Temporal intensity and likelihood — Examples

The extension by zero has unit jumps only at the simple event times of $\eta$, and $N_t$ is finite because $N_1$ is Poisson with mean $1$. The effective intensity ${\lambda }^{\ast }(t)={\mathbf{1}}_{(0,1]}(t)$ is deterministic Borel and hence predictable by [F4]. Direct integration gives $A_t={\int }_0^t{\lambda }^{\ast }(u)du=t\wedge 1$, so $A$ is continuous, nondecreasing, and bounded by $1$.

Let $\mathsf{H}$ be the space of finite counting measures on $(0,\mathrm{\infty })$ with the sigma-algebra generated by its Borel-set evaluation maps. Define the history random element by $H_s(B):=\eta (B\cap (0,s\wedge 1])$ for every Borel $B\subseteq (0,\mathrm{\infty })$. It is a finite counting measure, and it is measurable because each evaluation is a measurable count of the supplied point process; its generated sigma-algebra is the event-history filtration in the Given clause. For finite $u>0$, define $h\oplus u:=h+{\delta }_u$. For every Borel $B$, $(h\oplus u)(B)=h(B)+{\mathbf{1}}_B(u)$, so the update is jointly measurable in $h,u$ and appends exactly one event.

For $0\le s\le t$, the increment $N_t-N_s$ is the count on $(s,t]\cap (0,1]$, a Poisson variable of mean $\min (t,1)-\min (s,1)$. Refine a finite family of past Borel sets into the disjoint atoms of the finite Boolean algebra they generate. Every atom lies in $(0,s]$, so [F1] gives independence of its count vector from the future increment. These count cylinders generate the natural-history sigma-algebra, and [F5] extends independence to that sigma-algebra. Therefore $\mathbb{E}[N_t-N_s\mid {\mathcal{H}}_s]=\min (t,1)-\min (s,1)=A_t-A_s$. Thus $N-A$ is a martingale. Since $N_t$ has finite second moment by [F3] and $A_t\le 1$, this martingale is square-integrable and meets [F6]; $\mathbb{E}[A_T]\le 1$.

For $s\ge 0$ and $h\in \mathsf{H}$, supply the next-event kernel, viewed on $[0,\mathrm{\infty }]$ and supported on $(s,\mathrm{\infty }]$, by $K_{s,h}(C):={\int }_{C\cap (s,1]}{e}^{-(u-s)}du+e^{-(1-s{)}_{+}}{\mathbf{1}}_{\{\mathrm{\infty }\in C\}}$ for Borel $C\subseteq [0,\mathrm{\infty }]$, where $x_{+}:=\max (x,0)$. Its finite-time mass is $1-e^{-(1-s{)}_{+}}$, so each $K_{s,h}$ is a probability measure. For fixed $C$, the finite-time term is $e^s{\int }_{C\cap (s,1]}{e}^{-u}du$, a Borel function of $s$ because the tail integral is monotone on $[0,1]$ and zero for $s\ge 1$; the atom mass is also Borel. The finite-time density $f_{s,h}(u)={\mathbf{1}}_{\{s<u\le 1\}}{e}^{-(u-s)}$ is jointly measurable in $(s,h,u)$. By independent future increments, this is the conditional next-event law given $H_s$: its survivor for $u>s$ is $S_{s,h}(u)=e^{-((u\wedge 1)-s{)}_{+}}$, since the event of no point in $(s,u]\cap (0,1]$ has probability $e^{-((u\wedge 1)-s{)}_{+}}$. Thus $f_{s,h}\mathrm{/}{S}_{s,h}=1$ on $(s,1]$ and $0$ after $1$, agreeing almost everywhere with ${\lambda }^{\ast }(u\mid h_{u-})={\mathbf{1}}_{(0,1]}(u)$; this also agrees with [F7]. In particular, $K_{0,h}(\{\mathrm{\infty }\})=e^{-1}$ and $K_{s,h}(\{\mathrm{\infty }\})=1$ for $s\ge 1$.

Taking $u\to \mathrm{\infty }$ in step 3.1 gives $\mathrm{Pr}(T_1=\mathrm{\infty })=e^{-1}$, also the zero-count probability $\mathrm{Pr}(N_1=0)$ by [F3]. No events occur after time $1$, and $N_1$ is finite almost surely, so the process terminates and is non-explosive.

By [F1], the supplied process has total count with law $\mathrm{Pois}(\theta T)$, hence finite almost surely; at $\theta =0$ it is empty. For $\theta >0$, partition $(0,T]$ into $2^n$ equal half-open intervals. A cell count $C$ has mean ${\lambda }_n=\theta T\mathrm{/}{2}^n$, and ${\mathbf{1}}_{\{C\ge 2\}}\le C(C-1)\mathrm{/}2$ gives $P(C\ge 2)\le {\lambda }_n^2\mathrm{/}2$ by [F3]. Thus the probability that any cell contains two events is at most ${\theta }^2{T}^2\mathrm{/}{2}^{n+1}$. Any repeated event forces such a collision for every $n$, so its probability is zero. The supplied process is therefore simple. Extend its count by $N_t=\eta ((0,t\wedge T])$. For $0\le s<t$, $N_t-N_s$ is the count on $(s,t]\cap (0,T]$, independent of each finite cylinder of past Borel-set counts by [F1]. Indeed, refine any finite family of past Borel sets into the disjoint atoms of the finite Boolean algebra they generate. Each atom lies in $(0,s\wedge T]$ and is disjoint from $(s,t]\cap (0,T]$, so [F1] makes the atom-count vector independent of the future increment; the cylinder vector is a function of those atom counts. Such cylinders generate ${\mathcal{H}}_s$, so [F4] extends independence to the whole history sigma-algebra. Therefore $\mathbb{E}[N_t-N_s\mid {\mathcal{H}}_s]=\theta ((t\wedge T)-(s\wedge T))$, and $N_t-A_t$ is a martingale for $A_t=\theta (t\wedge T)$. The effective intensity is ${\lambda }^{\ast }(t)=\theta {\mathbf{1}}_{(0,T]}(t)$, which is deterministic Borel and predictable by [F5]; its integral is $A_t$, continuous and bounded by $\theta T$. Also $N_t$ has finite second moment by [F3]. Thus $N-A$ is square-integrable and the Poisson process meets the compensator contract [F6].

