Chapter 4, Problem 113SE

a.

To determine

Obtain $E\left(X\right)$ and $V\left(X\right)$.

Answer to Problem 113SE

The expression for $E\left(X\right)$ is:

$\underset{\_}{E\left(X\right)=\frac{p}{{\lambda }_1}+\frac{\left(1-p\right)}{{\lambda }_2}}$

The expression for $V\left(X\right)$ is:

$\underset{\_}{V\left(X\right)=\frac{2p}{{\lambda }_1^2}+\frac{2\left(1-p\right)}{{\lambda }_2^2}-{\left[\frac{p}{{\lambda }_1}+\frac{\left(1-p\right)}{{\lambda }_2}\right]}^2}$

Explanation of Solution

Given info:

The fusion of two exponential distributions for the modeling behavior is termed as “criticality level of the situation” X by the authors.

The probability density function of X is given below:

$f\left(x;{\lambda }_1,{\lambda }_2,p\right)=\{\begin{array}{l}p{\lambda }_1{e}^{-{\lambda }_{1}x}+\left(1-p\right){\lambda }_2{e}^{-{\lambda }_{2}x}x\ge 0\\ 0\text{Otherwise}\end{array}$

Calculation:

The mean of the exponential distribution is:

${\displaystyle \underset{0}{\overset{\infty }{\int }}x\lambda e^{-\lambda x}dx}=\frac{1}{\lambda }$

The variance of the exponential distribution is:

$\begin{array}{c}E\left(X^2\right)=V\left(X\right)+{\left[E\left(X\right)\right]}^2\\ =\frac{1}{{\lambda }^2}+\frac{1}{{\lambda }^2}\\ =\frac{2}{{\lambda }^2}\end{array}$

The expected mean is obtained as given below:

$\begin{array}{c}\mu =E\left(X\right)\\ ={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}xf\left(x\right)dx}\\ ={\displaystyle \underset{0}{\overset{\infty }{\int }}x\left[p{\lambda }_1{e}^{-{\lambda }_{1}x}+\left(1-p\right){\lambda }_2{e}^{-{\lambda }_{2}x}\right]dx}\\ ={\displaystyle \underset{0}{\overset{\infty }{\int }}xp{\lambda }_1{e}^{-{\lambda }_{1}x}dx}+{\displaystyle \underset{0}{\overset{\infty }{\int }}x\left(1-p\right){\lambda }_2{e}^{-{\lambda }_{2}x}dx}\\ =\frac{p}{{\lambda }_1}+\frac{1-p}{{\lambda }_2}\end{array}$

Thus, the expected mean for mixed exponential distributions is:

$\underset{\_}{E\left(X\right)=\frac{p}{{\lambda }_1}+\frac{\left(1-p\right)}{{\lambda }_2}}$

The variance of a continuous random variable is given as ${\sigma }^2=V\left(X\right)=E\left(X^2\right)-{\left(E\left(X\right)\right)}^2$.

Here,

$\begin{array}{c}E\left(X^2\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{x}^{2}f\left(x\right)dx}\\ ={\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^2\left[p{\lambda }_1{e}^{-{\lambda }_{1}x}+\left(1-p\right){\lambda }_2{e}^{-{\lambda }_{2}x}\right]dx}\\ ={\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^{2}p{\lambda }_1{e}^{-{\lambda }_{1}x}dx}+{\displaystyle \underset{0}{\overset{\infty }{\int }}{x}^2\left(1-p\right){\lambda }_2{e}^{-{\lambda }_{2}x}dx}\\ =\frac{2p}{{\lambda }_1^2}+\frac{2\left(1-p\right)}{{\lambda }_2^2}\end{array}$

Thus, the variance is:

$\begin{array}{c}V\left(X\right)=E\left(X^2\right)-{\left(E\left(X\right)\right)}^2\\ =\frac{2p}{{\lambda }_1^2}+\frac{2\left(1-p\right)}{{\lambda }_2^2}-{\left[\frac{p}{{\lambda }_1}+\frac{\left(1-p\right)}{{\lambda }_2}\right]}^2\end{array}$

Thus, the $\underset{\_}{V\left(X\right)=\frac{2p}{{\lambda }_1^2}+\frac{2\left(1-p\right)}{{\lambda }_2^2}-{\left[\frac{p}{{\lambda }_1}+\frac{\left(1-p\right)}{{\lambda }_2}\right]}^2}$.

b.

To determine

Find the cumulative distribution of X.

Answer to Problem 113SE

The cumulative distribution function of X is:

$\underset{\_}{F\left(x\right)=\{\begin{array}{l}0x\le 0\\ p\left(1-e^{-{\lambda }_{1}x}\right)+\left(1-p\right)\left(1-e^{-{\lambda }_{2}x}\right)x>0\end{array}}$

Explanation of Solution

Denote the cumulative distribution function (cdf) of X as $F\left(x\right)$. The cdf of X here, at a value x, is:

For $x\le 0$,

Since X takes the positive values from the interval $\left[0,\infty \right]$, then $F\left(x\right)=0$.

