According to Exercise 10, the minimum value of n independent random variables (each of which has an exponential density with mean \(\mu\)) is exponential density with mean \(\mu/n\).
In case of Exercise 11, mean \(\mu=1000\) is exponential life time of a single blurb. Minimum value of n independent random variables is life time of the blurb that burns out first. Expected life time of the blurb that burns out first is \(\mu/n=1000/100=10\) hours.
\(f_Z(z) = (1/2)\lambda e^{-\lambda |z|}\)
The exponential probability density function is defined by the following formula:
\(f_{X1}(x) = f_{X2}(x) = \begin{cases}\lambda e^{-\lambda x} & \text{where } x \geq 0, \\0, & \text{otherwise. } \\ \end{cases}\)
Since X1 and X2 are independent random variables we can calculate joint density function :
\(f(x_1, x_2) = f(x_1)f(x_2) = \lambda e^{-\lambda x_1}\lambda e^{-\lambda x_2} = \lambda^2 e^{-\lambda(x_1 + x_2)}\)
Since \(Z=X1−X2\) let \(X1=Z+X2\). Now, let’s substitute \(X1\) with \(Z+X2\) in the equation below:
\(f(z+x_2, x_2) = \lambda^2 e^{-\lambda(z+x_2 + x_2)}=\lambda^2 e^{-\lambda(z+2 x_2)}\)
In order to calculate f_Z(z) we have to integrate f(z+x_2, x_2) from \(-\infty\) to \(\infty\).
According to exponential probability density function equation, both \(X1\) and \(X2\) are greater or equal 0. It suggests that \(Z\) fall into interval \((-\infty;0]\) when \(X2\geq X1\) and \(Z\) fall into interval \([0;\infty)\) when \(X1\geq X2\).
So that, let’s calculate \(f_Z(z)\) in the interval \((-\infty;0]\)
\(f_Z(z) = \int_{-\infty}^0\lambda^2 e^{-\lambda(z+2 x_2)}dx_2 = \lambda^2 e^{-\lambda z}\int_{-\infty}^0e^{-2\lambda x_2}dx_2= \lambda^2 e^{-\lambda z}\int_{-\infty}^0e^{-2\lambda x_2}dx_2=\lambda^2 e^{-\lambda z}(\frac{ -1}{2\lambda}) \\\)
Now, let’s calculate \(f_Z(z)\) in the interval \([0;\infty)\)
\(f_Z(z) = \int_0^{\infty}\lambda^2 e^{-\lambda(z+2 x_2)}dx_2 = \lambda^2 e^{-\lambda z}\int_0^{\infty}e^{-2\lambda x_2}dx_2= \lambda^2 e^{-\lambda z}\int_0^{\infty}e^{-2\lambda x_2}dx_2=\lambda^2 e^{-\lambda z}(\frac{1}{2\lambda}) \\\)
Let’s summarize,
\(f(z) =\begin{cases}\frac{-\lambda}{2} (e^{-\lambda z}) & z < 0 \\\frac{\lambda}{2} (e^{-\lambda z}) & z \geq 0\end{cases}\)
Therefore, we proved that \(f_Z(z) = (1/2)\lambda e^{-\lambda |z|}\)
\(P(|X - \mu| \geq \epsilon) \leq \frac{\sigma^2}{\epsilon^2}\)
\(\epsilon=2\)
\(P(|X - 10| \geq 2) \leq \frac{\frac{100}{3}}{2^2}\leq \frac{25}{3}\)
\(\epsilon=5\)
\(P(|X - 10| \geq 5) \leq \frac{\frac{100}{3}}{5^2}\leq \frac{4}{3}\)
\(\epsilon=9\)
\(P(|X - 10| \geq 2) \leq \frac{\frac{100}{3}}{9^2}\leq \frac{100}{243}\)
\(\epsilon=20\)
\(P(|X - 10| \geq 2) \leq \frac{\frac{100}{3}}{20^2}\leq \frac{1}{12}\)