For a nonhomogeneous Poisson process with intensity function λ(t), t ≥ 0, where ∫0^∞

Question

For a nonhomogeneous Poisson process with intensity function λ(t), t ≥ 0, where [latex]\int^∞_0 λ(t)\, dt = ∞[/latex], let X₁, X₂, . . . denote the sequence of times at which events occur.

(a) Show that [latex]\int^{X_1}_0 λ(t)\, dt[/latex] is exponential with rate 1.
(b) Show that [latex]\int^{X_i}_{X_{i−1}} λ(t)\, dt, i \geqslant 1[/latex], are independent exponentials with rate 1, where X₀ = 0.
In words, independent of the past, the additional amount of hazard that must be experienced until an event occurs is exponential with rate 1.

Accepted Answer

Let [latex]m(t)=\int_{0}^{t} \lambda(s)\, d s[/latex], and let m^-1(t) be the inverse function. That is, m(m^-1(t))=t.

[latex]\begin{aligned}(a)\qquad P\left\{m\left(X_{1} ight)>x ight\} & =P\left\{X_{1}>m^{-1}(x) ight\} \& =P\left\{N\left(m^{-1}(x) ight)=0 ight\} \& =e^{-m\left(m^{-1}(x) ight)} \& =e^{-x}\end{aligned}[/latex]

[latex]\begin{aligned}(b)\qquad P\left\{m\left(X_{i} ight)-m\left(X_{i-1} ight) ight. & \left.>x|m\left(X_{1} ight), \ldots, m\left(X_{i-1} ight)-m\left(X_{i-2} ight) ight\} \& =P\left\{m\left(X_{i} ight)-m\left(X_{i-1} ight)>x|X_{1}, \ldots, X_{i-1} ight\} \& =P\left\{m\left(X_{i} ight)-m\left(X_{i-1} ight)>x|X_{i-1} ight\} \& =P\left\{m\left(X_{i} ight)-m\left(X_{i-1} ight)>x|m\left(X_{i-1} ight) ight\}\end{aligned}[/latex]

Now,

[latex]\begin{aligned}P & \left\{m\left(X_{i} ight)-m\left(X_{i-1} ight)>x|X_{i-1}=y ight\} \& =P\left\{\int_{y}^{X_{i}} \lambda(t) d t>x |X_{i-1}=y ight\} \& =P\left\{X_{i}>c| X_{i-1}=y ight\} \quad ext { where } \int_{y}^{c} \lambda(t) d t=x \& =P\left\{N(c)-N(y)=0|X_{i-1}=y ight\} \& =P\{N(c)-N(y)=0\} \& =\exp \left\{-\int_{y}^{c} \lambda(t) d t ight\} \& =e^{-x}\end{aligned}[/latex]

Question 11.EX.23: For a nonhomogeneous Poisson process with intensity function......

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