the renewal function and the elementary renewal theorem

Once you have a renewal process, the most basic quantity to ask for is: on average, how many renewals have happened by time t? That expected count, m(t) = E[N(t)], is the renewal function. It is the renewal-process answer to the question that, for a Poisson process, was simply lambda*t. For a general renewal process m(t) is usually NOT a straight line at short times — it bends as the gap distribution's shape makes early renewals more or less likely — but its long-run slope is the headline result.

The elementary renewal theorem says that as t grows large, m(t) / t tends to 1 / E[T], the reciprocal of the mean interarrival time. In plain words, the long-run average rate of renewals is one over the average gap — exactly what intuition demands, but now it is a theorem, holding for any interarrival distribution with finite mean. So if parts last on average 4 years, you replace them at a long-run rate of 1/4 per year, and the expected number of replacements by year t is approximately t/4 for large t. (A sharper companion result, the renewal theorem proper, even pins down the constant correction term, but the elementary version captures the essential rate.)

This is what makes renewal theory usable for planning: budgeting spare parts, sizing maintenance crews, or estimating lifetime costs all hinge on the long-run rate 1/E[T], which depends ONLY on the mean gap, not its full shape. Two honest caveats. First, the limit is asymptotic — early on, before the process settles, m(t)/t can differ noticeably from 1/E[T]. Second, the elementary theorem requires E[T] to be finite; if the gap distribution has such a heavy tail that its mean is infinite, the long-run rate collapses to zero and the usual planning intuition fails.