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?


1 Answer 1


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, 2015 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, 2015 at 16:38

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