Proof of Time Change of the Homogeneous Poisson Process

We verify the three conditions of Stochastic Process, Independent Increments, and Inhomogeneous Poisson Process for $N$. Throughout, $\mathbb{R}$ denotes the real numbers, and we use the properties of the mean function recorded in Stochastic Process, Independent Increments, and Inhomogeneous Poisson Process: $\Lambda(0)=0$ and $\Lambda$ is nondecreasing. The mean function of the rate-one homogeneous process $M$ is $u\mapsto u$ (the Riemann integral of the constant $1$, as noted in that definition and recorded in the statement), so condition 3 for $M$ reads: for all real $0\le u<v$, the increment $M_v-M_u$ has the Poisson distribution with parameter $v-u$.

Step 1 (process and initial value). For each $t\ge0$, $\Lambda(t)$ is a fixed nonnegative real number and $N_t=M_{\Lambda(t)}$ is a member of the family $M$, hence a random variable; so $N$ is a stochastic process on $(\Omega,\mathcal{F},P)$. Moreover $N_0=M_{\Lambda(0)}=M_0=0$ by condition 1 for $M$.

Step 2 (independent increments). Fix real $0\le t_0<t_1<\dots<t_r$ and set $u_i=\Lambda(t_i)$, so that $0\le u_0\le u_1\le\dots\le u_r$ and

Call the index $i$ degenerate if $u_{i-1}=u_i$; then $M_{u_i}-M_{u_{i-1}}=0$ identically on $\Omega$. List the distinct values among $u_0,\dots,u_r$ as $v_0<v_1<\dots<v_m$. Since the $u_i$ are nondecreasing, consecutive distinct values are adjacent in this list, so each nondegenerate increment equals $M_{v_j}-M_{v_{j-1}}$ for exactly one $j$, and distinct nondegenerate indices $i$ correspond to distinct $j$. By condition 2 for $M$ applied to the times $v_0<\dots<v_m$, the nondegenerate increments are independent.

The whole family of the $r$ increments is then independent: fix Borel sets $B_i$ and a set $S$ of indices. For a degenerate index $i$, the event $\{M_{u_i}-M_{u_{i-1}}\in B_i\}$ equals $\Omega$ if $0\in B_i$ and $\emptyset$ otherwise. If some degenerate event in $S$ is $\emptyset$, both sides of the product identity required by Independence of Events and of Random Variables vanish. Otherwise every degenerate event in $S$ is $\Omega$, with $P(\Omega)=1$, so the intersection over $S$ reduces to the intersection over the nondegenerate members of $S$, and the product identity follows from independence of the nondegenerate increments. Thus $N$ has independent increments.

Step 3 (increment distributions). Let $0\le s<t$. If $\Lambda(s)<\Lambda(t)$, condition 3 for $M$ gives that $N_t-N_s=M_{\Lambda(t)}-M_{\Lambda(s)}$ has the Poisson distribution with parameter $\Lambda(t)-\Lambda(s)$. If $\Lambda(s)=\Lambda(t)$, then $N_t-N_s=0$ identically, and its distribution is the unit mass at $0$, which is exactly the Poisson distribution with parameter $0=\Lambda(t)-\Lambda(s)$, as recorded in Poisson Distribution. In both cases condition 3 holds for $N$ with intensity $\lambda$.

Hence $N=(M_{\Lambda(t)})_{t\ge0}$ satisfies conditions 1–3 of Stochastic Process, Independent Increments, and Inhomogeneous Poisson Process and is an inhomogeneous Poisson process with intensity $\lambda$ on $(\Omega,\mathcal{F},P)$. $\blacksquare$