According to exercise 10:
Let \(X_1, X_2, . . . , X_n\) be \(n\) independent random variables each of which has an exponential density with mean \(\mu\).
Let \(M\) be the minimum value of the \(X_j\) .
Show that the density for \(M\) is exponential with mean \(\frac{\mu}{n}\).
Hint: Use cumulative distribution functions.
So, we have \(n=100\) exponential variables, each representing the lifetime of a lightbulb, where the average lightbulb has a lifetime of 1000 hours, which is distributed exponentially across the set of lightbulbs (independent and identically distributed.)
The time for the first lightbulb to burn out is thus the minimum of this set.
Because the lifetime of each lightbulb is exponentially distributed, we know that the probability that any individual lightbulb has not burned out by time \(M=m\) is \(Pr(X_i>m) = e^{-\lambda m}\) , where \(\lambda = \frac{1}{\mu} = \frac{1}{1000}\) .
We know that the probability that no lightbulb has burned out by time \(M=m\) is
\[\begin{aligned}
Pr(X_1 > m) \cdot Pr(X_2 > m) \cdot ... \cdot Pr(X_n > m) &= e^{-\lambda m} \cdot e^{-\lambda m} \cdot ... \cdot e^{-\lambda m} \\
&= (e^{-\lambda m})^n \\
&= e^{-\lambda m n} \\
&= e^{-\frac{m \cdot n}{\mu}} \\
&= e^{-\frac{m \cdot 100}{1000}} \\
&= e^{-\frac{m}{10}}
\end{aligned}\]
The above is the probability that no bulb has burned out by time \(M=m\).
Therefore the probability that the first bulb has burned out by time \(M=m\) is \(Pr(M \le m) = 1- e^{-\frac{m}{10}}\) .
This is an exponential distribution with \(\lambda = \frac{1}{\mu} = \frac {1}{10}\) .
Therefore, the expected time for the first of these bulbs to burn out is 10 hours.
We have \({\displaystyle f(x_1;\lambda )={\begin{cases}\lambda e^{-\lambda x_1}&x_1\geq 0,\\0&x_1<0.\end{cases}}}\) and \({\displaystyle f(x_2;\lambda )={\begin{cases}\lambda e^{-\lambda x_2}&x_2\geq 0,\\0&x_2<0.\end{cases}}}\) .
By independence, the joint PDF is \({\displaystyle f(x_1,x_2;\lambda )={\begin{cases}\lambda^2 e^{-\lambda x_1}e^{-\lambda x_2}&x_1,x_2\geq 0,\\0&(x_1<0) \vee (x_2<0).\end{cases}}}\) .
Let \(Z = X_1 - X_2\) . Because this is a difference, Z can be positive or negative.
We seek \(F_Z(z)=Pr(Z \le z)\) where \(-\infty < z < \infty\) .
In the case where \(z \le 0\), the CDF is:
\[
\begin{aligned}
F_Z(z) &= Pr(Z \le z) \\
&= Pr(X_1 - X_2 \le z) \\
&= Pr(X_2 \ge X_1 - z) \\
&= \int_0^\infty \int_{x_1-z}^\infty f_{X_1,X_2}(x_1,x_2)dx_2 dx_1 \\
&= \int_0^\infty \int_{x_1-z}^\infty \lambda^2 e^{-\lambda x_1}e^{-\lambda x_2}dx_2 dx_1 \\
&= \int_0^\infty \lambda e^{-\lambda x_1} \left[\int_{x_1-z}^\infty \lambda e^{-\lambda x_2}dx_2 \right]dx_1 \\
&= \int_0^\infty \lambda e^{-\lambda x_1} \left[-e^{-\lambda x_2} \right]_{x_1-z}^\infty dx_1 \\
&= \int_0^\infty \lambda e^{-\lambda x_1} \left[-0+e^{-\lambda {(x_1-z)}} \right] dx_1 \\
&= \int_0^\infty \lambda e^{-\lambda x_1} e^{-\lambda {(x_1-z)}} dx_1 \\
&= \int_0^\infty \lambda e^{-\lambda {(2x_1-z)}} dx_1 \\
&= e^{\lambda z}\int_0^\infty \lambda e^{-\lambda {(2x_1)}} dx_1 \\
&= e^{\lambda z} \cdot \frac{1}{2} \\
\end{aligned}
\]
Differentiating, \(f_Z(z) = \frac{\lambda e^{\lambda z}}{2}\) for \(z < 0\) .
