Finding a pdf of time until failure of a system

Question

Suppose a system consists of five components that work independently. Suppose the lifetime of each component is exponentially distributed with a mean lifetime of half a year.

Let $T$ be the time when the second failure occurs. What is the pdf of $T$?

Apparently, $T$ is a random variable and I'm not sure how to tackle this problem. I am mainly stuck with how to relate the expenontial distribution of the components with $T$. Do I just need to compute $$P(C_1 < t, C_2 <t, \dotsc, C_5 < t)?$$ Here $C_i$'s are the $i$-th component's lifetime.
A hint would be appreciate it.

Answer 1

Sketch for a solution: we have two possible scenarios. In the first one we can assume that the system halts when one of it components broke, and in the second that it doesn't.

The first scenario is simpler, in this case we can represent the time to halt as $H:=\min_j C_j$ and the exercise ask for the pdf of $H_1+H_2$, where the $H_k$ are independent and identically distributed with the same distribution each one as of $H$.

By the other hand if we order the $C_j$ components from shorter to longer lifetimes then the second scenario asks for the pdf of the second order statistic of the $C_j$.

Answer 2

Denote the pdf for the lifetime of the 5 components by $f(t)$. Then since all the components function independently, and assuming that the machine continues operating until all of them break down, the joint distribution of the breakdown times is simply

$$g_{C_1,..., C_5}(t_1,...,t_5)=f(t_1)f(t_2)f(t_3)f(t_4)f(t_5)$$

Denote by $P(t_1<T_1,..., t_5<T_5)=\prod_{i=1}^5\int_0^{T_i} dx_i~ g_{C_1,..., C_5}(t_1,..., t_5)$ the multivariate cdf of the joint distribution, which describes the probability of individual breakdowns happening before times $T_1,..., T_5$.

Assume for the moment that component $k$ broke down first and $\ell$ second, $k\neq \ell$, the second one before time $T_\ell$. Note that the breakdown time of the first component is $t_k<T_\ell$ and the rest of the breakdown times $t_{i}, i\in {1,2,3,4,5}\backslash\{k,\ell\}$ happen after $T_\ell$. Of course, since we want the probability distribution of the time of the second failure, we have to integrate over any allowed possibilities for all other times, while maintaining the above constraints. Hence, the probability density of the second breakdown to happen before time $T_\ell$ is

$$P(t_k<t_\ell,t_\ell<T_\ell, t_i>t_\ell)=\int_0^{T_{\ell}} dt_{\ell}\int_0^{t_\ell} dt_k \prod_{i\neq k,\ell}\left(\int_{t_\ell}^{\infty}dt_{i}\right)g_{C_1,..., C_5}(t_1,...,t_5)~~~ ,~~k\neq \ell$$

Now note that the probability is always the same, no matter which set of components that fail first, $k,\ell$ we choose. $\ell$ can be chosen in 5 ways and hence $k$ can be chosen in 4, and thus the pdf for the second breakdown to happen at time $T$ is just

$$5\cdot 4 \frac{dP(t_1<t_2, t_2<T, t_{3,4,5}>t_2)}{dT}=20f(T)F(T)(1-F(T))^3$$

where $F(t)$ is the cdf of the lifetime of the component and we picked wlog $k=1,\ell=2$. One can easily check that this distribution is properly normalized, for any $f$.