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.)
Exercise 10 states that given \(n\) independent random variables with an expontential density of mean \(\mu\) then the expected value for the density of the minimum value is \(\frac{\mu}{n}\). Adapting this to exercise 11 we have \(n=100\) lightbults and \(\mu=1000\) hours. Thus this distribution is expontential with mean \(\frac{1000}{100}=10\). Therefore the expected time for the first of the bulbs to burn out is 10 hours.
Assume that \(X_1\) and \(X_2\) are independent random variables, each having an exponential density with parameter \(\lambda\). Show that \(Z = X - Y\) has density \(f_Z(z)=\frac{1}{2}\lambda e^{-\lambda |z|}\)
The textbook provides us with a definition for when \(Z = X + Y\) but not when \(Z = X - Y\). This can be addressed by defining \(W = -Y\) giving us \(Z = X + W\)
The function defining \(W\) is \(f_W(x)=f_Y(-x)=\lambda e^{\lambda x}\)
Now \(f_Z(x)=\int f_X(z-w)f_W(w)dw = \int \lambda e^{-\lambda (z-w)} \lambda e^{\lambda w}\)
This simplifies to \(\int \lambda^2 e^{\lambda (z-2w)}\)
There are two situations to address here, when \(x < 0\) and \(x > 0\). Each situation causes half of the equation to drop out:
When \(x < 0\), then \(f_W(x)=0\) so \(\int \lambda^2 e^{\lambda z}\) and When \(x > 0\), then \(f_Z(x)=0\) so \(\int \lambda^2 e^{\lambda -2w}\)
Unfortunately, this is as far as I could get. I believe each integral will provide the appropriate portion of the density function, but I think I made a mistake in creating the integrals as my results do not make any sense.
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 Inequality states \(P(|X-\mu| \geq \epsilon) \leq \frac{\sigma^2}{\epsilon^2}\)
\(P(|X - 10| \geq 2) \leq \frac{\frac{100}{3}}{4}=8.333\)
This value must be logically bound at 1 as it is impossible for a probaiblity to be greater than 1.
\(P(|X - 10| \geq 5) \leq \frac{\frac{100}{3}}{25}=1.333\)
This value must be logically bound at 1 as it is impossible for a probability to be greater than 1.
\(P(|X - 10| \geq 9) \leq \frac{\frac{100}{3}}{81}=0.412\)
\(P(|X - 10| \geq 20) \leq \frac{\frac{100}{3}}{400}=0.083\)
Taking the last example, it states that the probability that \(X\) is greater than 30 or less than -10 is AT MOST \(7.6%\).