EXAMPLE 5.22

Service time for a certain type of bank transaction is a random variable having an exponential distribution with parameter $\lambda$ . Suppose $X_1$ and $X_2$ are service times for two different customers, assumed independent of each other. Consider the total service time $T_o=X_1+X_2$ for the two customers, also a statistic. The cdf of $T_o$ is, for $t\ge 0$ ,

$$\begin{array}{rl}{F}_{T_0}\left(t\right)& =P\left(X_1+X_2\le t\right)={\iint }_{\left\{\left(x_1,x_2\right):x_1+x_2\le t\right\}}f\left(x_1,x_2\right)d{x}_{1}d{x}_2\\ & ={\int }_0^t{\int }_0^{t-x_1}\lambda e^{-\lambda x_1}\cdot \lambda e^{-\lambda x_2}d{x}_{2}d{x}_1\\ & ={\int }_0^t\left[\lambda e^{-\lambda x_1}-\lambda e^{-\lambda t}\right]d{x}_1\\ & =1-e^{-\lambda t}-\lambda t{e}^{-\lambda t}.\end{array}$$

The region of integration is pictured in Figure 5.10.

Figure 5.10 Region of integration to obtain cdf of $T_o$ in Example 5.22

The pdf of $T_o$ is obtained by differentiating $F_{T_o}\left(t\right)$ :

$$f_{T_o}\left(t\right)=\{\begin{array}{cc}{\lambda }^{2}t{e}^{-\lambda t}& t\ge 0\\ 0& t<0\end{array}\text{\hspace{1em}(5.5)}$$

This is a gamma $\mathrm{pdf}\left(\alpha =2\text{and}\beta =1/\lambda \right)$ . The pdf of $\bar{X}=T_o/2$ is obtained from the relation $\{\bar{X}\le \bar{x}\}$ iff $\left\{T_o\le 2\bar{x}\right\}$ as

$$f_{\bar{X}}\left(\bar{x}\right)=\{\begin{array}{cc}4{\lambda }^2\bar{x}{e}^{-2\lambda \bar{x}}& \bar{x}\ge 0\\ 0& \bar{x}<0\end{array}\text{\hspace{1em}(5.6)}$$

The mean and variance of the underlying exponential distribution are $\mu =1/\lambda$ and ${\sigma }^2=1/{\lambda }^2$ . From Expressions (5.5) and (5.6), it can be verified that $E\left(\bar{X}\right)=1/\lambda$ , $V\left(\bar{X}\right)=1/\left(2{\lambda }^2\right),E\left(T_o\right)=2/\lambda$ , and $V\left(T_o\right)=2/{\lambda }^2$ . These results again suggest some general relationships between means and variances of $\bar{X},T_o$ , and the underlying distribution.