danh...@yahoo.com wrote: > The Poisson process part, in the long run, is justified by the > Palm-Khintchine theorem, which states that under some mild > conditions ( which are to technical for me to give here), failure > epochs of a system with a large number of independently failing > components form a Poisson process. When the system is new, the > time-to-failure distribution of each component determines the > failure rate of the system, and the process is not Poisson unless > the inter-failure times are exponential.
Thanks, Dan. I do get intuitively that a mish-mash of renewal processes looks Poissonian.
Rich Ulrich wrote: > I was never very good at remember citations for stuff that seems > pretty easy and obvious. > > Here is a treatment that should be easy enough to understand; and > HERE IT IS, posted online. Look at the headers for the message ID. > It can be recovered through Google-groups. > > Consider a simple system where the time to one failure is > > t1= N(10,1); thus for failure 2, 3, ... the SD increases > linearly, and the variance as the square of the number > t2= N(20,4) > t3= N(30,9) > .. > t25= N(250, 625) > > If you plot the first three of them, you will see that there if > very, very little overlap between "second failure" for a device and > "first failure", and not much for "third failure". But the peak is > lower at each successive failure, for that unit of time, and the > spread about each peak is increasing in width. For looking at the > times of 1, 2, or 3 failures, you can have a pretty exact picture by > simply overlaying the three curves. > > _________9________5________3___ ... (Number: height of peak) > > > Depending on how particular you are, you can't just overlay the > graphs to show "all failures" after the 3rd or 4th failure; you want > to add the densities. > > Clearly, by the time you have reached the time of the 25th failure > on-the-average, many devices will be on an earlier or later count, > because the SD is now 25, and the peak is no longer prominent.
Rich, I get that if you superpose the t1, t2, etc., you get a series of diminishing peaks. However, the time to failure of the architypal component is decaying exponential rather than normal. So the fact that the 2nd order response to a step function starts at zero with zero slope was a bit of a mystery.
I confided with a colleague, who didn't have a reference. However, he described that it is related to conditional probabilities (if I understood correctly) which to me means Bayes or Total Probability Law. I have yet to sit down and suss it out, but he indicated that one has to convolve t1, t2, t3, etc., where they are decaying exponential PDFs. I'll chime in if I get to that point. As a citation reference, I'm going to cite the manuals where I found the curves in (for now).