Let $\mathsf{H}$ be the space of finite counting measures on $(0,\mathrm{\infty })$ with its evaluation sigma-algebra, and define $H_s(B):=\eta (B\cap (0,s\wedge T])$ for Borel $B\subseteq (0,\mathrm{\infty })$. Each $H_s$ is finite and measurable by the evaluation measurability of $\eta$, and $\sigma (H_s)={\mathcal{H}}_s$. For finite $u>0$ put $h\oplus u=h+{\delta }_u$; the evaluation identity $(h\oplus u)(B)=h(B)+{\mathbf{1}}_B(u)$ shows that the update is jointly measurable and appends one event.

The specified initial history contributes $c_0{e}^{-\beta u}$ at time $u>0$; its past-event representation follows by factoring $e^{-\beta u}$ out of the nonnegative weighted sum. For a fixed finite observed pattern, the pre-event intensity at $t_i$ is therefore $\mu +c_0{e}^{-\beta t_i}+\alpha {\sum }_{j<i}{e}^{-\beta (t_i-t_j)}$. Integrating the baseline gives $\mu T$, and integrating the initial excitation gives $c_0(1-e^{-\beta T})\mathrm{/}\beta$ by [F10]. For each observed event time $t_i$, the integral ${\int }_{t_i}^T{e}^{-\beta (u-t_i)}du$ equals $\frac{1-e^{-\beta (T-t_i)}}{\beta }$ by [F10], also when $t_i=T$, when both sides are zero. Summing these terms gives the displayed cumulative hazard.

For $s\ge 0$ and $h\in \mathsf{H}$, supply the next-event kernel, supported on $(s,\mathrm{\infty }]$, by $K_{s,h}(C):={\int }_{C\cap (s,T]}\theta e^{-\theta (u-s)}du+e^{-\theta (T-s{)}_{+}}{\mathbf{1}}_{\{\mathrm{\infty }\in C\}}$ for Borel $C\subseteq [0,\mathrm{\infty }]$, where $x_{+}:=\max (x,0)$. Its finite-time mass is $1-e^{-\theta (T-s{)}_{+}}$, so each $K_{s,h}$ is a probability measure, including when $\theta =0$ or $s\ge T$. For fixed $C$, the finite-time term is $\theta e^{\theta s}{\int }_{C\cap (s,T]}{e}^{-\theta u}du$, a Borel function of $s$ because the tail integral is monotone on $[0,T]$ and zero for $s\ge T$; the atom mass is also Borel. The finite-time density $f_{s,h}(u)=\theta {\mathbf{1}}_{\{s<u\le T\}}{e}^{-\theta (u-s)}$ is jointly measurable in $(s,h,u)$. Independent increments give the same conditional survivor, $S_{s,h}(u)=e^{-\theta ((u\wedge T)-s{)}_{+}},u>s,$ so the kernel is compatible with the observed finite-window Poisson law and supplies its terminal continuation after $T$. Its hazard is $\theta$ on $s<u\le T$ and $0$ for $u>T$, agreeing with ${\lambda }^{\ast }(u\mid h_{u-})=\theta {\mathbf{1}}_{(0,T]}(u)$ (and is identically zero at $\theta =0$). Thus the homogeneous survivor formula used in the likelihood is exactly $e^{-\theta (u-s)}$ for $s<u\le T$.

Substitute the Hawkes pre-event values and integrated hazard from step 1.3 into [F7], conditioning on the same specified initial history for every parameter value. This gives exactly the displayed Hawkes likelihood ratio, including the initial-excitation contribution when $m=0$; for empty history $c_0=0$. The calculation is conditional on a compatible process law and proves no existence, non-explosion, or stationarity property for the specification.

Within the observation window, $0\le s<u\le T$, the probability of no event in $(s,u]$ is $e^{-\theta (u-s)}$, so the no-event hazard is $\theta$ there. Applying [F7] to the $m$ events gives exact ratio $e^T{\theta }^m{e}^{-\theta T}$, including $m=0$ by the empty-product and terminal-survival conventions.

If $m>0$ and $\theta >0$, the log-likelihood is $\mathrm{\ell }(\theta )=m\log \theta -\theta T$, with derivative ${\mathrm{\ell }}^{\mathrm{\prime }}(\theta )=m\mathrm{/}\theta -T$ by [F8]. It is positive for $0<\theta <m\mathrm{/}T$ and negative for $\theta >m\mathrm{/}T$. The mean value theorem [F9] therefore makes $\mathrm{\ell }$ increase up to $m\mathrm{/}T$ and decrease thereafter; at $\theta =0$ the likelihood is zero because $m>0$. Thus the unique maximum is $\widehat{\theta }=m\mathrm{/}T$.

If $m=0$, the likelihood is $e^{-\theta T}$, which strictly decreases on $\theta \ge 0$ because $T>0$ and the exponential is strictly increasing. Its maximum is attained at $\theta =0$. If zero were excluded from the parameter space, this case would have only a boundary supremum.