# The lifetime of two parallel connected diods

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.

• 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. – Abstraction Apr 5 '16 at 9:59
• Thank you so much for providing some ideas. What is the distribution of max of T1,T2 ? – Yilin Wang Apr 5 '16 at 20:06
• Distribution of $T=\max \{T_1,T_2\}$ is $$F_T(t)=P(T\leq t)=P(T_1\leq t\cap T_2\leq t)=P(T_1\leq t)P(T_2\leq t)=[F(t)]^2.$$ – Mick A Apr 6 '16 at 4:36
• @MickA You are awesome baby – Yilin Wang Apr 11 '16 at 6:17

## 1 Answer

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 $ET_\text{max}=2\int_0^1uF^{-1}(u)du$.