11. 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?
\(n = 100\) lighbulbs
mean = 1000
\(\lambda = \frac{1}{\text{mean}} = \frac{1}{1000}\)
And since each of the \(100\) lighbulbs has an exponential lifetime of \(1000\) hours, then the expected time for the first of them to burn out is: (similar to example 6.29 page 275 in the probability text)
$E() = = = = 10 $ hours.
14. 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 \(fZ(z) = \frac{1}{2} \lambda e^{-\lambda |z|}\)
The convolution theorem is: \(fZ(z) = \int_{-\infty}^{\infty} fX_1(x) \cdot fX_2(x+z) dx\)
We know that the density function of each of the random variable is:
\(fX_1(x_1) = \lambda e^{- \lambda x_1}\)
\(fX_2 (x_2) = \lambda e^{-\lambda x_2}\)
Substituting those densities in the convolution theorem:
$$ \[\begin{align*} f_Z(z) &= \int_{-\infty}^{\infty} \lambda e^{-\lambda x} \cdot \lambda e^{-\lambda (x+z)} dx \\ f_Z(z) &= \lambda ^2 \int_{-\infty}^{\infty} e^{-\lambda x} \cdot e^{-\lambda (x+z)} dx \\ f_Z(z) &= \lambda ^2 \int_{-\infty} ^{\infty} e^{- \lambda (2x+z)} dx \\ f_Z(z) &= \lambda ^2 e^{-\lambda z} \int_{- \infty} ^{\infty} e^{-2 \lambda x} dx \\ f_Z(z) &= \lambda ^2 e^{-\lambda z} [-\frac{1}{2 \lambda}e^{-2 \lambda x}]_{- \infty} ^{\infty} \\ f_Z(z) &= \frac{\lambda ^2}{2 \lambda} e^{-\lambda z} [e^{-\infty} - e^{\infty}] \\ f_Z(z) &= \frac{\lambda}{2} e^{-\lambda z} (0-0) \\ f_Z(z) &= \frac{\lambda}{2} e^{-\lambda z} \\ f_Z(z) &= \frac{1}{2} \lambda e^{- \lambda |z|} \end{align*}\] $$
Since \(e^{-\lambda z}\) doesn’t have \(x\) and the integration with respect to x, then \(e^{-\lambda z}\) is considered as a constant, so we can take it out of the integral.
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.
\(P(|X−10|≥2)\).
\(P(|X−10|≥5)\).
\(P(|X−10|≥9)\).
\(P(|X − 10| ≥ 20)\).
Chebyshev’s Inequality says (page 305):
Let \(X\) be a discrete random variable with expected value \(\mu = E(X)\), and let \(\epsilon > 0\) be any positive real number. Then
\(P(|X- \mu | ) \geq \epsilon ) \leq \frac{V(X)}{\epsilon ^2}\)
So the upper bound that we are looking for is \(\frac{V(X)}{\epsilon ^2}\)
And we can find the variance using: \(V(X) = E((X - \mu)^2)\)
But in this case \(V(X)\) is given, which is \(V(X)= \sigma ^2 = \frac{100}{3}\)
Using the example 8.1 page 306 from the probability text, The upper bound will be \(\frac{1}{k^2}\) where in this case there 2 standard deviations. And \(\epsilon = k \sigma\) \(\Rightarrow k = \frac{\epsilon}{\sigma} = \frac{2}{\sqrt{\frac{100}{3}}} = \frac{2 \sqrt{3}}{10}\)
#Setting the parameters:
k <- (2* sqrt(3)) / 10
#calculate the upper bound
UpperBound_2 <- format(1/ k^2 , digits=3)
#Displaying the upper bound
cat("The upper bound of the probability with 2 standard deviations using Chebyshev's Inequality is " , UpperBound_2, "\n")## The upper bound of the probability with 2 standard deviations using Chebyshev's Inequality is 8.33In this case there are 5 standard deviations. And \(\epsilon = k \sigma\) \(\Rightarrow k = \frac{\epsilon}{\sigma} = \frac{5}{\sqrt{\frac{100}{3}}} = \frac{5 \sqrt{3}}{10}\)
#Setting the parameters:
k <- (5* sqrt(3)) / 10
#calculate the upper bound
UpperBound_5 <- format(1/ k^2 , digits=3)
#Displaying the upper bound
cat("The upper bound of the probability with 5 standard deviations using Chebyshev's Inequality is " , UpperBound_5, "\n")## The upper bound of the probability with 5 standard deviations using Chebyshev's Inequality is 1.33In this case there are 9 standard deviations. And \(\epsilon = k \sigma\) \(\Rightarrow k = \frac{\epsilon}{\sigma} = \frac{9}{\sqrt{\frac{100}{3}}} = \frac{9 \sqrt{3}}{10}\)
#Setting the parameters:
k <- (9* sqrt(3)) / 10
#calculate the upper bound
UpperBound_9 <- format(1/ k^2 , digits=3)
#Displaying the upper bound
cat("The upper bound of the probability with 9 standard deviations using Chebyshev's Inequality is " , UpperBound_9, "\n")## The upper bound of the probability with 9 standard deviations using Chebyshev's Inequality is 0.412In this case there are 20 standard deviations. And \(\epsilon = k \sigma\) \(\Rightarrow k = \frac{\epsilon}{\sigma} = \frac{20}{\sqrt{\frac{100}{3}}} = \frac{20 \sqrt{3}}{10}\)
#Setting the parameters:
k <- (20* sqrt(3)) / 10
#calculate the upper bound
UpperBound_20 <- format(1/ k^2 , digits=3)
#Displaying the upper bound
cat("The upper bound of the probability with 20 standard deviations using Chebyshev's Inequality is " , UpperBound_20, "\n")## The upper bound of the probability with 20 standard deviations using Chebyshev's Inequality is 0.0833