I have a system consisting of components $S_1$ and $S_2$ whose lifetimes $T_1$ and $T_2$ follow the exponential distribution with parameter $\lambda$. At time $t=0$ the component $S_1$ is switched on and $S_2$ is kept off until $S_1$ fails (and is immediately switched on). What is the distribution of the lifetime of the system?

To me the logical solution would be $f(T_1,\lambda)+f(T_2,\lambda)$ where f is the probability density function $f(x,\lambda)=\lambda e^{-\lambda x}$.

$\Rightarrow$ $\lambda e^{-\lambda T_1}+\lambda e^{-\lambda T_2}$

but I have nothing to verify it with. Am I on the right track?


That is wrong, you can't just add density functions.

The CDF of the sum of any two independent continuous random variables can be computed using the law of total probability $$ F_{X+Y}(u)=P(X+Y\leq u)=\int_{-\infty}^\infty f_X(x)P(Y<u-x)dx. $$

Plug in the exponential distribution and you can compute the integral. Make sure you get the limits of the integral right ($X$ and $Y$ cannot be negative).

To get the density you take derivative with respect to $u$. If you do everything correctly you should get a Gamma distribution with parameters $2$ and $\lambda$.

  • $\begingroup$ I must be doing something wrong because my integral does not converge. I'm interpreting $P(Y<u-x)=F_y(Y<u-x)=1-e^{-\lambda (u-x)}$, is this correct? That gives $F_{X+Y}(u)=P(X+Y\leq u)=\int_{0}^\infty \lambda e^{-\lambda x}(1-e^{-\lambda (u-x)})$, which according to my calculations does not converge. $\endgroup$ – Markus K. Apr 1 '15 at 15:16
  • $\begingroup$ The integral limit is wrong, x cannot be greater than the sum: i.e. $0\leq x\leq u$. $\endgroup$ – QQQ Apr 1 '15 at 16:38

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