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The lifetime of a machine is modelled by an exponential random variable $X$ with $P(X>x) = e^{-\lambda x}, \lambda, x > 0$. This machine cannot be repaired. A maintenance crew checks this machine at times $T, 2T, 3T, ...., $ wjere $T$ is a specified length of time

  • Find the probability that the machine fails between the $(k-1)$th and the $k$th inspection, $k = 1, 2, ...$.
  • Find the expected value of the first time when the maintenance crew finds that the machine is down. [HINT: $\sum_{j = 1}^{\infty} jq^j = (1 - q)^{-2}q$ if $|q| < 1$.]

I did the first bit and correctly got the probability to be $e^{-\lambda kT}(e^{-\lambda T} - 1)$.

But I'm stuck as to what to do from here. I don't see how the summation would come into calculating the expected value?

EDIT:

Consider the events $A_k=\{(k-1)T < X \leq kT\}$ and denote $U$ the duration of the time when the machine is down before it is discovered to be down. Then use the formulas

$$EU = \sum_{k=1}^{\infty} E(U |A_k)P(A_k)$$

and

$$E(U|A_k) = \int_0^{T} P(U > u |A_k) du$$

to find the expected duration of the time when the machine is down before it is discovered to be down.

I'm still working through this bit, but my first question is, in the first equation, should that read $E(U)$? If not, what does $EU$ mean?

EDIT 2: In the answers, it also says $EU$, so what does this mean?

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  • $\begingroup$ How do you propose to calculate the expected value? $\endgroup$ – Jonathan Christensen Jan 21 '13 at 21:08
  • $\begingroup$ By definition, the expectation (for a discrete distribution) is the probability-weighted sum of values. So, whenever you see an expectation expressed as a summation, try to separate its terms into a product of a probability and a value. To connect this with the hint, notice that part of your formula for the probability involves a power $\exp(-\lambda k T)$ = $(\exp(-\lambda T))^k$. $\endgroup$ – whuber Jan 21 '13 at 21:08
  • $\begingroup$ @whuber That weighted sum values bit is the bit I don't get with expectation of conditional probability. I looked at it on Wikipedia and this says that $\mathbb{E}(X | Y = y) = \sum x P(X = x | Y = y)$ but I don't really understand what I'm adding up. I can't find an example $\endgroup$ – Kaish Jan 21 '13 at 21:13
  • $\begingroup$ In your comment I believe the left hand side is the expectation rather than the conditional probability. The Wikipedia article on expectation has examples. $\endgroup$ – whuber Jan 21 '13 at 21:24
  • $\begingroup$ @whuber Yeah sorry. I meant that to be expectation. Whilst you're online, could you have a quick look at the question in the edit please. It's very simple. $\endgroup$ – Kaish Jan 21 '13 at 21:45

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