Deriving the Multiplicative Intensity Likelihood

Why the likelihood has two parts

A counting-process path records both when events occur and how long the process is observed without another event. A likelihood based on that path must account for both kinds of information: each observed jump is supported by the intensity at its time, while every event-free interval contributes evidence that no jump occurred. The result is a product of intensity values at the jumps multiplied by an exponential involving the negative integrated intensity.

The derivation below treats a single counting process on a fixed observation interval $[0,\tau]$. The key idea extends to multiple processes, but the focus here is the likelihood contribution of one observed process. The earlier material on event-history likelihoods from intensities introduced the connection between intensities and event-time likelihoods; here we derive the compact counting-process expression and identify precisely what it represents.

Definition: Let $N(t)$, $0\leq t\leq\tau$, be a simple counting process with $N(0)=0$, and let $\lambda(t;\theta)$ be its conditional intensity under parameter $\theta$. On a path with jump times $0<T_1<\cdots<T_D\leq\tau$, the intensity-based likelihood kernel is $$ L(\theta)\ \propto\ \left\{\prod_{j=1}^{D}\lambda(T_j;\theta)\right\} \exp\left\{-\int_0^\tau \lambda(t;\theta)\,dt\right\}. $$ The proportionality omits factors that do not depend on $\theta$. The intensity is predictable: its value at a possible jump time is determined by information available just before that time.

Conditions for the path likelihood

The expression is a likelihood for the observed path under a model, not just an algebraic summary of event counts. The following conditions clarify its use.

Simple paths and finite activity. The process has unit jumps, no simultaneous jumps of this same process, and finitely many jumps on the observation interval with probability one.
An absolutely continuous compensator. The compensator has the form $\int_0^t\lambda(s;\theta)\,ds$, with a nonnegative, predictable intensity. Thus the model has no unrepresented fixed-time jump atoms.
A common observation scheme. The interval and any conditioning information, such as a predictable at-risk process, are treated as fixed or conditioned upon. If the law of that information also depends on $\theta$, its likelihood contribution cannot simply be discarded.
A common support. The model must assign a valid density to the observed path. In particular, an observed jump where the modeled intensity is zero makes the likelihood zero.

For a multiplicative intensity, write $\lambda(t;\theta)=Y(t)\alpha(t;\theta)$, where $Y(t)$ is a predictable exposure or at-risk indicator and $\alpha(t;\theta)$ is the rate while exposed. At an observed jump, $Y(T_j)=1$ under the usual convention that the process is at risk immediately before the jump. The integrated term then accumulates intensity only over exposed time.

Deriving the jump and no-jump contributions

Suppose the process has already followed a particular history up to a time $s$, and consider the next interval before the next jump. Conditional on the history, the probability of no jump from $s$ through $t$ is the survival probability associated with the conditional intensity along the no-jump continuation:

P(\text{no jump on }(s,t]\mid\text{history at }s) = \exp\left\{-\int_s^t\lambda(u;\theta)\,du\right\}.

If the next jump occurs at $t$, its conditional density is the probability of surviving without a jump up to $t$, multiplied by the instantaneous rate of a jump at $t$. Therefore its density is

\lambda(t;\theta) \exp\left\{-\int_s^t\lambda(u;\theta)\,du\right\}.

Apply this argument successively between observed jumps. The intervals are $(0,T_1]$, $(T_1,T_2]$, and so on, ending with the event-free interval $(T_D,\tau]$. Multiplying the conditional densities supplies one factor $\lambda(T_j;\theta)$ for every observed jump. The exponential terms combine because integrals over adjacent intervals add:

\prod_{j=1}^{D} \exp\left\{-\int_{T_{j-1}}^{T_j}\lambda(u;\theta)\,du\right\} \cdot \exp\left\{-\int_{T_D}^{\tau}\lambda(u;\theta)\,du\right\} = \exp\left\{-\int_0^\tau\lambda(u;\theta)\,du\right\}, \qquad T_0=0.

Consequently, the density for the ordered jump times is the stated product-integral expression. Equivalently, the log-likelihood, up to terms not depending on $\theta$, is

\ell(\theta) = \sum_{j=1}^{D}\log\lambda(T_j;\theta) - \int_0^\tau\lambda(t;\theta)\,dt.

This derivation assumes the conditional intensity describes the path as it unfolds, including any dependence on the observed history. The intensity at a jump is evaluated using the predictable, just-before-jump information. The final survival factor is essential: it includes the interval after the last observed event, not just the gaps between events.

Worked Example: Constant Intensity Over a Fixed Interval

Consider a hypothetical process observed for $\tau=8$ hours with three jumps at $1.2$, $3.5$, and $6.1$ hours. Assume a constant intensity $\lambda(t;\theta)=\theta$, measured in events per hour, and evaluate the likelihood kernel at $\theta=0.5$ events per hour.

The model is nonnegative and constant, the observation interval is fixed, and the three jumps are distinct. These assumptions satisfy the simple-path and continuous-intensity conditions for this example. The jump product is $(0.5)^3=0.125$, and the integrated intensity is $0.5\times 8=4$. Thus

L(0.5)\ \propto\ (0.5)(0.5)(0.5)\exp(-0.5\times 8) =0.125e^{-4} \approx 0.002289.

This is a density for the three ordered event times, not a probability that the process takes exactly those times. Its units are $(\text{events per hour})^3$, as appropriate for a density over three time coordinates. A parameter-free constant in the choice of reference measure would not change likelihood comparisons between values of $\theta$.

