# Renewal Process: laplace transform approach

1. Apr 24, 2013

### SantyClause

1. The problem statement, all variables and given/known data
Use the Laplace transform approach to find the renewal function for a renewal process with interrenewal p.d.f. as follows:

g(x) = (c^2)xe^(-cx) , x > 0

3. The attempt at a solution
M*(s) = G*(s)/(1-G*(s)) where M*(s) and G*(s) denote laplace transforms

I have that G*(s) = -c/(c+s)^2 - 1/(c+s) and when I plug that into the above equation I get something that isnt easily inverted :(

It should be noted taht I haven't used laplace transforms in YEARS and they were always very weird to me.

2. Apr 24, 2013

### Ray Vickson

You need to be more careful: M* and G* are Laplace transforms or WHAT? If $\tilde{f}(s)$ is the LT of f(x), then since f is the density of a 2-Erlang random variable (= a sum of two iid exponentials), we have that its LT is the product of the exponential LTs:
$$\tilde{f}(s) = \frac{c^2}{(s+c)^2}.$$ I don't know how you got your G*(s). If you thought you wanted the LT $\tilde{F}(s)$ of F(x) = cdf of f(x), you should have gotten
$$\tilde{F}(s) = \frac{1}{s} \tilde{f}(s) = \frac{c^2}{s(s+c)^2} = \frac{1}{s} - \frac{c}{(s+c)^2} - \frac{1}{c+s}.$$

Anyway, the renewal density m(t) has LT $\tilde{m}(s)$ given by
$$\tilde{m}(s) = \frac{\tilde{f}(s)}{1-\tilde{f}(s)},$$and the renewal function
$M(t) = \int_0^t m(\tau) \, d\tau$ has LT $$\tilde{M}(s) = \frac{1}{s} \tilde{m}(s).$$

3. Apr 25, 2013

### SantyClause

Ok I think I understand now. Of course the two things should have been off only by 1/s.

But for the following formula

we are supposed to use the density rather than the cdf?

4. Apr 25, 2013

### Ray Vickson

I'll let you figure out the reason, but I wrote exactly what I said and meant.