Proof of Existence of the Inhomogeneous Poisson Process

Throughout, $\mathbb{N}$ denotes the natural numbers, $\mathbb{N}_0=\mathbb{N}\cup\{0\}$, $\mathcal{B}(\mathbb{R})$ the Borel $\sigma$-algebra, and $\Lambda$ the mean function of $\lambda$ with its properties from Stochastic Process, Independent Increments, and Inhomogeneous Poisson Process: $\Lambda(0)=0$, $\Lambda$ is nondecreasing, and $\Lambda(t)-\Lambda(s)=\int_s^t\lambda(u)\,du\ge0$ for $0\le s\le t$. Set $\mu_k=\Lambda(k+1)-\Lambda(k)\ge0$ for $k\in\mathbb{N}_0$. We write $\pi_a(c)=\exp(-a)a^{c}/c!$ for $a\ge0$, $c\in\mathbb{N}_0$, for the probability mass function of the Poisson distribution with parameter $a$, with the conventions $0!=1$ and $x^{0}=1$ of that definition.

Step 1 ($\Lambda$ is continuous on every $[0,T]$). Fix $T>0$. The intensity $\lambda$ is continuous on $[0,T]$, so by the extreme value theorem it attains a maximum $M\ge0$ there. For $0\le x\le x'\le T$, the Riemann integral of $\lambda$ over $[x,x']$ lies below the upper sum for the one-interval partition, which is at most $M(x'-x)$; hence

$$0\le\Lambda(x')-\Lambda(x)\le M\,(x'-x).$$

Given $x\in[0,T]$ and $\varepsilon>0$, taking $\delta=\varepsilon/(M+1)$ shows that $\Lambda$ is continuous at $x$ (relative to $[0,T]$). Since $T$ was arbitrary, $\Lambda$ is continuous on $[0,\infty)$ in this sense.

Step 2 (the point distributions $\nu_k$). Fix $k\in\mathbb{N}_0$.

Case $\mu_k>0$. For $v\in(0,1)$ define

$$G_k(v)=\inf\{x\in[k,k+1]:\Lambda(x)-\Lambda(k)\ge v\mu_k\},$$

which exists by the greatest lower bound property: the set is bounded below by $k$ and contains $k+1$, since $\Lambda(k+1)-\Lambda(k)=\mu_k\ge v\mu_k$. Thus $G_k(v)\in[k,k+1]$. We claim: for $v\in(0,1)$ and $x\in[k,k+1]$,

$$G_k(v)\le x\iff v\mu_k\le\Lambda(x)-\Lambda(k).\tag{Q}$$

If $v\mu_k\le\Lambda(x)-\Lambda(k)$, then $x$ belongs to the defining set, so its infimum is at most $x$. Conversely, suppose $G_k(v)\le x$ and let $\varepsilon>0$. By continuity of $\Lambda$ at $G_k(v)$ (Step 1) choose $\delta>0$ with $\Lambda(x')-\Lambda(G_k(v))<\varepsilon$ for all $x'\in[k,k+1]$ with $x'-G_k(v)<\delta$ (such $x'$ satisfy $\Lambda(x')\ge\Lambda(G_k(v))$ when $x'\ge G_k(v)$, by monotonicity). Since $G_k(v)$ is the infimum, there is a member $x'$ of the defining set with $G_k(v)\le x'<G_k(v)+\delta$; then $v\mu_k\le\Lambda(x')-\Lambda(k)<\Lambda(G_k(v))-\Lambda(k)+\varepsilon$. As $\varepsilon>0$ was arbitrary, $v\mu_k\le\Lambda(G_k(v))-\Lambda(k)\le\Lambda(x)-\Lambda(k)$ by monotonicity. This proves (Q).

