The survival times $X$ and $Y$ for two components of a system are independent exponentially-distributed random variables. X has density $f_X(x) = \frac{2}{3}e^{-2x/3}$ and Y has density $f_Y(y)=\frac{1}{2}e^{-y/2}.$ The system works as long as either component continues to function. Find the probability density function for the survival time, $Z$, of the system.

My method:

1) Find the probability that both components fail, to end up with a cumulative distribution function for the survival of Z.

2) Differentiate the CDF to get the PDF of Z.

However, I do not know how to put Z in terms of X and Y. Certainly it is not $Z=X+Y$, since the system works when EITHER component is working. Would $Z=X+Y-XY$ be the correct way to relate the three variables?



Your general approach is correct, just note that:


$P(Z \le z)=P(X \le z, Y \le z)$

$F_Z(z)=F_X(z)F_Y(z) \\ f_Z(z)=\frac{d}{dz}F(z)=\frac{d}{dz}\left(F_X(z)F_Y(z)\right)=f_X(z)F_Y(z)+f_Y(z)F_X(z)$

May be computing $F_Z(z)$ directly and then differentiating be less liable to mistakes.

  • $\begingroup$ Since the system works as long as one component is functioning, it would take both components to fail to cause the whole system to fail. In that case, shouldn't $P(Z<z) = P(X<z, Y<z)$? $\endgroup$ – Luchia Dec 11 '13 at 19:18
  • $\begingroup$ @Luchia Yes you are right, my mistake. $\endgroup$ – hhsaffar Dec 11 '13 at 19:24
  • $\begingroup$ @Luchia I updated my answer. $\endgroup$ – hhsaffar Dec 11 '13 at 19:26
  • $\begingroup$ @Luchia And be careful when you are differentiating! $\endgroup$ – hhsaffar Dec 11 '13 at 19:35

If the components are independent then you multiply their CDFs and then differentiate

  • $\begingroup$ Yes, but then it will be in terms of X and Y, not Z. How can I get it to end up in terms of Z? $\endgroup$ – Luchia Dec 11 '13 at 17:19
  • $\begingroup$ Let z be the failure time then substitute z into each CDF and multiply $\endgroup$ – user76844 Dec 12 '13 at 0:17

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