A company buys 100 lightbulbs, each of which has an exponential lifetime of 1000 hours What is the expected time until the first lightbulb burns out?
Solution
\(P(M \le m)=1-e^\frac{-nm}{\mu}\)
For this particular problem, n=100 and \(\mu\)=1000
\(E(M) = \int_{0}^{\infty}x \frac{n}{\mu}e^{-\frac{nx}{\mu}}dx = -\frac{e^{\frac{n}{\mu} x}(\frac{n}{\mu}x+1)}{\frac{n}{\mu}}|^{\infty}_{0} = \frac{\mu}{n} = 10\)
The expected time until the first lightbulb burns out is 10 hours.
Assume that \(X\) and \(Y\) are independent random variables both with exponential density given by \(\lambda\) Show that \(Z = X-Y\) has density:
\(f_{Z}(z) = \frac{1}{2}\lambda e^{-\lambda |z|}\)
Solution
\(Z = X - Y \implies X = Z + Y\)
Then \(f_{Z}(z) = \int_{-\infty}^{\infty}f_{X}(z+y)f_{Y}(y)dy\)
We know that \(f_{X}(x)\) and \(f_{Y}(x)\) are 0 when \(x<0\)
We also know that the range of possible values for \(Z\) is \((-\infty, \infty)\)
If \(z\leq 0\), then \((z+y)\) and \(y\) must both be positive, or else \(f_{X}(x)\) and \(f_{Y}(x)\) are 0
This implies \(y > -z\)
So \(f_{Z}(z) = \int_{-z}^{\infty}f_{X}(z+y)f_{Y}(y)dy = \int_{-z}^{\infty}\lambda^{2}e^{-\lambda(z + 2y)}dy = \frac{-\lambda}{2}e^{-\lambda(z+2y)}|_{-z}^{\infty}\)
\(= \frac{\lambda}{2}e^{\lambda z} = \frac{\lambda}{2}e^{-\lambda |z|}\)
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:
\(P\left ( \left |X-10 \right |\geq 2 \right )\)
\(P\left ( \left |X-10 \right |\geq 5 \right )\)
\(P\left ( \left |X-10 \right |\geq 9 \right )\)
\(P\left ( \left |X-10 \right |\geq 20 \right )\)
Solution
\(P\left ( \left |X-10 \right |\geq 2 \right ) \leq \frac{\frac{100}{3}}{2^{2}} = \frac{100}{12} \implies\) the upper bound is 1
\(P\left ( \left |X-10 \right |\geq 5 \right ) \leq \frac{100}{3\cdot 5^{2}} = \frac{4}{3} \implies\) upper bound is 1
\(P\left ( \left |X-10 \right |\geq 9 \right ) \leq \frac{100}{3\cdot 9^{2}} = \frac{100}{243} \implies\) upper bound is 0.4115
\(P\left ( \left |X-10 \right |\geq 20 \right ) \leq \frac{100}{3\cdot 20^{2}} = \frac{1}{12} \implies\) upper bound is 0.08333