Proof of Jump Times of the Homogeneous Poisson Process: Finiteness and Exponential Interarrival Law

Step 1: pathwise facts and measurability. Let $c$ be any counting path and $k\ge1$. We claim $\{t:\ c(t)\ge k\}=[\tau_k(c),\infty)$ when the left set is nonempty. Indeed, if $s>\tau_k(c)$ there is $t\in[\tau_k(c),s]$ with $c(t)\ge k$, so $c(s)\ge k$ by monotonicity; and $c(\tau_k(c))=\inf\{c(s):s>\tau_k(c)\}\ge k$ by right-continuity. Consequently, for every $u\ge0$,

Applying this to the paths of $Y$: $\{\tau_k\le u\}=\{Y_u\ge k\}$ is an event for every $u$, so each $\tau_k$ is measurable as an extended-real random variable (measurability here meaning that $\{\tau_k\le u\}$ is an event for every real $u$). Next, if $\tau_k(c)=\tau_{k+1}(c)=t<\infty$ then $c(t)\ge k+1$ while $c(s)\le k-1$ for all $s<t$, so $c(t)-c(t-)\ge2$, contradicting unit jumps; hence $\tau_k(c)<\tau_{k+1}(c)$ whenever $\tau_k(c)<\infty$. Also $\tau_1(c)>0$: if $\tau_1(c)=0$ then $c(0)\ge1$ by the display, contradicting $c(0)=0$. Divergence is a sure statement: for any real $M\ge0$, $c(M)$ is a nonnegative integer $k_0$, and then $\tau_{k_0+1}(c)>M$ by the display. Finally, finiteness: $Y_u$ has the Poisson distribution with parameter $u$ (it is the increment $Y_u-Y_0$, and $Y_0=0$ since paths are counting paths), so

where $e$ is the real exponential function. Since for $u\ge0$ every term of the defining series of the exponential is nonnegative, $e^{u}\ge u^{j+1}/(j+1)!$, so each term satisfies $u^je^{-u}\le (j+1)!/u\to0$ as $u\to\infty$, and $P(\tau_k=\infty)\le\inf_{n}P(\tau_k>n)=0$. Intersecting the countably many almost sure events $\{\tau_k<\infty\}$ over $k$ and combining with the sure statements above proves part (a).

Step 2: two elementary estimates. First, $e^{-\mu}\ge1-\mu$ for all $\mu\ge0$: for $\mu\ge1$ this is trivial since $1-\mu\le0\le e^{-\mu}$, and for $0\le\mu<1$ the geometric series gives $\frac{1}{1-\mu}=\sum_{j\ge0}\mu^j\ge\sum_{j\ge0}\frac{\mu^j}{j!}=e^{\mu}$ (the series representation of the exponential), so $e^{-\mu}=1/e^{\mu}\ge1-\mu$ by the basic properties of the exponential. Consequently, for $\mu\ge0$, a variable $K$ with the Poisson distribution with parameter $\mu$ satisfies

Second, for $0<\delta\le1$ we have $1-\delta\le e^{-\delta}\le1-\delta+\tfrac{\delta^2}{2}$: the lower bound is the first estimate, and for the upper bound the defining series gives

where the regrouping of the absolutely convergent series into consecutive pairs is legitimate and each bracket is nonnegative because $\delta^{2i+2}/(2i+2)!\le\delta^{2i+1}/(2i+1)!$ for $0<\delta\le1$. Hence $\tfrac{\delta}{2}\le\delta-\tfrac{\delta^2}{2}\le1-e^{-\delta}\le\delta$, and for any $v\ge0$,

Step 3: fresh start at a deterministic time. Fix $r\ge0$ and set $\hat{Y}_u=Y_{r+u}-Y_r$ for $u\ge0$. Every path of $\hat{Y}$ is a counting path (it starts at $0$, is nondecreasing, right-continuous, integer-valued, and inherits unit jumps from the path of $Y$). The process $\hat{Y}$ has independent increments and $\hat{Y}_{u'}-\hat{Y}_u=Y_{r+u'}-Y_{r+u}$ has the Poisson distribution with parameter $u'-u$, so $\hat{Y}$ is a homogeneous Poisson process with rate $1$. Moreover, $Y$ has independent increments and $Y_0=0$ surely (its paths are counting paths), so part (b) of the grouping and fresh-start lemma shows that the $\sigma$-algebra $\sigma(\hat{Y}_u:u\ge0)$ generated by all post-$r$ increments is independent of $\mathcal{F}^Y_r=\sigma(Y_s:s\le r)$.

Step 4: the survival identity by induction on $k$. For $k=1$: $P(\xi_1>u_1)=P(\tau_1>u_1)=P(Y_{u_1}=0)=e^{-u_1}$ by Step 1. Let $k\ge2$ and assume the identity for $k-1$ jumps, for every rate-$1$ homogeneous Poisson process with counting paths. Fix $u_1,\dots,u_k\ge0$ and $0<\delta\le\min(1,u_2)$ if $u_2>0$ (the case $u_2=0$ follows from the case $u_2>0$ by monotone convergence of probabilities along $u_2\downarrow0$, since the events increase; we henceforth take $u_2>0$). Anchor a grid at $u_1$: let $t_j=u_1+j\delta$ for $j\ge0$, and define the disjoint events