By (Q), $G_k$ is nondecreasing, and for every real $x$ the set $\{v\in(0,1):G_k(v)\le x\}$ equals: $\emptyset$ for $x<k$; the interval $\{v\in(0,1):v\le(\Lambda(x)-\Lambda(k))/\mu_k\}$ for $x\in[k,k+1]$; all of $(0,1)$ for $x>k+1$. In every case $\{v:G_k(v)>x\}$ is an interval contained in $(0,1)$, hence a Borel set. Let $(\Omega_1,\mathcal{F}_1,L)$ be the probability space with $\Omega_1=(0,1)$, $\mathcal{F}_1=\{B\in\mathcal{B}(\mathbb{R}):B\subseteq(0,1)\}$, and $L$ the restriction of Lebesgue measure, as in Existence of Independent Sequences with Prescribed Distributions. By the generator criterion of Measurable Function and Real-Valued Measurable Function, $G_k:\Omega_1\to\mathbb{R}$ is measurable. Define

$$\nu_k(B)=L\bigl(G_k^{-1}(B)\bigr)\qquad(B\in\mathcal{B}(\mathbb{R})).$$

Since preimages commute with countable disjoint unions, $\nu_k$ inherits countable additivity from the measure $L$, and $\nu_k(\mathbb{R})=L((0,1))=1$ by Existence of Lebesgue Measure on the Real Line; so $\nu_k$ is a probability measure. For $k\le a\le b\le k+1$, writing $F(x)=(\Lambda(x)-\Lambda(k))/\mu_k\in[0,1]$, the sets $\{G_k\le a\}\subseteq\{G_k\le b\}$ have $L$-measures $F(a)$ and $F(b)$ (each is an interval of the form $(0,c]$ or $(0,c)$ intersected with $(0,1)$ with $c=F(a)$ or $F(b)$, of Lebesgue measure $c$ by Existence of Lebesgue Measure on the Real Line); by finite additivity,

$$\nu_k\bigl((a,b]\bigr)=F(b)-F(a)=\frac{\Lambda(b)-\Lambda(a)}{\mu_k}.\tag{E}$$

Case $\mu_k=0$. Let $\nu_k$ be the unit mass at $k$: $\nu_k(B)=1$ if $k\in B$ and $0$ otherwise. This is a probability measure (in any sequence of pairwise disjoint Borel sets, at most one contains $k$). Note that $\nu_k\bigl((a,b]\bigr)=0$ for every interval $(a,b]$ with $a\ge k$, since $k\notin(a,b]$.

In both cases, for every interval $(a,b]$ with $k\le a\le b\le k+1$:

$$\mu_k\,\nu_k\bigl((a,b]\bigr)=\Lambda(b)-\Lambda(a).\tag{E'}$$

Indeed, for $\mu_k>0$ this is (E); for $\mu_k=0$ both sides vanish, the right one because $0\le\Lambda(b)-\Lambda(a)\le\Lambda(k+1)-\Lambda(k)=\mu_k=0$ by monotonicity.

Step 3 (one independent family for all blocks). Every $j\in\mathbb{N}$ has a unique representation $j=2^{k}(2i+1)$ with $k,i\in\mathbb{N}_0$: existence follows by strong induction ($j$ odd gives $k=0$; $j$ even gives $j=2j'$ with $j'<j$, and prepending one factor $2$ to a representation of $j'$ represents $j$); uniqueness holds because $2^{k}(2i+1)=2^{k'}(2i'+1)$ with $k<k'$ would make the odd number $2i+1$ equal to the even number $2^{k'-k}(2i'+1)$. Define a sequence of probability measures $(\rho_j)_{j\in\mathbb{N}}$ by: $\rho_j=$ the Poisson distribution with parameter $\mu_k$ if $j=2^{k}$ (that is, $i=0$), and $\rho_j=\nu_k$ if $j=2^{k}(2i+1)$ with $i\ge1$. By Existence of Independent Sequences with Prescribed Distributions there are a probability space $(\Omega,\mathcal{F},P)$ and an independent sequence $(X_j)_{j\in\mathbb{N}}$ on it with $X_j$ of distribution $\rho_j$. Put

$$K_k=X_{2^{k}},\qquad U_{k,i}=X_{2^{k}(2i+1)}\quad(k\in\mathbb{N}_0,\ i\in\mathbb{N}),$$

