Consider two diods connected in parallel. Suppose the life times T1,T2 of each diod have the same distribution function F.What is the life time T of the whole system? What is the distribution of T?

My intuition was that T= Max (T1,T2) because it is parallel, even if one failed, the other would still work. As a result, the distribution of T is just F. Can anyone please help me out? Many thanks.

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    $\begingroup$ Is system considered "alive" when one diode fails? (For example, if failed diode starts conducting current in both directions, the answer would probably be "no") Also, distribution of $\max(T_1,T_2)$ isn't $F$ in general case. $\endgroup$ – Abstraction Apr 5 '16 at 9:59
  • $\begingroup$ Thank you so much for providing some ideas. What is the distribution of max of T1,T2 ? $\endgroup$ – Yilin Wang Apr 5 '16 at 20:06
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    $\begingroup$ Distribution of $T=\max \{T_1,T_2\}$ is $$F_T(t)=P(T\leq t)=P(T_1\leq t\cap T_2\leq t)=P(T_1\leq t)P(T_2\leq t)=[F(t)]^2.$$ $\endgroup$ – Mick A Apr 6 '16 at 4:36
  • $\begingroup$ @MickA You are awesome baby $\endgroup$ – Yilin Wang Apr 11 '16 at 6:17

It's probably better to expand Mick A comment.

First of all, the comment is right when $T_1$ and $T_2$ are independent. Note that in real system that's usually not the case since there are events fatal for both diodes. For independent model, as it was shown, $F_{\text{max}}(t) = F(t)^2$ thus probability density $p_{\text{max}}(t) = 2p(t)F(t)$ and expectation of lifetime ("average lifetime") $ET_{\text{max}} = 2\int_0^{+\infty}tp(t)F(t)dt$ (if average lifetime of a single diode exists - it should - then this integral converges since it's less than $2\int_0^{+\infty}tp(t)dt = 2ET$). There seems to be no reasonable way to express $ET_\text{max}$ in terms of moments of $T$ in general case (see this question for uniform distribution case, this and this for Gaussian (normal)).

One possible trick for computing $ET_\text{max}$ is taking $u=F(t)$, then $ET_\text{max}=2\int_0^1uF^{-1}(u)du$.


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