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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

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    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

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