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Suppose we have $n$ batteries which has a lifetime that is exponentially distributed with parameter $\lambda$. Each battery's lifetime is independent.

If we initially put 2 batteries on and every time a battery fails, we replace it with another battery until there is only one working battery.

What is the probability that we can use these $n$ more than $x$ years. And what is the expectation of the total time until there is only one battery left working.


I realize that we need to use the memoryless property of exponential distribution r.vs but I kind of got stuck there.

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2 Answers 2

up vote 3 down vote accepted

Hints:

The failure of the batteries is a Poisson process

Having two batteries on at the same time doubles the intensity of the process

You can use the $n$ batteries up to the point at which there have been $n-1$ failures

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Could you please explain a little bit more about this? Can I simply assume that the process in the question can be modeled as a Gamma distribution random variable with parameter $(n-1,\lambda)$? @Henry –  geraldgreen Nov 17 '11 at 19:30
    
@John:You have to double the intensity, so the second parameter is $2\lambda$ –  Henry Nov 19 '11 at 10:01

I believe whenever you change a battery, the remaining one just acts like it has been freshly changed too, according to the memoryless property. Thus the question can be simplified into: How long could $(n-1)$ batteries last. Because batteries' lifespan are all idd, the probability of their sum should be: $$ P(x)=\int_0^\infty \cdots \int_0^{t_i} \prod_{i=1}^{n-1} f(t_i-x)f(x)dx\,. $$

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Not quite - the distribution of failure times of a pair of batteries is the minimum of two exponential distributions with rate $\lambda$, which is another exponential with rate $2\lambda$. –  Chris Taylor Nov 17 '11 at 8:32

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