e^{-\lambda x}, & \mbox{if } x\ge 0\end{cases}\)
Therefore, \(f(z)=\int _{ 0 }^{ \infty } e^{-\lambda x} \lambda e^{-\lambda (z-x)}dx\\\)
Simplified, \(f(z)=\int _{ 0 }^{ \infty } \lambda e^{-\lambda x} * \lambda e^{-\lambda x} * \lambda e^{\lambda z}dx\\\)
Pull out constant, \(f(z)=\lambda e^{\lambda z}\int _{ 0 }^{ \infty } \lambda e^{-2\lambda x} dx\\\)
Integrate, \(f(z)=\lambda e^{\lambda z} (-\frac{1}{2}e^{-2\lambda x})\\\)
Evaluate, \(f(z)=\frac{\lambda}{2} e^{\lambda |z|}\\\)\[ f(x_1)=(1/2)\lambda e^{-\lambda x_1}\\ f(x_2)=(1/2)\lambda e^{-\lambda x_2} \] Please note that we can assume x1 and x2 to be greater than or equal to zero
Since x1 and x2 are independent, we can compute the joint density function as follows:
\[ f(x_1, x_2)=f(x_1)f(x_2)\\ =(\lambda e^{-\lambda x_1})(\lambda e^{-\lambda x_2})\\ =\lambda^{2}e^{x_1+x_2} \]
We know z=x1+x2 but lets pick some variable h to be x2 such that we have z=x1+h. We can re-work our new equation into x1=z+h. The purpose of this is to use the Jacobian for the transformation of variables.
We compute the jacobian as follows:
\[ J=\frac{\delta(x_1, x_2)}{\delta(z,v)}\\ =\begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix}\\ =1 \]
We can now develop our joint probability distribution with our transformed variables. We will do some substitution a little later.
\[ g(z, v)=\lambda^{-\lambda(z+v+v)}\\ =\lambda^{-\lambda(z+2v)} \]
Please note that v>0 and z is bounded from -infinity to infinity
Recall our substitution
\[ z=x_1-v\\ v=x_1-z \] Some additional conditions
\[ v>-z,-\infty <z<\infty\\ v>0,z>0 \]
With our transformed PDF, we can use the conditions we have derived to assemble our integration. Rather than going through every step of the integration, I will highlight the process. Remove the lambda squared since it is a constant, then apply u substitution where u is x+2v. After u substitution, we need to evaluate the resulting improper integral by taking the limit as some arbitrary upper bound goes to infinity.
For z greater than -infinity but less than 0
\[ \int _{ -x }^{ \infty }g(z,v)dv\\ =\int _{ -x }^{ \infty }{ \lambda ^{ { 2 } } } { e }^{ -\lambda (x+2v) }dv\\ =\frac{1}{2}e^{\lambda x} \] for z greater than 0 \[ \int _{ x }^{ \infty }g(z,v)dv\\ \int _{ x }^{ \infty }{ \lambda ^{ { 2 } } } { e }^{ -\lambda (x+2v) }dv\\ =\frac{1}{2}e^{-\lambda x} \] The probability density function is \[ g(z) =\frac{1}{2}e^{\lambda x}\ ,-\infty<z<\infty\\ =\frac{1}{2}e^{-\lambda x}\ ,0<z<\infty \] Now we can combine our equations with our problem statment \[ f_z(Z)=(1/2)\lambda e^{-\lambda |z|} \\ =(1/2)\lambda e^{-\lambda |x_1+x_2|} \\ \]
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.
Answer
Chebyshev Inequality: \(P(|X-\mu|\ge\epsilon) \le \frac{\sigma^2}{\epsilon^2}\) or, per example 8.4, \(P(|X-\mu|\ge k\sigma) \le \frac{1}{k^2}\).
requiring a solution , \(\mu=10\) and \(\sigma = \sqrt{\frac{100}{3}} = \frac{10}{\sqrt{3}}\).
If \(\epsilon = k\sigma\), then \(k=\frac{\epsilon}{\sigma} = \frac{\epsilon\sqrt{3}}{10}\).
Let \(u\) be upper bound in Chebyshev’s Inequality, then \(u = \frac{1}{k^2} = \frac{1}{(\epsilon\sqrt{3}/10)^2} = \frac{100}{3\epsilon^2}\).
\[ P(|X-10|\ge2)=P(|X-10|\ge.2\sqrt3 \frac{10}{\sqrt3})\le\frac{1}{(.2\sqrt3)^{2}}\le8.33 \]
#a.) P(|X-10|>=2)
a<- (100/3)/(2^2)
a## [1] 8.333333\[ P(|X-10|\ge5)=P(|X-10|\ge.5\sqrt3 \frac{10}{\sqrt3})\le\frac{1}{(.5\sqrt3)^{2}}\le1.33 \]
\(\epsilon=5\), the upper bound is \(u= \frac{100}{3\times 5^2} = \frac{4}{3} \approx 1.3333\). Due to the fact that probability cannot be greater than \(1\), the upper bound is \(1\).
#b.) P(|X-10|>=2)
b<- (100/3)/(5^2)
b## [1] 1.333333\[ P(|X-10|\ge9)=P(|X-10|\ge.9\sqrt3 \frac{10}{\sqrt3})\le\frac{1}{(.9\sqrt3)^{2}}\le0.412 \]
#c.) P(|X-10|>=9)
c<-(100/3)/(9^2)
c## [1] 0.4115226\[ P(|X-10|\ge20)=P(|X-10|\ge2\sqrt3 \frac{10}{\sqrt3})\le\frac{1}{(2\sqrt3)^{2}}\le0.083 \]
#d.) P(|X-10|>=20)
d<-(100/3)/(20^2)
d## [1] 0.08333333