Statistics for Maintenance Modelling: Renewal Processes

# Question on Asymptotic Formulae¤

What is the result for M(t) when 2 = 0 and when 2 = 2 (the exponential distribution)? Can you make sense of these results?

1. When 2 = 0, we have simply a lifetime that is always , and in this limiting case, the number of renewals by time t is
   t / - (1/2). By starting with a new component, rather than one halfway through its life, we have half a renewal less to make.
2. With large 2, we have on average more replacements to make, because some can fail very quickly. For the exponential distribution, 2 = 2 and and M(t) = t /, so the asymptotic formula is exactly correct. We have M(t) = t / because the exponential distribution is memoryless - the failure rate does not depend on when the previous failure occurred, and there is a constant hazard of failure of 1/. This point is quite deep. A renewal process for the exponential distribution is identical to an asynchronous renewal process.