By independent increments, $P(G_j)=e^{-t_{j-1}}\cdot\delta e^{-\delta}$. On $G_j$ we have $\tau_1\in(t_{j-1},t_j]$, $\xi_1>u_1$, and no jump of $Y$ in $(\tau_1,t_j]$. Let $E=\{\xi_1>u_1,\dots,\xi_k>u_k\}$. Since $E\subseteq\{\xi_1>u_1,\ \tau_1<\infty\}=\bigsqcup_{j\ge1}\{Y_{t_{j-1}}=0,\ Y_{t_j}\ge1\}$ and $\{Y_{t_{j-1}}=0,Y_{t_j}\ge1\}\setminus G_j\subseteq\{Y_{t_{j-1}}=0,\ Y_{t_j}-Y_{t_{j-1}}\ge2\}$, Step 2 gives

Now fix $j$ and let $\hat{Y}=\hat{Y}^{(j)}$ be the fresh process of Step 3 at time $r=t_j$, with jump times $\hat{\tau}_i$ and interarrival times $\hat{\xi}_i$. The events in $\hat{\xi}_1,\dots,\hat{\xi}_{k-1}$ used below belong to $\sigma(\hat{Y}_u:u\ge0)$: by Step 1 applied to the counting paths of $\hat{Y}$, $\{\hat{\tau}_i\le u\}=\{\hat{Y}_u\ge i\}$, so each $\hat{\tau}_i$ is measurable with respect to $\sigma(\hat{Y}_u:u\ge0)$, and hence so is every event formed from the $\hat{\xi}_i$ as in the theorem statement. On $G_j$, the jumps of $Y$ after $t_j$ are exactly the jumps of $\hat{Y}$ shifted by $t_j$, so $\tau_2=t_j+\hat{\tau}_1$ and $\xi_i=\hat{\xi}_{i-1}$ for $i\ge3$, while $\xi_2=\tau_2-\tau_1\in[\hat{\tau}_1,\hat{\tau}_1+\delta)$ because $\tau_1\in(t_j-\delta,t_j]$. Hence, on $G_j$,

Since $G_j\in\mathcal{F}^Y_{t_j}$ and, by Step 3, every event of $\sigma(\hat{Y}_u:u\ge0)$ is independent of $G_j$, the induction hypothesis applied to $\hat{Y}$ gives

Summing over $j\ge1$, using $\sum_jP(G_j)=\delta e^{-\delta}\,\frac{e^{-u_1}}{1-e^{-\delta}}$, the inclusion $\bigsqcup_j\big(G_j\cap\{\xi_2>u_2,\dots,\xi_k>u_k\}\big)\subseteq E$ (valid since $G_j\subseteq\{\xi_1>u_1\}$), and the $2\delta e^{-u_1}$ error bound above for the reverse inclusion,

As $\delta\downarrow0$ we have $\frac{\delta}{1-e^{-\delta}}\to1$ (from $\delta-\tfrac{\delta^2}{2}\le1-e^{-\delta}\le\delta$ in Step 2) and $e^{\pm\delta}\to1$, so both bounds converge to $e^{-(u_1+\dots+u_k)}$ and the sandwich yields $P(E)=e^{-(u_1+\dots+u_k)}$, completing the induction.

Step 5: independence. Taking all $u_i=0$ in the survival identity gives $P(\xi_1>0,\dots,\xi_k>0)=1$; in particular $P(\xi_i>0)=1$ for each $i$, and taking $u_j=0$ for $j\neq i$ gives $P(\xi_i>u)=e^{-u}$ for all $u\ge0$. We must show that $\xi_1,\dots,\xi_k$ are independent, that is,

for all Borel sets $B_1,\dots,B_k$.

First, this factorization holds whenever every $B_i$ is either $\mathbb{R}$ or a ray $(u_i,\infty)$ with $u_i$ real. Since each $\xi_i\ge0$ everywhere by construction, a ray with $u_i<0$ gives $\{\xi_i\in(u_i,\infty)\}=\Omega=\{\xi_i\in\mathbb{R}\}$, so we may take every ray to have $u_i\ge0$; and intersecting with, or removing, the probability-one events $\{\xi_j>0\}$ in the coordinates with $B_j=\mathbb{R}$ changes no probability (an intersection with a probability-one event has the same probability). The survival identity, with $u_j=0$ in those coordinates, together with the marginal formula, then gives

Next we upgrade coordinate by coordinate. Fix $n\in\{0,\dots,k-1\}$ and suppose the factorization holds whenever $B_1,\dots,B_n$ are arbitrary Borel sets and each of $B_{n+1},\dots,B_k$ is a ray or $\mathbb{R}$ (the case $n=0$ was just proved). Fix such a choice with the $(n{+}1)$-st coordinate left free, and consider, as functions of a Borel set $B$,

Both are countably additive with $\mu(\mathbb{R})=\nu(\mathbb{R})$ (the induction hypothesis with $B_{n+1}=\mathbb{R}$), and they agree when $B$ is a ray (the induction hypothesis again). The collection $\{B\text{ Borel}:\mu(B)=\nu(B)\}$ contains $\mathbb{R}$, is closed under proper differences and increasing unions (countable additivity), and contains the rays; the rays together with $\mathbb{R}$ form a $\pi$-system generating the Borel $\sigma$-algebra (every interval $(a,b]$ is a difference $(a,\infty)\setminus(b,\infty)$ of rays, $[b,\infty)=\bigcap_n(b-\tfrac1n,\infty)$, and every open set is a countable union of open intervals $(a,b)=(a,\infty)\setminus[b,\infty)$ with rational endpoints), so by Dynkin's $\pi$-$\lambda$ theorem the collection contains every Borel set. Hence the factorization holds with $B_1,\dots,B_{n+1}$ arbitrary Borel sets and the remaining coordinates rays or $\mathbb{R}$. After $k$ such steps the factorization holds for arbitrary Borel sets, which is the independence of $\xi_1,\dots,\xi_k$, with the stated survival functions. Since $k$ was arbitrary, part (b) is proved. $\blacksquare$