and let $I_k=\{2^{k}(2i+1):i\in\mathbb{N}_0\}$, so that the sets $I_k$, $k\in\mathbb{N}_0$, are pairwise disjoint by uniqueness of the representation. For each $k$, the block family $(K_k,U_{k,1},U_{k,2},\dots)$ is independent (every finite subfamily of an independent family is independent, by Independence of Events and of Random Variables), $K_k$ has the Poisson distribution with parameter $\mu_k$, and each $U_{k,i}$ has distribution $\nu_k$. Let $\widetilde K_k$ be the truncation of $K_k$ as in Thinning: Cell Counts of a Poisson Number of Independent Points, and for a Borel set $A$ let

$$C_k(A)=\sum_{i=1}^{\widetilde K_k}\mathbf{1}_{\{U_{k,i}\in A\}}$$

be the block-$k$ cell count of $A$, where $\mathbf{1}_{E}$ is $1$ on $E$ and $0$ off $E$ and an empty sum is $0$. By Part 1 of Thinning: Cell Counts of a Poisson Number of Independent Points, each $C_k(A)$ is an $\mathbb{N}_0$-valued random variable, measurable with respect to the generated $\sigma$-algebra $\mathcal{G}_k=\sigma(K_k,(U_{k,i})_{i\in\mathbb{N}})$; and $\mathcal{G}_k=\sigma(X_j:j\in I_k)$, since the two generating families coincide. By Grouping Lemma for Independent Random Variables applied to $(X_j)_{j\in\mathbb{N}}$ and the disjoint blocks $(I_k)_{k\in\mathbb{N}_0}$, the family $(\mathcal{G}_k)_{k\in\mathbb{N}_0}$ is independent.

Step 4 (the process). For $t\ge0$ define

$$N_t=\sum_{k\in\mathbb{N}_0,\;k<t}C_k\bigl((k,\min(t,k+1)]\bigr).$$

By the Archimedean property there is an integer exceeding $t$, so the outer sum has finitely many terms; each term is an $\mathbb{N}_0$-valued random variable by Step 3, and a finite sum of $\mathbb{N}_0$-valued random variables is again one (its level sets are finite unions, over decompositions of the value, of finite intersections of level sets of the summands). Hence $N=(N_t)_{t\ge0}$ is a stochastic process on $(\Omega,\mathcal{F},P)$ in the sense of Stochastic Process, Independent Increments, and Inhomogeneous Poisson Process. For $t=0$ the index set of the sum is empty, so $N_0=0$ identically; this is condition 1 of Stochastic Process, Independent Increments, and Inhomogeneous Poisson Process.

Step 5 (increments are sums of cell counts). Let $0\le s<t$ and $k<t$, $k\in\mathbb{N}_0$. If $k<s$, then $(k,\min(s,k+1)]\subseteq(k,\min(t,k+1)]$ and, pointwise on $\Omega$, subtracting indicator sums gives the count of the set difference

$$(k,\min(t,k+1)]\setminus(k,\min(s,k+1)]=(s,t]\cap(k,k+1];$$

if $s\le k<t$, the term for $k$ is absent from $N_s$ and $(k,\min(t,k+1)]=(s,t]\cap(k,k+1]$. In all cases, for every $\omega$,

$$N_t-N_s=\sum_{k\in\mathbb{N}_0,\;k<t}C_k\bigl((s,t]\cap(k,k+1]\bigr).\tag{I}$$

Step 6 (finite-dimensional computation). Fix $r\in\mathbb{N}$ and real $0\le t_0<t_1<\dots<t_r$, and fix an integer $m\ge t_r$ (Archimedean property). For $1\le i\le r$ and $0\le k\le m$ define the clamp $c_{k,i}=\min(\max(k,t_{i-1}),t_i)$, and for $1\le i\le r$ and $0\le k\le m-1$ the cell and associated quantities

$$J_{k,i}=(t_{i-1},t_i]\cap(k,k+1],\qquad C_{k,i}=C_k(J_{k,i}),\qquad p_{k,i}=\nu_k(J_{k,i}),\qquad \ell_{k,i}=\mu_k\,p_{k,i}.$$