In the case where \(z > 0\), the CDF is:
\[ \begin{aligned} F_Z(z) &= Pr(Z \le z) \\ &= 1-Pr(Z > z) \\ &= 1-Pr(X_1 - X_2 > z) \\ &= Pr(X_2 \ge X_1 + z) \\ &= \int_0^\infty \int_0^{x_1+z} f_{X_1,X_2}(x_1,x_2)dx_2 dx_1 \\ &= \int_0^\infty \int_0^{x_1+z} \lambda^2 e^{-\lambda x_1}e^{-\lambda x_2}dx_2 dx_1 \\ &= \int_0^\infty \lambda e^{-\lambda x_1} \left[\int_0^{x_1+z} \lambda e^{-\lambda x_2}dx_2 \right]dx_1 \\ &= \int_0^\infty \lambda e^{-\lambda x_1} \left[1-e^{-\lambda x_2} \right]_0^{x_1+z} dx_1 \\ &= \int_0^\infty \lambda e^{-\lambda x_1} \left[1-e^{-\lambda {(x_1+z)}}-(1-1) \right] dx_1 \\ &= \int_0^\infty \lambda e^{-\lambda x_1} \left[1-e^{-\lambda {(x_1+z)}} \right] dx_1 \\ &= \int_0^\infty \lambda e^{-\lambda x_1} -\lambda e^{-\lambda x_1}e^{-\lambda {(x_1+z)}} dx_1 \\ &= \int_0^\infty \lambda e^{-\lambda x_1} -\lambda e^{-2\lambda x_1}e^{-\lambda z} dx_1 \\ &= \int_0^\infty \lambda e^{-\lambda x_1} -\int_0^\infty \lambda e^{-2\lambda x_1}e^{-\lambda z} dx_1 \\ &= 1 -e^{-\lambda z}\int_0^\infty \lambda e^{-2\lambda x_1} dx_1 \\ &= 1-e^{-\lambda z}\int_0^\infty \lambda e^{-2\lambda x_1} dx_1 \\ &= 1-e^{-\lambda z} \frac{1}{2} \\ &= 1-\int_0^\infty \lambda e^{-\lambda x_1} e^{-\lambda {(x_1-z)}} dx_1 \\ &= 1-\int_0^\infty \lambda e^{-\lambda {(2x_1-z)}} dx_1 \\ &= 1-e^{\lambda z}\int_0^\infty \lambda e^{-\lambda {(2x_1)}} dx_1 \\ &= 1-e^{-\lambda z} \cdot \frac{1}{2} \\ \end{aligned} \]
Differentiating, \(f_Z(z) = \frac{\lambda e^{-\lambda z}}{2}\) for \(z \ge 0\) .
Thus, the density for all z is \(f_Z(z) = \frac{\lambda e^{-\lambda |z|}}{2}\)
This is the Laplace distribution, also known as the “Double Exponential” distribution.
Chebychev’s Inequality states that under the above conditions, \(Pr(|X-\mu| \ge \epsilon) \le \frac{\sigma^2}{\epsilon^2} , \forall \epsilon>0\).
If \(\epsilon = k\cdot \sigma\), this can be rewritten as
\(Pr(|X-\mu| \ge k \cdot \sigma) \le \frac{\sigma^2}{k^2 \cdot \sigma^2}=\frac{1}{k^2} , \forall \epsilon>0\)
\(\epsilon = 2 = k\cdot \sigma = k \cdot \sqrt{\frac{100}{3}}\) \(\epsilon^2 = 2^2=4=k^2 \cdot \frac{100}{3}\) \(\frac{1}{k^2} = \frac{100}{3\cdot4}=\frac{100}{12}=8.33333\)
Because probabilities cannot be greater than 1, the upper bound is 1.
\(\epsilon = 5 = k\cdot \sigma = k \cdot \sqrt{\frac{100}{3}}\) \(\epsilon^2 = 5^2=25=k^2 \cdot \frac{100}{3}\) \(\frac{1}{k^2} = \frac{100}{3\cdot25}=\frac{100}{75}=1.33333\)
Because probabilities cannot be greater than 1, the upper bound is 1.
\(\epsilon = 9 = k\cdot \sigma = k \cdot \sqrt{\frac{100}{3}}\) \(\epsilon^2 = 9^2=81=k^2 \cdot \frac{100}{3}\) \(\frac{1}{k^2} = \frac{100}{3\cdot81}=\frac{100}{243} \approx 0.4115\)
Thus, the upper bound is \(\frac{100}{243} \approx 0.4115\).
\(\epsilon = 20 = k\cdot \sigma = k \cdot \sqrt{\frac{100}{3}}\) \(\epsilon^2 = 20^2=400=k^2 \cdot \frac{100}{3}\) \(\frac{1}{k^2} = \frac{100}{3\cdot400}=\frac{100}{1200} \approx 0.083333\)
Thus, the upper bound is \(\frac{1}{12} \approx 0.083333\).