EXAMPLE 5.11

If $X_1,\dots ,X_n$ represent the lifetimes of $n$ components, the components operate independently of one another, and each lifetime is exponentially distributed with parameter $\lambda$, then for $x_1\ge 0,x_2\ge 0,\dots ,x_n\ge 0$,

$$f\left(x_1,x_2,\dots ,x_n\right)=\left(\lambda e^{-\lambda x_1}\right)\cdot \left(\lambda e^{-\lambda x_2}\right)\cdot \cdots \cdot \left(\lambda e^{-\lambda x_n}\right)={\lambda }^n{e}^{-\lambda \sum x_i}$$

Suppose a system consisting of these components will fail as soon as a single component fails. Let $T$ represent system lifetime. Then the probability that the system lasts past time $t$ is

$$\begin{array}{rl}P\left(T>t\right)& =P\left(X_1>t,\dots ,X_n>t\right)\\ & ={\int }_t^{\infty }\dots {\int }_t^{\infty }f\left(x_1,\dots ,x_n\right)d{x}_1\dots d{x}_n\\ & =\left({\int }_t^{\infty }\lambda e^{-\lambda x_1}d{x}_1\right)\dots \left({\int }_t^{\infty }\lambda e^{-\lambda x_n}d{x}_n\right)\\ & ={\left(e^{-\lambda t}\right)}^n=e^{-n\lambda t}.\end{array}$$

Therefore,

$$P\left(\text{ system lifetime }\le t\right)=1-e^{-n\lambda t}\text{ for }t\ge 0$$

which shows that system lifetime has an exponential distribution with parameter $n\lambda$ ; the expected value of system lifetime is $1/n\lambda$.

A variation on the foregoing scenario appeared in the article “A Method for Correlating Field Life Degradation with Reliability Prediction for Electronic Modules” (Quality and Reliability Engr. Intl., 2005: 715-726). The investigators considered a circuit card with $n$ soldered chip resistors. The failure time of a card is the minimum of the individual solder connection failure times (mileages here). It was assumed that the solder connection failure mileages were independent, that failure mileage would exceed $t$ if and only if the shear strength of a connection exceeded a threshold $d$, and that each shear strength was normally distributed with a mean value and standard deviation that depended on the value of mileage $t:\mu \left(t\right)=a_1-a_{2}t$ and $\sigma \left(t\right)=a_3+a_{4}t$ (a weld’s shear strength typically deteriorates and becomes more variable as mileage increases). Then the probability that the failure mileage of a card exceeds $t$ is

$$P\left(T>t\right)={\left(1-\Phi \left(\frac{d-(a_1-a_{2}t}{a_3+a_{4}t}\right)\right)}^n$$

The cited article suggested values for $d$ and the $a_i$ ’s based on data. In contrast to the exponential scenario, normality of individual lifetimes does not imply normality of system lifetime.