1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Renewal Process: laplace transform approach

  1. Apr 24, 2013 #1
    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. jcsd
  3. Apr 24, 2013 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    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:
    [tex] \tilde{f}(s) = \frac{c^2}{(s+c)^2}.[/tex] 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
    [tex] \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}.[/tex]

    Anyway, the renewal density m(t) has LT ##\tilde{m}(s)## given by
    [tex] \tilde{m}(s) = \frac{\tilde{f}(s)}{1-\tilde{f}(s)},[/tex]and the renewal function
    ##M(t) = \int_0^t m(\tau) \, d\tau## has LT [tex]\tilde{M}(s) = \frac{1}{s} \tilde{m}(s).[/tex]
     
  4. Apr 25, 2013 #3
    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?
     
  5. Apr 25, 2013 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    I'll let you figure out the reason, but I wrote exactly what I said and meant.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Renewal Process: laplace transform approach
  1. Laplace Transform (Replies: 1)

  2. Laplace Transforms (Replies: 4)

  3. Laplace transformation (Replies: 2)

  4. Laplace transform (Replies: 3)

Loading...