renewal equation for $\sim U(0,1)$ interarrivals should be


how can this be solved?

can I make substitution $y=t-s$ to get

$m(t)=t+\int_0^t{m(y)f(y)dy}$ if all I know is that t>1 ?

What will make this substitution valid?

  • $\begingroup$ When $t>1$ the first term in the renewal equation for $U(0,1)$ is NOT $t$, it's 1. (The first term is the cdf.) $\endgroup$ – Dan Stowell Oct 14 '14 at 15:21

You need to correctly do the substitution $y=t-s$ to get


And now differentiate in $t$:


Which you can solve easily. You'll need an initial value of $m(t)$ which you can deduce from the limit condition $m(t)/t$ for renewal processes.

By the way you should check to see if you've written things down right. There should be a density function inside the integral: $m(t-s)f(s)$ and the density is 0 when $s>1$.

  • $\begingroup$ Alex, your answer follows the logic presented in the book, however, it's only applicable for t less than 1, for t (0,1) for example the last equation takes form $m'(t) = 1 - e^{t-1} + m(t)$ although I don't understand why $\endgroup$ – dark blue Oct 3 '14 at 11:48
  • $\begingroup$ You're dealing with a uniform distribution right? So inside the integral there's a density $f(s)=1$ for $s$ in $[0,1]$ and 0 elsewhere. Try incorporating this $\endgroup$ – Alex R. Oct 3 '14 at 14:40
  • $\begingroup$ Alex, I'm sorry, but I don't understand. I understand the f(s) part, but I still don't know how to express m(t) in terms of other $m(t-1)$ or $m(t+1)$: $m(t)= 1 + \int_0^t{m(t-s)f(s)ds}$. What you are writing: m'=1+m does not seem to work for t > 1 $\endgroup$ – dark blue Oct 3 '14 at 17:55

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