Renewal Process: laplace transform approach

[SantyClause]

Homework Statement

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

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 isn't easily inverted :(

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

 

[Ray Vickson]

Homework Statement

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

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 isn't easily inverted :(

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

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} = <br /> \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).

 

[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

Anyway, the renewal density m(t) has LT $\tilde{m}(s)$ given by
\tilde{m}(s) = \frac{\tilde{f}(s)}{1-\tilde{f}(s)},[/tex]

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

 

[Ray Vickson]

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?

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