# Distribution of a lifetime of a system

I have a quick question which I can't figure out how to start. I actually do not understand how to model the probabilities. can anyone help? Thanks. Here it is:

A system will function as long as at least one of the three components functions. When all three components are functioning, the distribution of the life of each is exponential with parameter $\frac{\lambda}{3}$. When only two are functioning, the distribution of the life of each is exponential with parameter $\frac{\lambda}{2}$; and when only one is functioning, the distribution of life is exponential with parameter $\lambda$.

(a) What is the distribution of the lifetime of the system?

(b) Suppose now that only one component (of the three components) is used and it is replaced when it fails. What is the distribution of the lifetime of such a system?

With regards to modeling the probability distributions, do the phrases in the given text mean that: $$f_{X_1|X_2,X_3}(x_1|x_2,x_3)=f_{X_2|X_1,X_3}(x_2|x_1,x_3)=f_{X_3|X_1,X_2}(x_3|x_1,x_2)=\frac{\lambda}{3}e^{\frac{-\lambda x_i}{3}}$$ $$f_{X_1|X_2}(x_1|x_2)=f_{X_1|X_3}(x_1|x_3)=f_{X_2|X_1}(x_2|x_1)=f_{X_2|X_3}(x_2|x_3)=f_{X_3|X_1}(x_3|x_1)=f_{X_3|X_2}(x_3|x_2)=\frac{\lambda}{2}e^{\frac{-\lambda x_i}{2}}$$ and $$f_{X_1}(x_1)=f_{X_2}(x_2)=f_{X_3}(x_3)=\lambda e^{-\lambda x_i}$$

• Hint: Exponential distributions are "without memory", that is once the first component dies, we can assume that we start with two fresh components remaining. – Hagen von Eitzen Jan 9 '14 at 15:28
• Yes, I know that. But my problem is more of the phrase "the distribution of the life of each is exponential etc." That is, for the case of "all three are functioning", is each $X_i$ modeled as a conditional prob. e.g., $X_1|X_2,X_3$? – math_stat_enthusiast Jan 9 '14 at 15:41
• Yes, that is right. You can think of three resistors that will sometimes fails, and as each one fails the remaining will receive more current, rising their own probability of failing. – Zado Jan 9 '14 at 15:53

## 1 Answer

The point is that the failure rate of resistors is constant at $\lambda$ regardless of how many are operating. You have a Poisson process where in each time interval $dt$ the chance of a failure is $\lambda dt$ The system fails if there have been three or more events, so at time $t$ the average number of failures is $\lambda t$ and the chance of failure by time $t$ is $\sum_{k=3}^\infty \frac {(\lambda t)^3}{k!}e^{-\lambda t}$

As I read it, you now need four failures with the same failure rate as above. So change $3$ to $4$. If you keep replacing components, it never fails.

• I don't get it. How did you go about the answer? And is this for (a) or (b)? Sorry, I find it hard to concretely understand the question. :( – math_stat_enthusiast Jan 10 '14 at 14:21
• The first paragraph is for a, the second for b. As there are three resistors before the first failure and each has failure rate $\lambda 3$, the total failure rate is $\lambda$. That doesn't change as they fail. You need three resistor failures for system failure. Does that make sense? – Ross Millikan Jan 10 '14 at 15:26
• I still don't get why. Doesn't it suppose to matter how many are still functioning since the individual distributions are actually dependent on how many are still working? – math_stat_enthusiast Jan 10 '14 at 15:40
• It is because the failure rate of each one changes depending on the number still working. If the failure rate were constant regardless of how many were working, failures would come 3 times as fast at the start as at the end. The constants were chosen to make it so. – Ross Millikan Jan 10 '14 at 15:52
• @Studentmath: that is correct. That way, the probability of failure in any interval of time is constant, even if there is a failure before or during. In an interval $dt$ it is always $\lambda dt$ That is how I justify the Poisson distribution-the failures are independent. It is just like you have one component with failure rate $\lambda^{-1}$ that is instantly repaired on failure. – Ross Millikan Jun 4 '14 at 3:41