When $J_{k,i}\ne\emptyset$ it equals the interval $(c_{k,i},c_{k+1,i}]$ with $k\le c_{k,i}\le c_{k+1,i}\le k+1$, so by (E') $\ell_{k,i}=\Lambda(c_{k+1,i})-\Lambda(c_{k,i})$; when $J_{k,i}=\emptyset$ we have $c_{k,i}=c_{k+1,i}$ and $p_{k,i}=0$, so the same identity holds, both sides being $0$. Since $c_{0,i}=t_{i-1}$ (as $t_{i-1}\ge0$) and $c_{m,i}=t_i$ (as $m\ge t_i$), telescoping gives

\sum_{k=0}^{m-1}\ell_{k,i}=\Lambda(t_i)-\Lambda(t_{i-1})=:a_i.\tag{T}$$ By (I), pointwise on $\Omega$,

D_i:=N_{t_i}-N_{t_{i-1}}=\sum_{k=0}^{m-1}C_{k,i}\qquad(1\le i\le r),

since blocks with $t_i\le k<m$ have empty cells and contribute the zero count. For each fixed $k$, the cells $J_{k,1},\dots,J_{k,r}$ are pairwise disjoint Borel sets, so Part 2 of [Thinning: Cell Counts of a Poisson Number of Independent Points](/theorems/00076d48-0649-4f07-b086-49621627fd37?v=ad5d94b7-9c7a-485b-b150-0d36db17d6b0) applied to the block family of Step 3 gives, for all $(n_{k,1},\dots,n_{k,r})\in\mathbb{N}_0^{r}$,

P\Bigl(\bigcap_{i=1}^{r}{C_{k,i}=n_{k,i}}\Bigr)=\prod_{i=1}^{r}\pi_{\ell_{k,i}}(n_{k,i}).

Each event $\bigcap_{i}\{C_{k,i}=n_{k,i}\}$ lies in $\mathcal{G}_k$ ($\sigma$-algebras are closed under finite intersections, and each level set lies in $\mathcal{G}_k$ by Step 3). Since the $(\mathcal{G}_k)_{k}$ are independent (Step 3), the defining product formula of [Sigma-Algebra Generated by Random Variables and Independence of Sigma-Algebras](/theorems/45cbc037-1cc0-4685-b6f8-c85362ca91c5?v=2972f869-5f7e-4958-a4c2-4e269734b251) yields, for every array $(n_{k,i})\in\mathbb{N}_0^{m\times r}$,

P\Bigl(\bigcap_{k=0}^{m-1}\bigcap_{i=1}^{r}{C_{k,i}=n_{k,i}}\Bigr)=\prod_{k=0}^{m-1}\prod_{i=1}^{r}\pi_{\ell_{k,i}}(n_{k,i}).\tag{A}

**Step 7 (the increments are independent Poisson variables).** Fix $(d_1,\dots,d_r)\in\mathbb{N}_0^{r}$. As every $\omega$ realizes exactly one value array $(C_{k,i}(\omega))$, the event $\bigcap_{i}\{D_i=d_i\}$ is the disjoint union, over the finitely many arrays $(n_{k,i})\in\mathbb{N}_0^{m\times r}$ with column sums $\sum_{k}n_{k,i}=d_i$ for every $i$, of the events in (A). Summing (A) over these arrays, factoring the finite sum over the product of the $r$ independent column-composition sets, and using finite distributivity,

P\Bigl(\bigcap_{i=1}^{r}{D_i=d_i}\Bigr)=\prod_{i=1}^{r}\ \sum_{\substack{(n_0,\dots,n_{m-1})\in\mathbb{N}0^{m}\ n_0+\dots+n{m-1}=d_i}}\ \prod_{k=0}^{m-1}\exp(-\ell_{k,i})\frac{\ell_{k,i}^{,n_k}}{n_k!}.

