1.1 PS1 Answer

From Ex10, we know that, let \(X_{1}, X_{2}, \cdots, X_{n}\) be \(n\) independent random variables each of which has an exponential density with mean \(\mu\) and let \(M\) be the minimum value of the \(X_{j}\), the density for \(M\) is exponential with mean \(\frac{\mu}{n}\).

By using this property, when \(X_{i}\sim Exp(\lambda_{i})\), we have \(min(X_{i})\sim Exp(\Sigma \lambda_{i})\)

For Ex11, from the given information, we have \(n=100\) exponential variables where each represents the lifetime of one lightbulb with the average as \(1000\) hours. Assuming independence, the expected time for the first lightbulb to burnout is the minimum of lifetime of this set of \(100\) lightbulbs.

For any individual lightbulb, mean \(=1000\) and \(E[x]=\frac{1}{\lambda}=\frac{1}{1000}\)

For \(min(X_{i})\), new mean \(= mean^{'}=\frac{mean}{n}=\frac{1000}{100}=10=\frac{1}{\lambda^{'}}\). Thus, \(\lambda^{'}=\frac{1}{10}\) and \(E[min(X_{i})]=\frac{1}{\lambda^{'}}=10\).

Therefore, the expected time for the first of these 100 lightbulbs to burn out is 10 hours.

2.1 PS2 Answer

As both \(X_{1}\) and \(X_{2}\) are independent exponential variables, both are evaluated on the interval \([0,\infty)\).

We have \(f_{X_{1}}(x_{1})=f_{X_{2}}(x_{2})= \left\{\begin{matrix}\lambda e^{-\lambda x} & x \geq 0\\ 0 & otherwise \end{matrix}\right.\)

\[f_{Z}(z)= f_{X_{1}+(-X_{2})}(z) = \int_{-\infty}^{\infty} f_{x_{1}}(x_{1})f_{-x_{2}}(z-x_{1})dx_{1}\] \[=\int_{-\infty}^{\infty} f_{x_{1}}(x_{1})f_{x_{2}}(x_{1}-z)dx_{1}\] \[=\int_{-\infty}^{\infty} \lambda e^{-\lambda (x_{1})} \lambda e^{-\lambda (x_{1}-z)} dx_{1}\] \[=\int_{-\infty}^{\infty} \lambda^{2} e^{\lambda (z-2x_{1})} dx_{1}\]

For \(z \geq 0\) and \(x_{2}=x_{1}-z \geq 0,\) then \(x_{1}\geq z\,\). For \(z<0\) and \(x_{2}=x_{1}-z \geq 0,\) then \(x_{1}\geq 0\).

• For \(z<0,\, x_{1}\geq 0\),

\[f_{Z}(z)= f_{X_{1}+(-X_{2})}(z) = \int_{-\infty}^{\infty} f_{x_{1}}(x_{1})f_{x_{2}}(x_{1}-z)dx_{1}\] \[=\int_{-\infty}^{\infty} \lambda^{2} e^{\lambda (z-2x_{1})} dx_{1}\] \[= \int_{0}^{\infty} \lambda e^{\lambda z} \lambda e^{-2\lambda x_{1}} dx_{1}\] \[=-\frac{\lambda}{2} e^{\lambda z} \int_{0}^{\infty} -2\lambda e^{-2\lambda x_{1}} dx_{1}\] \[=-\frac{\lambda}{2} e^{\lambda z} \left [ e^{-2\lambda x_{1}} \right ]^{\infty}_{0}\] \[=-\frac{\lambda}{2} e^{\lambda z}[0-1]\] \[=\frac{\lambda}{2} e^{\lambda z}\]

• For \(z \geq 0,\, x_{1}\geq z\),

\[f_{Z}(z)= f_{X_{1}+(-X_{2})}(z) = \int_{-\infty}^{\infty} f_{x_{1}}(x_{1})f_{x_{2}}(x_{1}-z)dx_{1}\] \[=\int_{-\infty}^{\infty} \lambda^{2} e^{\lambda (z-2x_{1})} dx_{1}\]

\[= \int_{z}^{\infty} \lambda e^{\lambda z} \lambda e^{-2\lambda x_{1}} dx_{1}\] \[=-\frac{\lambda}{2} e^{\lambda z} \int_{z}^{\infty} -2\lambda e^{-2\lambda x_{1}} dx_{1}\] \[=-\frac{\lambda}{2} e^{\lambda z} \left [ e^{-2\lambda x_{1}} \right ]^{\infty}_{z}\] \[=-\frac{\lambda}{2} e^{\lambda z}[0-e^{-2\lambda z}]\] \[=\frac{\lambda}{2} e^{-\lambda z}\]

Combine the above results, we have

\[f_{Z}(z)= \left\{\begin{matrix} \frac{\lambda}{2} e^{\lambda z} & z < 0\\ \frac{\lambda}{2} e^{-\lambda z} & z\geq 0 \end{matrix}\right.\] \[f_{Z}(z)=\frac{1}{2}\lambda e^{-\lambda \left | z \right |}\]

3.1 PS3 Answer

From textbook page306 Example 8.1, we have the following information:

Let \(X\) by any random variable with \(E(X)=\mu\) and \(V(X)=\sigma^{2}\). Then, if \(\epsilon=k\sigma\), Chebyshev’s Inequality states that \[P(\left | X-\mu \right | \geq k\sigma) \leq \frac{\sigma^{2}}{k^{2}\sigma^{2}}=\frac{1}{k^{2}}\]