For $x>0$,

$\begin{array}{c}F\left(x\right)={\displaystyle \underset{0}{\overset{x}{\int }}f\left(x\right)dx}\\ ={\displaystyle \underset{0}{\overset{x}{\int }}\left[p{\lambda }_1{e}^{-{\lambda }_{1}x}+\left(1-p\right){\lambda }_2{e}^{-{\lambda }_{2}x}\right]dx}\\ ={\displaystyle \underset{0}{\overset{x}{\int }}p{\lambda }_1{e}^{-{\lambda }_{1}x}dx}+{\displaystyle \underset{0}{\overset{x}{\int }}\left(1-p\right){\lambda }_2{e}^{-{\lambda }_{2}x}dx}\\ =p{\lambda }_1{\displaystyle \underset{0}{\overset{x}{\int }}{e}^{-{\lambda }_{1}x}dx}+\left(1-p\right){\lambda }_2{\displaystyle \underset{0}{\overset{x}{\int }}{e}^{-{\lambda }_{2}x}dx}\\ =p{\lambda }_1{\frac{e^{-{\lambda }_{1}x}}{-{\lambda }_1}]}_0^x+\left(1-p\right){\lambda }_2{\frac{e^{-{\lambda }_{2}x}}{-{\lambda }_2}]}_0^x\end{array}$

$\begin{array}{c}=p{\lambda }_1\left(\frac{e^{-{\lambda }_{1}x}}{-{\lambda }_1}+\frac{e^0}{{\lambda }_1}\right)+\left(1-p\right){\lambda }_2\left(\frac{e^{-{\lambda }_{2}x}}{-{\lambda }_2}+\frac{e^0}{{\lambda }_2}\right)\\ =p\frac{{\lambda }_1}{{\lambda }_1}\left(-e^{-{\lambda }_{1}x}+1\right)+\left(1-p\right)\frac{{\lambda }_2}{{\lambda }_2}\left(-e^{-{\lambda }_{2}x}+1\right)\\ =p\left(1-e^{-{\lambda }_{1}x}\right)+\left(1-p\right)\left(1-e^{-{\lambda }_{2}x}\right)\end{array}$

Hence, the cumulative distribution function of X is:

$\underset{\_}{F\left(x\right)=\{\begin{array}{l}0x\le 0\\ p\left(1-e^{-{\lambda }_{1}x}\right)+\left(1-p\right)\left(1-e^{-{\lambda }_{2}x}\right)x>0\end{array}}$

c.

To determine

Find the value of $P\left(X>0.01\right)$.

Answer to Problem 113SE

The value of $P\left(X>0.01\right)$ is 0.403.

Explanation of Solution

Calculation:

The value of $P\left(X>0.01\right)$ is obtained as shown below:

Substitute $p=0.5$${\lambda }_1=40$ and ${\lambda }_2=200$

$\begin{array}{c}P\left(X>0.01\right)=1-P\left(X\le 0.01\right)\\ =1-F\left(0.01\right)\\ =1-\left[0.5\left(1-e^{-40\left(0.01\right)}\right)+\left(1-0.5\right)\left(1-e^{-200\left(0.01\right)}\right)\right]\\ =1-\left[0.5\left(1-e^{-0.4}\right)+\left(1-0.5\right)\left(1-e^{-2}\right)\right]\\ =1-0.5+0.5e^{-0.4}-0.5+0.5e^{-2}\end{array}$

$\begin{array}{c}=0.5e^{-0.4}+0.5e^{-2}\\ =0.5\left(0.67\right)+0.06767\\ =0.335+0.06767\\ =0.403\end{array}$

Thus, the value of $P\left(X>0.01\right)$ is 0.403.

d.

To determine

Find the probability that the X is within one standard deviation from its mean value.

Answer to Problem 113SE

The probability that the X is within one standard deviation from its mean value is 0.879.

Explanation of Solution

Calculation:

From part (a), the expected mean for two exponential distributions is:

$E\left(X\right)=\frac{p}{{\lambda }_1}+\frac{\left(1-p\right)}{{\lambda }_2}$

Substitute $p=0.5$${\lambda }_1=40$ and ${\lambda }_2=200$

$\begin{array}{c}E\left(X\right)=\frac{0.5}{40}+\frac{\left(1-0.50\right)}{200}\\ =0.0125+0.0025\\ =0.015\end{array}$

The standard deviation of X is given by:

$\begin{array}{c}\sigma =\sqrt{V\left(X\right)}\\ =\sqrt{\frac{2\left(0.5\right)}{{40}^2}+\frac{2\left(1-0.5\right)}{{200}^2}-{\left[0.015\right]}^2}\\ =\sqrt{0.000625+0.000025-{\left[0.015\right]}^2}\end{array}$

$\begin{array}{c}=\sqrt{0.00065-0.000225}\\ =\sqrt{0.000425}\\ =0.0206\end{array}$

The probability that the X is within one standard deviation from its mean value is obtained as shown below:

$\begin{array}{c}P\left(\left|X-\mu \right|\le \sigma \right)=P\left(\mu -\sigma \le X\le \mu +\sigma \right)\\ =P\left(0.015-0.0206\le X\le 0.015+0.0206\right)\\ =P\left(-0.0056\le X\le 0.0356\right)\end{array}$

Here, the random variable X do not takes negative value.

