Use the renewal equation to show that the renewal function of a Poisson process












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Use the renewal equation to show that the renewal function of a
Poisson process with rate $lambda > 0$ is $m(t) = lambda t$.










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    $begingroup$


    Use the renewal equation to show that the renewal function of a
    Poisson process with rate $lambda > 0$ is $m(t) = lambda t$.










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      $begingroup$


      Use the renewal equation to show that the renewal function of a
      Poisson process with rate $lambda > 0$ is $m(t) = lambda t$.










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      $endgroup$




      Use the renewal equation to show that the renewal function of a
      Poisson process with rate $lambda > 0$ is $m(t) = lambda t$.







      stochastic-processes poisson-process






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      edited Dec 21 '18 at 17:55









      Math1000

      19.4k31746




      19.4k31746










      asked Dec 5 '15 at 5:15









      zxc123zxc123

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          $begingroup$

          The interrenewal distribution is exponential with rate $lambda$, so $F(t)=1-e^{-lambda t}$ and so the renewal equation gives us
          begin{align}
          m(t) &= F(t) + Fstar m(t)\
          &= 1-e^{-lambda t} + int_0^t m(t-s)lambda e^{-lambda(s)} mathsf ds\
          &= 1-e^{-lambda t} + lambda e^{-lambda t}int_0^t m(x)e^{lambda x} mathsf dx
          end{align}
          (with the substitution $x=t-s$). Multiplying by $e^{lambda t}$ yields
          $$e^{lambda t}m(t) = e^{lambda t} -1 +lambdaint_0^t m(x)e^{lambda x} mathsf dx, $$
          and after differentiating with respect to $t$ we have
          $$lambda e^{lambda t}m(t) + e^{lambda t}m'(t) = lambda e^{lambda t}+lambda e^{lambda t}m(t). $$ Therefore $m'(t)=lambda$, and as $m(0)=0$, it follows that $m(t)=lambda t$.






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            $begingroup$

            The interrenewal distribution is exponential with rate $lambda$, so $F(t)=1-e^{-lambda t}$ and so the renewal equation gives us
            begin{align}
            m(t) &= F(t) + Fstar m(t)\
            &= 1-e^{-lambda t} + int_0^t m(t-s)lambda e^{-lambda(s)} mathsf ds\
            &= 1-e^{-lambda t} + lambda e^{-lambda t}int_0^t m(x)e^{lambda x} mathsf dx
            end{align}
            (with the substitution $x=t-s$). Multiplying by $e^{lambda t}$ yields
            $$e^{lambda t}m(t) = e^{lambda t} -1 +lambdaint_0^t m(x)e^{lambda x} mathsf dx, $$
            and after differentiating with respect to $t$ we have
            $$lambda e^{lambda t}m(t) + e^{lambda t}m'(t) = lambda e^{lambda t}+lambda e^{lambda t}m(t). $$ Therefore $m'(t)=lambda$, and as $m(0)=0$, it follows that $m(t)=lambda t$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              The interrenewal distribution is exponential with rate $lambda$, so $F(t)=1-e^{-lambda t}$ and so the renewal equation gives us
              begin{align}
              m(t) &= F(t) + Fstar m(t)\
              &= 1-e^{-lambda t} + int_0^t m(t-s)lambda e^{-lambda(s)} mathsf ds\
              &= 1-e^{-lambda t} + lambda e^{-lambda t}int_0^t m(x)e^{lambda x} mathsf dx
              end{align}
              (with the substitution $x=t-s$). Multiplying by $e^{lambda t}$ yields
              $$e^{lambda t}m(t) = e^{lambda t} -1 +lambdaint_0^t m(x)e^{lambda x} mathsf dx, $$
              and after differentiating with respect to $t$ we have
              $$lambda e^{lambda t}m(t) + e^{lambda t}m'(t) = lambda e^{lambda t}+lambda e^{lambda t}m(t). $$ Therefore $m'(t)=lambda$, and as $m(0)=0$, it follows that $m(t)=lambda t$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                The interrenewal distribution is exponential with rate $lambda$, so $F(t)=1-e^{-lambda t}$ and so the renewal equation gives us
                begin{align}
                m(t) &= F(t) + Fstar m(t)\
                &= 1-e^{-lambda t} + int_0^t m(t-s)lambda e^{-lambda(s)} mathsf ds\
                &= 1-e^{-lambda t} + lambda e^{-lambda t}int_0^t m(x)e^{lambda x} mathsf dx
                end{align}
                (with the substitution $x=t-s$). Multiplying by $e^{lambda t}$ yields
                $$e^{lambda t}m(t) = e^{lambda t} -1 +lambdaint_0^t m(x)e^{lambda x} mathsf dx, $$
                and after differentiating with respect to $t$ we have
                $$lambda e^{lambda t}m(t) + e^{lambda t}m'(t) = lambda e^{lambda t}+lambda e^{lambda t}m(t). $$ Therefore $m'(t)=lambda$, and as $m(0)=0$, it follows that $m(t)=lambda t$.






                share|cite|improve this answer









                $endgroup$



                The interrenewal distribution is exponential with rate $lambda$, so $F(t)=1-e^{-lambda t}$ and so the renewal equation gives us
                begin{align}
                m(t) &= F(t) + Fstar m(t)\
                &= 1-e^{-lambda t} + int_0^t m(t-s)lambda e^{-lambda(s)} mathsf ds\
                &= 1-e^{-lambda t} + lambda e^{-lambda t}int_0^t m(x)e^{lambda x} mathsf dx
                end{align}
                (with the substitution $x=t-s$). Multiplying by $e^{lambda t}$ yields
                $$e^{lambda t}m(t) = e^{lambda t} -1 +lambdaint_0^t m(x)e^{lambda x} mathsf dx, $$
                and after differentiating with respect to $t$ we have
                $$lambda e^{lambda t}m(t) + e^{lambda t}m'(t) = lambda e^{lambda t}+lambda e^{lambda t}m(t). $$ Therefore $m'(t)=lambda$, and as $m(0)=0$, it follows that $m(t)=lambda t$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Feb 23 '16 at 0:48









                Math1000Math1000

                19.4k31746




                19.4k31746






























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