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Let $\{N(t):t\in\mathbb R_+\}$ be a homogeneous Poisson process with rate $\lambda>0$. Let $T_0=0$ and $T_{n+1} = \inf\{t>T_n: N(t)=N(T_n)+1\}$ be the arrival times. The residual time $Y_t$ is defined as the waiting time until the next arrival, given an observer arrives at time $t$, $Y_t := T_{N(t)+1}-t$. The age process $A_t$ describes the elapsed time since the last arrival, $A_t := t-T_{N(t)}$. The duration of an interarrival time seen by some observer arriving at time $t$ is then $L_t:= Y_t+A_t = T_{N(t)+1}-T_{N(t)}.$

  1. Show that $Y_t$ has an exponential distribution with rate $\lambda$.
  2. Derive the distribution of $A_t$.
  3. Show that the interarrival distribution differs from an exponential distribution (the so-called inspection paradox). Specifically, show that $L_t$ has density $$ f_{L_t}(x) = \lambda(\lambda x)e^{-\lambda x}\cdot\mathsf 1_{(0,t)}(x) + \lambda(1+\lambda t)e^{-\lambda x}\cdot\mathsf 1_{[t,\infty)}(x). $$ Hint: Consider the two cases $0<x<t$ and $x\geqslant t$ separately, and treat them similarly, using the total probability formula.

By the memoryless property of the exponential distribution, we see that for $x>0$, $\mathbb P(Y_t>x) = e^{-\lambda x}$, so $Y_t$ has density $f_{Y_t}(x) = \lambda e^{-\lambda x}\cdot\mathsf 1_{(0,\infty)}(x)$. Now, clearly $\mathbb P(A_t>t)=0$, so the distribution function of $A_t$ is $$F_{A_t}(x) = (1-e^{-\lambda x})\cdot\mathsf 1_{[0,t)}(x) + \mathsf 1_{[t,\infty)}(x), $$ that is, $A_t$ has a mixture distribution of an exponential distribution on $(0,t)$ and a "jump" of magnitude $e^{-\lambda t}$ at $x=t$.

For $0<x<t$ we obtain the density of $L_t$ by convolution: \begin{align} f_{L_t}(x) &= \int_{\mathbb R} f_{Y_t}(y)f_{A_t}(x-y)\mathsf 1_{(0,t)}(x)\ \mathsf dy\\ &= \int_0^x \lambda e^{-\lambda y}\lambda e^{-\lambda(x-y)}\ \mathsf dy\cdot\mathsf 1_{(0,t)}(x)\\ &= \lambda^2 e^{-\lambda x}\int_0^x\ \mathsf dy\cdot\mathsf 1_{(0,t)}(x)\\\\ &= \lambda(\lambda x)e^{-\lambda x}\cdot\mathsf 1_{(0,t)}(x). \end{align} But I don't see how to find the density of $L_t$ for $x\geqslant t$. How can I show that this is $\lambda(1+\lambda t)e^{-\lambda x}\cdot\mathsf 1_{[t\infty)}(x)$?

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Math1000
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    The argument with minimal calculation is that the complementary CDF conditioned on an arrival at time $\gt t$ looks exactly like the complementary CDF of $Y_t$. Working backwards this implies the complementary CDF changes $\propto $ that of $Y_t$ for arrival time $\gt t$ and in particular the derivative is $\propto$ the exponential density. For any selection of $t$ you need the constant factor $(1+\lambda t)$ so that when you compute the integral of $f_{L_t}(x)$ over $(0,\infty)$, that it in fact integrates to one (that's what the hint is getting at with total probability). – user8675309 Jan 13 '20 at 08:15
  • @user8675309 Thanks, I have figured it out and posted a solution. – Math1000 Jan 13 '20 at 08:38

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Based on @user8675309's comment, we have that conditioned on $\{x>t\}$, the complementary distribution function of $L_t$ is proportional to the complementary distribution function of $Y_t$, and so the derivative is proportional to the exponential density with parameter $\lambda$. In other words, for $x>t$ we have that $f_{L_t}(x)\propto f_{Y_t}(x) = \lambda e^{-\lambda x}$, so $f_{L_t}(x) = C\lambda e^{-\lambda x}$ for some $C>0$. Integrating the density of $L_t$ from $0$ to $t$ yields $$ \int_0^t f_{L_t}(x) \ \mathsf dx = \int_0^t\lambda(\lambda x)e^{-\lambda x}\ \mathsf dx = 1-(1+\lambda t)e^{-\lambda t} , $$ so for the density to integrate to $1$ over $(0,\infty)$ we must have $$ \int_t^\infty C\lambda e^{-\lambda x}\ \mathsf dx = (1+\lambda t)e^{-\lambda t}. $$ But $$\int_t^\infty C\lambda e^{-\lambda x}\ \mathsf dx = Ce^{-\lambda t}, $$ so it follows that $C=1+\lambda t$. Hence the density of $L_t$, conditioned on $\{x>t\}$, is $f_{L_t}(x) = \lambda(1+\lambda t)e^{-\lambda x}$.

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