$\begin{array}{c}P\left(X<0.0356\right)=F\left(0.0356\right)\\ =\left[0.5\left(1-e^{-40\left(0.0356\right)}\right)+\left(1-0.5\right)\left(1-e^{-200\left(0.0356\right)}\right)\right]\\ =\left[0.5\left(1-e^{-1.424}\right)+\left(1-0.5\right)\left(1-e^{-7.12}\right)\right]\end{array}$

$\begin{array}{c}=1-0.5e^{-1.424}-0.5e^{-7.12}\\ =1-0.1204-0.000404\\ =0.879\end{array}$

Thus, the probability that the X is within one standard deviation from its mean value is 0.879.

e.

To determine

Find the coefficient of variation for an exponential random variable.

Explain about the value of coefficient of variation for a hyper exponential distribution.

Answer to Problem 113SE

The coefficient of variation for an exponential random variable is 1.

Explanation of Solution

Calculation:

It is given that the Coefficient of variation for a random variable X is:

$CV=\frac{\sigma }{\mu }$

From part (a), $\mu =\frac{1}{\lambda }$ and $\sigma =\frac{1}{\lambda }$ for single exponential distribution.

On simplifying,

$\begin{array}{c}CV=\frac{\frac{1}{\lambda }}{\frac{1}{\lambda }}\\ =1\end{array}$

The coefficient of variation for random variable X in hyper exponential distribution is:

$\begin{array}{c}CV=\frac{\sqrt{E\left(X^2\right)-{\mu }^2}}{\mu }\\ =\sqrt{\frac{E\left(X^2\right)-{\mu }^2}{{\mu }^2}}\\ =\sqrt{\frac{E\left(X^2\right)}{{\mu }^2}-\frac{{\mu }^2}{{\mu }^2}}\\ =\sqrt{\frac{\frac{2p}{{\lambda }_1^2}+\frac{2\left(1-p\right)}{{\lambda }_2^2}}{{\left[\frac{p}{{\lambda }_1}+\frac{\left(1-p\right)}{{\lambda }_2}\right]}^2}-1}\\ =\sqrt{\frac{\frac{2p{\lambda }_2^2+2\left(1-p\right){\lambda }_1^2}{{\lambda }_1^2+{\lambda }_2^2}}{\frac{{\left(p{\lambda }_2+\left(1-p\right){\lambda }_1\right)}^2}{{\lambda }_1^2+{\lambda }_2^2}}-1}\\ =\sqrt{\frac{2p{\lambda }_2^2+2\left(1-p\right){\lambda }_1^2}{{\left(p{\lambda }_2+\left(1-p\right){\lambda }_1\right)}^2}-1}\end{array}$

Denote $r=\frac{p{\lambda }_2^2+2\left(1-p\right){\lambda }_1^2}{{\left(p{\lambda }_2+\left(1-p\right){\lambda }_1\right)}^2}$

Then,

$CV=\sqrt{2r-1}$

The algebraic expression shows that the value of $r>1$ when ${\lambda }_1\ne {\lambda }_2$, the value of coefficient of variation is greater than 1.

Thus, the value of coefficient of variation is greater than 1.

f.

To determine

Find the coefficient of variation for an Erlang distribution.

Answer to Problem 113SE

The coefficient of variation for an Erlang distribution is $\underset{\_}{\frac{1}{\sqrt{n}}}$.

Explanation of Solution

Calculation:

The expected mean for the Erlang distribution is:

$E\left(X\right)=\frac{n}{\lambda }$

The standard deviation for the Erlang distribution is:

$V\left(X\right)=\frac{\sqrt{n}}{\lambda }$

The coefficient of variation for an Erlang distribution is given by:

$\begin{array}{c}CV=\frac{\frac{\sqrt{n}}{\lambda }}{\frac{n}{\lambda }}\\ =\frac{\sqrt{n}}{\lambda }\times \frac{\lambda }{n}\\ =\frac{1}{\sqrt{n}}\end{array}$

Thus, the coefficient of variation for the an Erlang distribution is $\underset{\_}{\frac{1}{\sqrt{n}}\left(<1\right)}$ for $n>1$.