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Let us consider a machine with two parts X and Y. The time before part X goes out of service has an Exponential distribution with $\lambda = 1$ and the time before part Y goes out of service has an Exponential distribution with $\lambda = 0.05$ (its life expectation is 20 time units). When the part X breaks it is replaced in 1 time unit and the whole machine is not working during this time (hence the part Y is not ageing while the part X is being replaced). When part Y goes out of service, the whole machine is considered dead. What is the expectation of the random variable T, which stands for the time before death of the whole machine?

I have tried to consider conditioning on the lifespan of the part Y (I denoted this random variable T_Y) and on the number of times the part X breaks before the part Y goes out of service (N). $$ E(T) = \sum_{k=0}^{\infty}\int_{0}^{\infty}E(T|T_Y, N)P(N=k|T_Y)P(T_Y=t_{y})dt_{y} $$ But I am unsure of the formula. Also I would say that $E(T|T_{Y}, N) = t_{y} + k$ but I don't see why exactly. And what would be $ P(N=k|T_{y})$?

user581250
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Since the exponential distribution is memoryless, each time part $X$ is replaced, we have a new Bernoulli trial, where success means that part $Y$ fails before part $X.$ From the distribution of the difference of two independent exponentially-distributed random variables, (see this question) we calculate that the probability of success is $$p={.05\over1.05}$$ and we know that the expected number of trials until success is $${1\over p}=21$$

That means, that on average, part $X$ fails $20$ times before part $Y$ fails, so $$E(T)=20+E(Y)=20+20=40$$

just as we would have guessed.

saulspatz
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  • Thank you for your response. Can I ask if I understand correctly? The number of fails of the part $X$ is independent from the time before the part $Y$ fails and has a geometric distribution (a sum of those Bernoulli trials you mentioned) with a probability of success $p = 1/21$? – user581250 Aug 06 '18 at 20:22
  • So if we were to count the variance of the time before death it could be something like $Var(T) = 1-p/(p^2) + 1/(\lambda^2) = 20*21 +20^2$? – user581250 Aug 06 '18 at 20:27
  • "The number of fails of the part $X$ is independent from the time before the part $Y$ fails." I don't think that this is what you mean; I can't make sense of it. I think that if $N$ is the number of failures of part $X$ before the failure of part $Y,$ then $N$ has the geometric distribution you describe. – saulspatz Aug 06 '18 at 20:52
  • I'm not sure about the variance. If $Z$ is the lifetime of part $Y$ then $T=Z+N$ But surely $Z$ and $N$ are correlated, so you can't add up the variances. This is really a different question, so you really should ask a different question if you want to pursue it. I don't think I know how to do it, at least not without a lot of work. – saulspatz Aug 06 '18 at 20:58
  • You are right, I can not compute the variance as a sum of variances. – user581250 Aug 06 '18 at 21:14