Laplace functional for a point process of exceedances
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Let $$mathit N_n(cdot)=sum_{i=1}^n varepsilon_{frac in}(cdot)mathit I_{(mathit X_i>u_n)}$$ be a point process of exceedances
where $(mathit X_n)$ is a sequence of random variables and $(u_n)$ is a sequence of real numbers.
How do you calculate $$int_{mathit E}g,dN_n$$ for a non-negative measurable function $g$ if we assume state space $E=(0,1]$ ?
I know that in general for a standard point process $N=sum_{i=1}^infty varepsilon_{X_i}$, the integral will be equal to $sum_{i=1}^infty g(X_i)$. However, I am having trouble to determine the above integral.
The most I could get is $int_E g,dN_n=sum_{i=1}^n g(frac in)mathit I_{(mathit X_i>u_n)}$. Is it true?
Any insight will be greatly appreciated. Thank you!
real-analysis probability lebesgue-integral laplace-transform
asked Jan 18 at 17:27
Yosua Kanichi SusiloYosua Kanichi Susilo
284
$endgroup$
add a comment |
$begingroup$
Let $$mathit N_n(cdot)=sum_{i=1}^n varepsilon_{frac in}(cdot)mathit I_{(mathit X_i>u_n)}$$ be a point process of exceedances
where $(mathit X_n)$ is a sequence of random variables and $(u_n)$ is a sequence of real numbers.
How do you calculate $$int_{mathit E}g,dN_n$$ for a non-negative measurable function $g$ if we assume state space $E=(0,1]$ ?
I know that in general for a standard point process $N=sum_{i=1}^infty varepsilon_{X_i}$, the integral will be equal to $sum_{i=1}^infty g(X_i)$. However, I am having trouble to determine the above integral.
The most I could get is $int_E g,dN_n=sum_{i=1}^n g(frac in)mathit I_{(mathit X_i>u_n)}$. Is it true?
Any insight will be greatly appreciated. Thank you!
real-analysis probability lebesgue-integral laplace-transform
asked Jan 18 at 17:27
Yosua Kanichi SusiloYosua Kanichi Susilo
284
$endgroup$
$begingroup$
After several thoughts, the above expression gave the result I sought in the end. Thank you for those who tried to help!
$endgroup$
– Yosua Kanichi Susilo
Jan 21 at 11:50
add a comment |
$begingroup$
Let $$mathit N_n(cdot)=sum_{i=1}^n varepsilon_{frac in}(cdot)mathit I_{(mathit X_i>u_n)}$$ be a point process of exceedances
where $(mathit X_n)$ is a sequence of random variables and $(u_n)$ is a sequence of real numbers.
How do you calculate $$int_{mathit E}g,dN_n$$ for a non-negative measurable function $g$ if we assume state space $E=(0,1]$ ?
I know that in general for a standard point process $N=sum_{i=1}^infty varepsilon_{X_i}$, the integral will be equal to $sum_{i=1}^infty g(X_i)$. However, I am having trouble to determine the above integral.
The most I could get is $int_E g,dN_n=sum_{i=1}^n g(frac in)mathit I_{(mathit X_i>u_n)}$. Is it true?
Any insight will be greatly appreciated. Thank you!
real-analysis probability lebesgue-integral laplace-transform
asked Jan 18 at 17:27
Yosua Kanichi SusiloYosua Kanichi Susilo
284
$endgroup$
Let $$mathit N_n(cdot)=sum_{i=1}^n varepsilon_{frac in}(cdot)mathit I_{(mathit X_i>u_n)}$$ be a point process of exceedances
where $(mathit X_n)$ is a sequence of random variables and $(u_n)$ is a sequence of real numbers.
How do you calculate $$int_{mathit E}g,dN_n$$ for a non-negative measurable function $g$ if we assume state space $E=(0,1]$ ?
I know that in general for a standard point process $N=sum_{i=1}^infty varepsilon_{X_i}$, the integral will be equal to $sum_{i=1}^infty g(X_i)$. However, I am having trouble to determine the above integral.
The most I could get is $int_E g,dN_n=sum_{i=1}^n g(frac in)mathit I_{(mathit X_i>u_n)}$. Is it true?
Any insight will be greatly appreciated. Thank you!
real-analysis probability lebesgue-integral laplace-transform
real-analysis probability lebesgue-integral laplace-transform
asked Jan 18 at 17:27
Yosua Kanichi SusiloYosua Kanichi Susilo
284
asked Jan 18 at 17:27
Yosua Kanichi SusiloYosua Kanichi Susilo
284
edited Jan 21 at 10:53
Yosua Kanichi Susilo
asked Jan 18 at 17:27
Yosua Kanichi SusiloYosua Kanichi Susilo
284
asked Jan 18 at 17:27
Yosua Kanichi SusiloYosua Kanichi Susilo
284
asked Jan 18 at 17:27
Yosua Kanichi SusiloYosua Kanichi Susilo
284
284
$begingroup$
After several thoughts, the above expression gave the result I sought in the end. Thank you for those who tried to help!
$endgroup$
– Yosua Kanichi Susilo
Jan 21 at 11:50
add a comment |
$begingroup$
After several thoughts, the above expression gave the result I sought in the end. Thank you for those who tried to help!
$endgroup$
– Yosua Kanichi Susilo
Jan 21 at 11:50
$begingroup$
After several thoughts, the above expression gave the result I sought in the end. Thank you for those who tried to help!
$endgroup$
– Yosua Kanichi Susilo
Jan 21 at 11:50
$begingroup$
After several thoughts, the above expression gave the result I sought in the end. Thank you for those who tried to help!
$endgroup$
– Yosua Kanichi Susilo
Jan 21 at 11:50
add a comment |
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$begingroup$
After several thoughts, the above expression gave the result I sought in the end. Thank you for those who tried to help!
$endgroup$
– Yosua Kanichi Susilo
Jan 21 at 11:50