Proof of Kolmogorov Forward Equations for the Inhomogeneous Poisson Process

Throughout, $\mathbb{N}$ denotes the natural numbers, $\mathbb{N}_0=\mathbb{N}\cup\{0\}$ the nonnegative integers as in the statement, and $\mathbb{R}$ the real numbers; $\lambda$, $\Lambda$, and $N$ are as in the statement, and we use the properties of $\Lambda$ recorded in Stochastic Process, Independent Increments, and Inhomogeneous Poisson Process: $\Lambda(0)=0$, $\Lambda$ is nondecreasing, and $\Lambda(t)-\Lambda(s)=\int_s^t\lambda(u)\,du$. From Basic Properties of the Exponential Function we use that $\exp$ is positive, $\exp'=\exp$, $\exp(u+v)=\exp(u)\exp(v)$, and the defining series.

Step 1 (Claim 1 and initial values). For $t>0$, condition 3 of Stochastic Process, Independent Increments, and Inhomogeneous Poisson Process with $s=0$ together with condition 1 gives that $N_t=N_t-N_0$ has the Poisson distribution with parameter $\Lambda(t)-\Lambda(0)=\Lambda(t)$; hence, since $\{k\}$ is a Borel set,

$$p_k(t)=P(N_t\in\{k\})=P_{\Lambda(t)}(\{k\})=\exp(-\Lambda(t))\,\frac{\Lambda(t)^{k}}{k!}.$$

For $t=0$: $N_0=0$, so $p_0(0)=1$ and $p_k(0)=0$ for $k\ge1$, which agrees with the displayed formula since $\Lambda(0)=0$, $\exp(0)=1$, and the conventions $0!=1$ and $0^{0}=1$ of Poisson Distribution, while $0^{k}=0$ for $k\ge1$. This proves Claim 1 and the stated initial values.

Step 2 (derivative of $\Lambda$). By Fundamental Theorem of Calculus, Part I in One Dimension applied on $[0,T]$ for any $T>t$, $\Lambda$ has derivative $\Lambda'(t)=\lambda(t)$ at every $t>0$; in particular $\Lambda$ is continuous and, since $\lambda$ is continuous, $\Lambda$ is a $C^1$ map on the open set $(0,\infty)$. At $t=0$ we use a direct squeeze: for $h>0$, let $m_h$ and $M_h$ be the minimum and maximum of $\lambda$ on $[0,h]$ (Extreme Value Theorem on a Compact Interval); every lower and upper sum of $\lambda$ on $[0,h]$ lies between $m_h\,h$ and $M_h\,h$, so $m_h\le\Lambda(h)/h\le M_h$, and by continuity of $\lambda$ at $0$ both bounds tend to $\lambda(0)$ as $h\downarrow0$; hence $\Lambda(h)/h\to\lambda(0)$ and $\Lambda(h)\to0$.

Step 3 (Claim 2 for $t>0$). Fix $t>0$. By the one-dimensional chain rule and $\exp'=\exp$, the function $t\mapsto\exp(-\Lambda(t))$ has derivative $-\lambda(t)\exp(-\Lambda(t))$. By the product rule for real functions — $(fg)'(x)=f'(x)g(x)+f(x)g'(x)$, from the factorization $f(x+s)g(x+s)-f(x)g(x)=(f(x+s)-f(x))g(x+s)+f(x)(g(x+s)-g(x))$ and limit arithmetic (Derivative at an Interior Point) — and induction on $k$, the function $t\mapsto\Lambda(t)^{k}$ has derivative $k\,\Lambda(t)^{k-1}\lambda(t)$ for $k\ge1$. Hence for $k\ge1$, using $k/k!=1/(k-1)!$ (Factorial of a Natural Number and the convention $0!=1$),

$$p_k'(t)=\exp(-\Lambda(t))\Bigl(-\lambda(t)\,\frac{\Lambda(t)^{k}}{k!}+\lambda(t)\,\frac{k\,\Lambda(t)^{k-1}}{k!}\Bigr)=\lambda(t)\bigl(p_{k-1}(t)-p_k(t)\bigr),$$

and for $k=0$, $p_0'(t)=-\lambda(t)\,p_0(t)$.

Step 4 (Claim 2 at $t=0$). From the defining series, $|\exp(-u)-1+u|\le u^{2}$ for $0\le u\le1$: the remainder is $\sum_{j\ge2}(-u)^{j}/j!$, bounded in absolute value by $u^{2}\sum_{j\ge2}1/j!\le u^{2}$. By Step 2, $u=\Lambda(h)\to0$ as $h\downarrow0$, so for small $h>0$,

$$\frac{p_0(h)-p_0(0)}{h}=\frac{\exp(-\Lambda(h))-1}{h}=-\frac{\Lambda(h)}{h}+\frac{R(h)}{h},\qquad|R(h)|\le\Lambda(h)^{2},$$

and $\Lambda(h)^{2}/h=(\Lambda(h)/h)\cdot\Lambda(h)\to\lambda(0)\cdot0=0$; hence the one-sided derivative is $p_0'(0)=-\lambda(0)=-\lambda(0)\,p_0(0)$. For $k\ge1$, $p_k(0)=0$ and

$$\frac{p_k(h)-p_k(0)}{h}=\exp(-\Lambda(h))\cdot\frac{1}{k!}\cdot\frac{\Lambda(h)}{h}\cdot\Lambda(h)^{k-1}\ \longrightarrow\ \begin{cases}\lambda(0), & k=1,\\ 0, & k\ge2,\end{cases}$$

which equals $\lambda(0)\bigl(p_{k-1}(0)-p_k(0)\bigr)$ in both cases. This proves Claim 2.

Step 5 (Claim 3). Let $(q_k)_{k\in\mathbb{N}_0}$ be as in Claim 3 and set $r_k=q_k-p_k$; each $r_k$ is differentiable in the same sense with $r_k(0)=0$, and by subtracting the two systems,

$$r_0'=-\lambda\,r_0,\qquad r_k'=\lambda\,r_{k-1}-\lambda\,r_k\quad(k\ge1).$$

We show $r_k\equiv0$ by induction on $k$. In the base case $k=0$, and in the inductive step where $r_{k-1}\equiv0$, the relevant equation is $r_k'=-\lambda\,r_k$ on all of $[0,\infty)$. Consider $\varphi(t)=r_k(t)\exp(\Lambda(t))$. By the product and chain rules exactly as in Step 3, for every $t>0$,

$$\varphi'(t)=\bigl(r_k'(t)+\lambda(t)\,r_k(t)\bigr)\exp(\Lambda(t))=0 .$$

Fix $t>0$. The function $\varphi$ is continuous on $[0,t]$ (differentiable functions are continuous, one-sidedly at the endpoints, directly from the difference-quotient limit) and differentiable on $(0,t)$ with zero derivative, so Mean Value Theorem in One Dimension gives $\xi\in(0,t)$ with $\varphi(t)-\varphi(0)=\varphi'(\xi)\,t=0$. Hence $\varphi(t)=\varphi(0)=r_k(0)\exp(0)=0$ for all $t\ge0$, and since $\exp(\Lambda(t))>0$ we get $r_k\equiv0$. By induction, $q_k=p_k$ for every $k\in\mathbb{N}_0$. $\blacksquare$