1. ## Lifetime Distribution of a parallel system

Suppose under normal operating conditions the operating time in days until failure of a certain type of component has an exponential distribution with mean Lamda(> 0). Suppose a system comprising of three such components works if at least one of the components works. You may assume that the random variables representing lifetimes of different components in the system to be independent.

(i)Find the life time distribution of the system.
(ii)Without finding the mean prove that the mean life time of the system is at least as large as that of a single component
(iii)Prove (ii) if mean is replaced by median
(iv)Find the variance of the life time of the system and compare it with that of a single component.

I ve got the answer to (i). But m stuck wid the rest.

Can someone pls help me to approach parts ii, iii and iv ?

1. The lifetime $\displaystyle X$ of a single component has cumulative distribution function $\displaystyle F(t)=\mathrm P(X\leq t)=1-\mathrm e^{-\lambda t}$.

The lifetime $\displaystyle T$ of the system has cumulative distribution function $\displaystyle R(t)=\mathrm P(T\leq t)=(1-\mathrm e^{-\lambda t})^3=F(t)^3$.

Notice since $\displaystyle 0\leq F(t)\leq 1$ that $\displaystyle R(t)\leq F(t)$ for all $\displaystyle t\geq0$.

2. The mean lifetime of a single component is $\displaystyle \mathrm E(X)=\int_0^\infty\mathrm P(X>t)\mathrm dt=\int_0^\infty \mathrm e^{-\lambda t}\mathrm dt=\frac1\lambda$.

The mean lifetime of the system is $\displaystyle \mathrm E(T)=\int_0^\infty\mathrm P(T>t)\mathrm dt$.

But $\displaystyle \mathrm P(T>t)=1-R(t)\geq 1-F(t)=\mathrm P(X>t)$ for all $\displaystyle t\geq0$.

Thus $\displaystyle \mathrm E(T)\geq\mathrm E(X)$.
3. The median lifetime of a single component is $\displaystyle m$ where $\displaystyle F(m)=\frac12$.

The median lifetime of the system is $\displaystyle M$ where $\displaystyle R(M)=\frac12$.

But $\displaystyle R(m)=F(m)^3=\frac18<R(M)$, so that $\displaystyle m<M$.

4. It is well known that $\displaystyle \mathrm{Var}(X)=\frac1{\lambda^2}$.

Now $\displaystyle \mathrm E(T)=\int_0^\infty(1-R(t))\mathrm dt=\int_0^\infty(3\mathrm e^{-\lambda t}-3\mathrm e^{-2\lambda t}+\mathrm e^{-3\lambda t})\mathrm dt=\frac3\lambda-\frac3{2\lambda}+\frac1{3\lambda}=\frac{11}{6\lamb da}$.

Also $\displaystyle \mathrm E(T^2)=\int_0^\infty 2t(1-R(t))\mathrm dt=\int_0^\infty 2t(3\mathrm e^{-\lambda t}-3\mathrm e^{-2\lambda t}+\mathrm e^{-3\lambda t})\mathrm dt$,

and so $\displaystyle \mathrm E(T^2)=-\frac{\mathrm d}{\mathrm d\lambda}\int_0^\infty (6\mathrm e^{-\lambda t}-3\mathrm e^{-2\lambda t}+\mathrm {\textstyle\frac23}e^{-3\lambda t})\mathrm dt=-\frac{\mathrm d}{\mathrm d\lambda}\left(\frac6\lambda-\frac3{2\lambda}+\frac2{9\lambda}\right)=\frac{85} {18\lambda^2}$.

Therefore $\displaystyle \mathrm{Var}(T)=\frac{85}{18\lambda^2}-\left(\frac{11}{6\lambda}\right)^2=\frac{49}{36\la mbda^2}=\left({\textstyle\frac76}\right)^2\mathrm{ Var}(X)$.