By [Basic Properties of the Exponential Function](/theorems/745eee19-f7d5-46cb-93c0-bea412b66c84?v=627637d7-8718-4c1a-afc1-3f1747730725), $\prod_k\exp(-\ell_{k,i})=\exp(-\sum_k\ell_{k,i})=\exp(-a_i)$ using (T); and by [Multinomial Theorem](/theorems/5fa77c58-46b2-4838-a0f0-5a7aa48501e2?v=ae51f06e-4c33-49c2-a5ca-da27d87e381d),

\sum_{\substack{n\in\mathbb{N}0^{m}\ n_0+\dots+n{m-1}=d_i}}\prod_{k=0}^{m-1}\frac{\ell_{k,i}^{,n_k}}{n_k!}=\frac{1}{d_i!}\sum_{\substack{n\in\mathbb{N}0^{m}\ n_0+\dots+n{m-1}=d_i}}\frac{d_i!}{n_0!\cdots n_{m-1}!}\prod_{k=0}^{m-1}\ell_{k,i}^{,n_k}=\frac{(\ell_{0,i}+\dots+\ell_{m-1,i})^{d_i}}{d_i!}=\frac{a_i^{,d_i}}{d_i!}.

$$Therefore$$

P\Bigl(\bigcap_{i=1}^{r}{D_i=d_i}\Bigr)=\prod_{i=1}^{r}\pi_{a_i}(d_i).

Each $g_i=\pi_{a_i}$ maps $\mathbb{N}_0$ into $[0,1]$ with $\sum_{c=0}^{\infty}g_i(c)=1$ (total-mass computation of [Poisson Distribution](/theorems/13339035-fea9-4910-b87a-d176a80a7f3b?v=7fc3df59-5000-4178-8d23-5def5d23b0d3)), and each $D_i$ takes all its values in $\mathbb{N}_0$. By [Factorized Joint Probability Mass Function Implies Independence](/theorems/ff1d24d1-04c8-46a2-b7b5-c819b44d6ec5?v=2dfc8e4f-17a7-47fa-9c36-37d7246665c9), the increments $D_1,\dots,D_r$ are independent, and for every Borel set $B$, $P(D_i\in B)=\sum_{c\in B\cap\mathbb{N}_0}\pi_{a_i}(c)$, which is the Poisson distribution with parameter $a_i=\Lambda(t_i)-\Lambda(t_{i-1})$ evaluated at $B$, by [Poisson Distribution](/theorems/13339035-fea9-4910-b87a-d176a80a7f3b?v=7fc3df59-5000-4178-8d23-5def5d23b0d3). Since the grid was arbitrary, $N$ has independent increments (condition 2 of [Stochastic Process, Independent Increments, and Inhomogeneous Poisson Process](/theorems/a9f96609-3aca-49f0-a7d8-c261334d5a0b?v=0280e5e4-70fc-4c52-b45c-ae60fba1263e)); and for $0\le s<t$, the case $r=1$ with $t_0=s$, $t_1=t$ shows that $N_t-N_s$ has the Poisson distribution with parameter $\Lambda(t)-\Lambda(s)$ (condition 3). With Step 4 this proves that $N$ is an inhomogeneous Poisson process with intensity $\lambda$ on $(\Omega,\mathcal{F},P)$. **Step 8 (homogeneous case).** Let $\theta\ge0$ be real and let $\lambda$ be the constant function with value $\theta$. It is nonnegative, and it is continuous on every $[0,T]$ (for any $\varepsilon>0$ every $\delta>0$ works in the definition of [continuity](/theorems/57c56cbb-67a0-4576-829b-60e2eaf9ee96?v=35b25058-b3a6-4e40-8314-d8efabe1b539&show_redacted=true)); so it is an intensity function. By the theorem just proved there is an inhomogeneous Poisson process $N$ with this intensity, and by [Stochastic Process, Independent Increments, and Inhomogeneous Poisson Process](/theorems/a9f96609-3aca-49f0-a7d8-c261334d5a0b?v=0280e5e4-70fc-4c52-b45c-ae60fba1263e) such a process is precisely a homogeneous Poisson process with rate $\theta$. $\blacksquare$