Question 11.EX.23: For a nonhomogeneous Poisson process with intensity function......
For a nonhomogeneous Poisson process with intensity function λ(t), t ≥ 0, where \int^∞_0 λ(t)\, dt = ∞, let X1, X2, . . . denote the sequence of times at which events occur.
(a) Show that \int^{X_1}_0 λ(t)\, dt is exponential with rate 1.
(b) Show that \int^{X_i}_{X_{i−1}} λ(t)\, dt, i \geqslant 1, 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.
Let m(t)=\int_{0}^{t} \lambda(s)\, d s, and let m-1(t) be the inverse function. That is, m(m-1(t))=t.
\begin{aligned}(a)\qquad P\left\{m\left(X_{1}\right)>x\right\} & =P\left\{X_{1}>m^{-1}(x)\right\} \\& =P\left\{N\left(m^{-1}(x)\right)=0\right\} \\& =e^{-m\left(m^{-1}(x)\right)} \\& =e^{-x}\end{aligned}
\begin{aligned}(b)\qquad P\left\{m\left(X_{i}\right)-m\left(X_{i-1}\right)\right. & \left.>x|m\left(X_{1}\right), \ldots, m\left(X_{i-1}\right)-m\left(X_{i-2}\right)\right\} \\& =P\left\{m\left(X_{i}\right)-m\left(X_{i-1}\right)>x|X_{1}, \ldots, X_{i-1}\right\} \\& =P\left\{m\left(X_{i}\right)-m\left(X_{i-1}\right)>x|X_{i-1}\right\} \\& =P\left\{m\left(X_{i}\right)-m\left(X_{i-1}\right)>x|m\left(X_{i-1}\right)\right\}\end{aligned}
Now,
\begin{aligned}P & \left\{m\left(X_{i}\right)-m\left(X_{i-1}\right)>x|X_{i-1}=y\right\} \\& =P\left\{\int_{y}^{X_{i}} \lambda(t) d t>x |X_{i-1}=y\right\} \\& =P\left\{X_{i}>c| X_{i-1}=y\right\} \quad \text { where } \int_{y}^{c} \lambda(t) d t=x \\& =P\left\{N(c)-N(y)=0|X_{i-1}=y\right\} \\& =P\{N(c)-N(y)=0\} \\& =\exp \left\{-\int_{y}^{c} \lambda(t) d t\right\} \\& =e^{-x}\end{aligned}
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