Chapter 6.8, Problem 18E

To determine

To Calculate: The standard deviation for a given exponential distribution function.

Answer to Problem 18E

The standard deviation of an exponential distribution is equal to its mean $\text{μ}$ .

Explanation of Solution

Given Information: The random variable is exponentially distributed with mean $\text{μ}$ .

Method Used: The standard deviation for a probability distribution function $f\left(x\right)$ is given by

  $\sigma ={\left[\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left(x-\text{μ}\right)}^{2}f\left(x\right)dx\right]}^{\frac{1}{2}}$

Calculation: The mean of the exponential distribution is $\text{μ}$ . So, the probability density function $f\left(x\right)$ is

  $f\left(x\right)=\frac{1}{\text{μ}}{e}^{-\frac{x}{\text{μ}}}$ if $x\ge 0$

  $f\left(x\right)=0$ if $x<0$

Therefore,

  $\sigma ={\left[\underset{0}{\overset{\infty }{{\displaystyle \int }}}{\left(x-\text{μ}\right)}^2\cdot \frac{1}{\text{μ}}{e}^{-\frac{x}{\text{μ}}}dx\right]}^{\frac{1}{2}}$

  $={\left\{{\left[{\left(x-\text{μ}\right)}^2\cdot \frac{1}{\text{μ}}\left(-\text{μ}\right)\cdot e^{-\frac{x}{\text{μ}}}\right]}_0^{\infty }-\underset{0}{\overset{\infty }{{\displaystyle \int }}}2\left(x-\text{μ}\right)\cdot \frac{1}{\text{μ}}\left(-\text{μ}\right)\cdot e^{-\frac{x}{\text{μ}}}\right\}}^{\frac{1}{2}}$

  $={\left[{\text{μ}}^2+2\underset{0}{\overset{\infty }{{\displaystyle \int }}}\left(x-\text{μ}\right)e^{-\frac{x}{\text{μ}}}dx\right]}^{\frac{1}{2}}$

  $={\left[{\text{μ}}^2+2\left\{{\left[\left(x-\text{μ}\right)\cdot \left(-\text{μ}\right)e^{-\frac{x}{\text{μ}}}\right]}_0^{\infty }-\underset{0}{\overset{\infty }{{\displaystyle \int }}}\left(-\text{μ}\right)e^{-\frac{x}{\text{μ}}}dx\right\}\right]}^{\frac{1}{2}}$

  $={\left[{\text{μ}}^2-2{\text{μ}}^2+2\text{μ}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{e}^{-\frac{x}{\text{μ}}}dx\right]}^{\frac{1}{2}}$

  $={\left[-{\text{μ}}^2+2{\text{μ}}^2\right]}^{\frac{1}{2}}$

  $=\text{μ}$

Therefore, the standard deviation of an exponential distribution is equal to its mean $\mu$ .