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.)
From Exercise 10 The minimum of \(\mathrm{n}\) independent random variables, each following an exponential distribution with mean \(\mu\), has an exponential distribution with mean \(\mu / \mathrm{n}\). \[ M=\min (X 1, X 2, \ldots, X n) \]
The probability that the minimum value is greater than a certain value \(x(P(M>x))\) as the probability that all \(n\) individual variables are greater than \(x\). \[ \begin{gathered} P(\min (X 1, X 2, \ldots, X n)>x)=P(X 1>x, \ldots, X n>x) \$=\mathrm{P}(\mathrm{X} 1>\mathrm{x}) \ldots \mathrm{P}(\mathrm{Xn}>\mathrm{x}) \$ \\ =e^{-\lambda_1 x} \ldots e^{-\lambda_n x}=e^{-\left(\lambda_1+\cdots+\lambda_n\right) x} \end{gathered} \]
The continuous distribution function of the minimum is an exponential with parameter \(\lambda_1+\cdots+\lambda_n\)
The probability that the minimum is \(X_{\mathrm{j}}\) is given by \(P\left(X_j<X_i\right.\) for \(\left.i \neq j\right)=\int_0^{\infty} P\left(X_j<X_i\right.\) for \(i \neq j) \lambda_j e^{-\lambda_j x} d x\) \[ \begin{aligned} & =\int_0^{\infty} \lambda_j e^{-\lambda_j} \prod_{i \neq j} P\left(X_i>x\right) d x=\lambda_i \int_0^{\infty} e^{-\left(\lambda_1+\cdots+\lambda_n\right) x} d x=\frac{\lambda_j}{\lambda_1+\cdots+\lambda_n}\left[-e^{-\left(\lambda_1+\cdots+\lambda_n\right) x}\right]_0^{\infty} \\ = & \frac{\lambda_j}{\lambda_1+\cdots+\lambda_n} \text { since } \lambda_1=\cdots=\lambda_n=\lambda, P\left(X_j<X_i \text { for } i \neq j\right)=\frac{1}{n} \end{aligned} \]
The minimum of \(n\) exponential random variables indeed follow an exponential distribution with a mean of \(\mu / \mathrm{n}\).
The lifetime expectancy of any bulb i \[ E\left[X_i\right]=\frac{1}{\lambda_i}=1000, \lambda_i=\frac{1}{1000} \]
As we are given 100 bulbs, \(n=100\) \[ \lambda=n \lambda_i=100 * \frac{1}{1000}=\frac{1}{10} \]
The density for the minimum value of \(\mathrm{Xi}\) \[ E\left[X_i \min \right]=\frac{1}{\lambda}=\frac{1}{\frac{1}{10}}=10 \] `
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. \[ f_Z(z)=(1 / 2) \lambda e^{-\lambda|z|} \]
As we are given that each independent random variable has an exponential density then \[ \begin{aligned} & f\left(X_1\right)=\lambda * e^{-\lambda * x_1} \\ & f\left(X_2\right)=\lambda * e^{-\lambda * x_2} \end{aligned} \]
Using the densities \[ \begin{aligned} f_Z(Z)=\int_{-\infty}^{+\infty} f_{X 2}\left(x_1-Z\right) f_{X 1}\left(x_1\right) d x_1=\int_0^{\infty} \lambda * e^{-\lambda * x_1} \lambda * e^{-\lambda *\left(x_1-z\right)} d x_1 \\ \quad=\int_0^{\infty} \lambda^2 * e^{-2 \lambda * x_1} e^{\lambda * z} d x_1=\int_0^{\infty} \lambda^2 * e^{\lambda\left(z-2 * x_1\right)} d x_1=(1 / 2) \lambda e^{-\lambda|z|} \end{aligned} \]
Let \(X\) be a continuous random variable with mean \(\mu=10\) and variance \(\sigma^2=100 /\) Using Chebyshev’s Inequality, find an upper bound for the following probabilities
Chebyshev’s Inequality \(\quad P(|X-\mu| \geq k \sigma) \leq \frac{1}{k^2}\) (a) \(P(|X-10| \geq 2)\)
An upper bound is 1. \(\sigma^2=\frac{100}{3}=>\sigma=\frac{10}{\sqrt{3}}\) From the problem a, \(k \sigma=2=>k=\frac{2}{\sigma}=\frac{2 * \sqrt{3}}{10}\) From the Chebyshev’s Inequality, \(\frac{1}{k^2}=\frac{100}{12}=8.333\) We cannot accept that, \(P(|X-\mu| \geq k \sigma) \leq 8.333\) Since the highest value of probability is 1 , the upper bound will be 1 . \[ (|X-\mu| \geq 2) \leq 1 \]
# Calculate the upper bound and round to two decimal places
µ<-10
σ <-sqrt(100/3)
kσ<-2
k<-2/σ
upper_a<-1/k^2
round(upper_a,2)## [1] 8.33(b). \(P(|X-10| \geq 5)\)
An upper bound is \(1: \quad k \sigma=5 \Rightarrow k=\frac{5}{\sigma}=\frac{5 * \sqrt{3}}{10}=\frac{\sqrt{3}}{2}\) From the Chebyshev’s Inequality, \(\frac{1}{k^2}=\frac{4}{3}=1.333\) We cannot accept that, \(P(|X-\mu| \geq k \sigma) \leq 1.333\) Since the highest value of probability is 1 , the upper bound will be 1 . \[ P(|X-\mu| \geq 5) \leq 1 \]
# Calculate the upper bound and round to two decimal places
kσ<-5
k<-5/σ
upper_a<-1/k^2
round(upper_a,2)## [1] 1.33(c). \(P(|X-10| \geq 9)\)
An upper bound is 100/243: \(k \sigma=9=>k=\frac{9}{\sigma}=\frac{9 * \sqrt{3}}{10}\) From the Chebyshev’s Inequality: \(\quad \frac{1}{k^2}=\frac{100}{243}\) The \(1 / k^{\wedge} 2\) is less than one: \(\quad P(|X-\mu| \geq 9) \leq \frac{100}{243}\)
# Calculate the upper bound and round to two decimal places
kσ<-9
k<-9/σ
upper_a<-1/k^2
round(upper_a,2)## [1] 0.41(d). \(P(|X-10| \geq 20)\)
An upper bound is 1/12: \(k \sigma=20=>k=\frac{20}{\sigma}=\frac{20 * \sqrt{3}}{10}=2 \sqrt{3}\) From the Chebyshev’s Inequality: \(\frac{1}{k^2}=\frac{1}{12}\) The \(1 / k^{\wedge} 2\) is less than one: \(P(|X-\mu| \geq 20) \leq \frac{1}{12}\)
# Calculate the upper bound and round to two decimal places
kσ<-20
k<-20/σ
upper_a<-1/k^2
round(upper_a,2)## [1] 0.08