The multiplicative intensity form

Substitute $\lambda(t;\theta)=Y(t)\alpha(t;\theta)$ into the path likelihood. For a predictable at-risk indicator, the expression becomes

L(\theta)\ \propto\ \left\{\prod_{j=1}^{D}Y(T_j)\alpha(T_j;\theta)\right\} \exp\left\{-\int_0^\tau Y(t)\alpha(t;\theta)\,dt\right\}.

On a valid observed path, $Y(T_j)=1$ at each jump, so those factors equal one. The remaining product evaluates the rate at the event times, while the integral is the total modeled intensity over the time at risk. If $Y(t)=0$ during some interval, that interval contributes neither a jump factor nor integrated exposure.

This form also shows why event counts alone are generally insufficient. Even with the same number of jumps, their timing can change the intensity factors when $\alpha(t;\theta)$ varies over time or depends on the observed history. The integrated-intensity term, in turn, reflects how much opportunity for events the path contained.

Worked Example: Constant Rate With Risk-Set Exposure

Suppose a hypothetical study observes a process with constant event rate $\beta$ per person-year while at risk. Across the observed histories, total time at risk is 10 person-years and there are four distinct jumps, all occurring while the process is at risk. Evaluate the likelihood kernel at $\beta=0.3$ events per person-year.

The assumed intensity is $\lambda(t;\beta)=Y(t)\beta$. The risk indicator is predictable, the jumps occur only when $Y=1$, and the total exposure is finite. The jump product is $\beta^4$; the integrated intensity is $\beta$ times total exposure. Therefore

L(0.3)\ \propto\ (0.3)^4\exp\{-(0.3)(10)\} =0.0081e^{-3} \approx 0.0004033.

The likelihood kernel has units $(\text{events per person-year})^4$, the density units for four event times under this exposure-time representation. The calculation uses 10 person-years rather than the calendar duration because the process contributes intensity only while at risk.

Worked Example: Piecewise-Constant Rate

In a hypothetical observation, the process is exposed for 5 hours in an early period and 4 hours in a later period. There are two jumps during the early exposure and one during the later exposure. Assume rates $\alpha_1=0.2$ events per hour and $\alpha_2=0.5$ events per hour in the respective periods. The risk process is known and predictable, and all three jumps occur during exposed time.

The jump product contains two early-period rates and one later-period rate. The integrated intensity is $5\alpha_1+4\alpha_2$. Substitution gives

L\ \propto\ \alpha_1^2\alpha_2 \exp\{-(5\alpha_1+4\alpha_2)\} =(0.2)^2(0.5)\exp\{-(5(0.2)+4(0.5))\} =0.02e^{-3} \approx 0.0009957.

The five-hour and four-hour exposures are kept separate because the rates differ by period. Replacing them with total exposure alone would lose the distinction between the two rate contributions.

Checking the construction and comparing approaches

Several checks can reveal a faulty likelihood before any estimation is attempted.

Check each observed jump against the intensity. If $\lambda(T_j;\theta)=0$ for an observed jump, the likelihood is zero at that parameter value. This may be a legitimate model implication or a sign that the model, event coding, or risk indicator is wrong.
Check that exposure is integrated over the full window. Omitting time after the final jump removes evidence from the final event-free interval. Integrating when $Y(t)=0$ incorrectly counts time when the process was not exposed.
Check jump structure. Simultaneous jumps of a process with unit jumps violate the simple-process setup. A model for grouped or marked events may be needed rather than treating tied observations as ordinary separate jumps.
Check the intensity scale. If intensity is measured in events per hour, integrating over hours gives a dimensionless cumulative intensity. Inconsistent time units can distort the exponential term.
Check what has been conditioned upon. If observation or covariate processes have a parameter-dependent distribution, conditioning them away may omit part of the joint likelihood.

The product-integral expression is equivalent to multiplying conditional event-time densities across successive gaps; its advantage is a concise representation of the whole path. A likelihood based only on the total count can be simpler, but generally discards timing information. For a constant intensity over a fixed interval, the count-based likelihood has a factorial term when the count is observed without event times; that term does not depend on the rate. It is not the same data representation as the density for ordered event times. A discretized Bernoulli construction can approximate the continuous-time likelihood using short intervals, but it is an approximation and requires intervals small enough that multiple jumps within a time step are negligible.

These distinctions matter when choosing what data the likelihood represents. The path likelihood uses exact jump times and event-free exposure. It does not automatically provide a model for parameter-dependent entry, censoring, or covariate distributions; those require appropriate additional likelihood contributions or justified conditioning. Likewise, if events can occur simultaneously or at fixed-time atoms, the continuous-intensity derivation must be adapted rather than applied unchanged.

Key takeaway: Each observed jump contributes its predictable intensity at that time, and all observed event-free exposure contributes the factor $\exp\{-\int_0^\tau\lambda(t;\theta)\,dt\}$. In a multiplicative intensity model, the at-risk process determines where that integrated intensity accumulates.

Check Your Understanding

Answer each question, then compare with the solutions below.

A process has two observed jumps and constant intensity $\theta$ over 6 hours. Write its likelihood kernel.
For intensity $\lambda(t;\theta)=Y(t)\theta$, what role does the at-risk indicator play in the integrated term?
A model assigns zero intensity at an observed jump. What is the likelihood contribution at that parameter value?
Why is the final event-free interval included even if no event occurs after the last observed jump?

Solutions. (1) The kernel is $\theta^2\exp(-6\theta)$, because there are two jump factors and integrated intensity $6\theta$. (2) It restricts accumulated intensity to times when the process is at risk: the integral is $\theta\int_0^\tau Y(t)\,dt$. (3) The jump product contains a zero factor, so the likelihood is zero. (4) No event through the end of observation contributes survival information, represented by the integrated intensity from the last jump to $\tau$.