Use the renewal equation to show that the renewal function of a Poisson process
$begingroup$
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
edited Dec 21 '18 at 17:55
Math1000
19.1k31745
asked Dec 5 '15 at 5:15
zxc123zxc123
62
$endgroup$
add a comment |
$begingroup$
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
edited Dec 21 '18 at 17:55
Math1000
19.1k31745
asked Dec 5 '15 at 5:15
zxc123zxc123
62
$endgroup$
add a comment |
$begingroup$
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
edited Dec 21 '18 at 17:55
Math1000
19.1k31745
asked Dec 5 '15 at 5:15
zxc123zxc123
62
$endgroup$
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
stochastic-processes poisson-process
edited Dec 21 '18 at 17:55
Math1000
19.1k31745
asked Dec 5 '15 at 5:15
zxc123zxc123
62
edited Dec 21 '18 at 17:55
Math1000
19.1k31745
asked Dec 5 '15 at 5:15
zxc123zxc123
62
edited Dec 21 '18 at 17:55
Math1000
19.1k31745
edited Dec 21 '18 at 17:55
Math1000
19.1k31745
edited Dec 21 '18 at 17:55
Math1000
19.1k31745
19.1k31745
asked Dec 5 '15 at 5:15
zxc123zxc123
62
asked Dec 5 '15 at 5:15
zxc123zxc123
62
asked Dec 5 '15 at 5:15
zxc123zxc123
62
62
add a comment |
add a comment |
1 Answer
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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 $m(t)=lambda t$.
answered Feb 23 '16 at 0:48
Math1000Math1000
19.1k31745
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1 Answer
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1 Answer
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oldest
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active
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votes
$begingroup$
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 $m(t)=lambda t$.
answered Feb 23 '16 at 0:48
Math1000Math1000
19.1k31745
$endgroup$
add a comment |
$begingroup$
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 $m(t)=lambda t$.
answered Feb 23 '16 at 0:48
Math1000Math1000
19.1k31745
$endgroup$
add a comment |
$begingroup$
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 $m(t)=lambda t$.
answered Feb 23 '16 at 0:48
Math1000Math1000
19.1k31745
$endgroup$
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 $m(t)=lambda t$.
answered Feb 23 '16 at 0:48
Math1000Math1000
19.1k31745
answered Feb 23 '16 at 0:48
Math1000Math1000
19.1k31745
answered Feb 23 '16 at 0:48
Math1000Math1000
19.1k31745
answered Feb 23 '16 at 0:48
Math1000Math1000
19.1k31745
19.1k31745
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add a comment |
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