# Expected value for a Poisson Process

A machine works for an exponentially distributed time with rate μ and then fails. A repair crew checks the machine at times distributed according to a Poisson process with rate λ; if the machine is found to have failed then it is immediately replaced. Find the expected time between replacements of machines.

What I have so far:

I believe that if M represents the failure time of the machine and T represents the time when the machine is checked, then the probability of the machine being replaced is:

P(M < T) = μ/(μ+λ)

Since both M and T are exponentially distributed. So, that got me into thinking that this as Exponential(μ+λ) distributed, and so the expected time for the machine to be fixed is just (μ+λ)/μ.

Another thought was to find the expected time of the failure of the machine (1/μ) and then the expected time of checking the machine conditioned on the machine having failed, which I think is just 1/λ by the no memory property.

However, I feel like I do not understand what I am doing and am just aimlessly trying to solve the question. I can't seem to find a similar problem or example in my textbook (I'm on Ch 5 of "Introduction to Probability Models" by Ross, 10ed), and if anyone can help me with the problem or refer me to some good resources, I would be extremely grateful.

Thank you!

Quick edit: It looks like the question asks for the expected time between replacements, so now I feel like I did things completely wrong :(

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$\frac{\mu+\lambda}{\mu}$ can't be right. This is unitless, and you need units of time, like $\frac{1}{\mu}$ or $\frac{1}{\lambda}$. – Ross Millikan Mar 9 '11 at 4:56
@Ross: Thank you, I feel dumb. μ and λ are rates, so any times would need to be expressed by 1/μ and 1/λ. How about this: The expected time until failure is 1/μ, and the expected time until the machine is checked is 1/λ. Since the event that the mechanic checks the machine is exponential with rate λ, and that it is memoryless, the time after the machine breaks and is checked is just 1/λ, and so the mean time until the machine is replaced is just 1/μ + 1/λ. – user7990 Mar 9 '11 at 5:19
I agree with that. It is just what you were saying in the paragraph that starts with "Another thought". I think the point of the exercise is to make sure you know the expected time of an exponential distribution and that a Poisson process is another name for the same thing. – Ross Millikan Mar 9 '11 at 5:40
Yes, I am confused by the relationship between the exponential distribution and the Poisson process. I need to review. – user7990 Mar 9 '11 at 5:57

The time $T$ between two successive replacements is the sum of the lifetime $D$ of the machine and the elapsed time $R$ between the moment when it fails and the moment when it is replaced. By hypothesis $D$ is exponential with parameter $\mu$ and $R$ is exponential with parameter $\lambda$. Hence $E(T)=E(D)+E(R)=1/\mu+1/\lambda$, as suggested in one paragraph of your post.
Note that the random variable $T$ is never exponentially distributed. Rather, when $\lambda\ne\mu$, its probability density function $f_T$ is defined on $t\ge0$ by $\displaystyle f_T(t)=\frac{\lambda\mu}{\lambda-\mu}(\mathrm{e}^{-\mu t}-\mathrm{e}^{-\lambda t})$ and, in the degenerate case $\lambda=\mu$, $f_T(t)=\lambda^2t\mathrm{e}^{-\lambda t}.$
PEV: No.   – Did May 28 '11 at 18:53