Why Change Measure by Comparing Intensities?
An event-history model specifies how quickly events occur given the information available just before each time. Changing the intensity changes the probability assigned to event paths: paths with events in times or types favoured by the new intensity gain relative weight, while paths with long event-free intervals may lose or gain weight according to the cumulative rates. A likelihood-ratio process makes this reweighting explicit at every time, not only at the end of follow-up.
This is useful for comparing event-process models, expressing a new event law relative to a reference law, and proving that a proposed change of intensity is mathematically valid. The key caution is that the pathwise likelihood formula alone does not guarantee a probability measure: its density must have expectation one.
Definition and Conditions
Here each intensity is the rate of jumps conditional on the information immediately before time \(t\); if an at-risk indicator is used, it is included in the intensity. Assume the two laws use the same initial distribution, share the same filtration and event-history structure, and have no simultaneous jumps across the counted event types. The rates must be nonnegative, predictable, and locally integrable, and the processes must be nonexplosive on the interval. These conditions allow the usual event-history likelihood construction.
For \(Q\) to be absolutely continuous with respect to \(P\), a path possible under \(Q\) must not require a jump where \(P\) assigns zero intensity. In particular, \(\mu_j(t)=0\) wherever \(\lambda_j(t)=0\), up to the relevant predictable measure. If both are zero, that event type contributes no jump or cumulative-rate difference there. When \(\lambda_j>0\), the pathwise likelihood ratio is
The product over jumps multiplies the relative rates at observed events. The exponential term accounts for the different event-free probabilities between jumps. Equivalently, where the ratios are defined, the process solves
This stochastic differential equation identifies \(L\) as a nonnegative local martingale under \(P\). To use \(L_\tau\) as a probability density, one needs the stronger property \(E_P(L_\tau)=1\). A nonnegative local martingale has expectation at most its initial value; it need not preserve expectation. One sufficient condition on a finite interval is that the intensities are bounded and the ratios \(\mu_j/\lambda_j\) are bounded on the region where \(\lambda_j>0\), with \(\mu_j=0\) where \(\lambda_j=0\). These bounds make the stochastic-exponential integrands bounded and their predictable quadratic variation finite. More generally, other integrability conditions can establish the same unit-expectation property.
Why the Formula Has This Form
For a path with jumps at specified times, the event-history likelihood under an intensity is the product of the intensity at each jump times the exponential of minus the integrated intensity. This is the multiplicative intensity likelihood. Taking the likelihood under the proposed rates and dividing by the likelihood under the reference rates gives the displayed formula: jump contributions divide to \(\mu_j/\lambda_j\), and the integrated-rate terms divide to \(\exp\{-\int(\mu_j-\lambda_j)\}\). The shared initial distribution cancels.
A short stochastic-calculus check gives the same result. At a jump of type \(j\), the product in \(L_t\) changes by the factor \(\mu_j/\lambda_j\). Between jumps, its derivative is \(-L_{t-}\sum_j(\mu_j-\lambda_j)\). Thus
The equality follows because \(\lambda_j(\mu_j/\lambda_j-1)=\mu_j-\lambda_j\). The final expression is a stochastic integral against the compensated counting processes under \(P\), hence a local martingale under the stated local conditions. If it is a true martingale with \(E_P(L_\tau)=1\), define \(Q\) on \(\mathcal{F}_\tau\) by \(Q(A)=E_P[L_\tau I(A)]\). The density process is then \(L_t=E_P(L_\tau\mid\mathcal{F}_t)\).
Under this change of measure, the compensated processes have the proposed rates: \(N_j(t)-\int_0^t\mu_j(s)\,ds\) is a local martingale under \(Q\). One way to verify the claim is to apply the product rule to \(L_t\) and this proposed compensated process. The drift terms cancel because the change in intensity is \(\mu_j-\lambda_j\); the remaining terms are local-martingale terms. This is the counting-process change-of-measure result. It requires the density and support conditions above, not merely a formal substitution of rates.
Worked Path Calculations
Worked Example: One Event Before Censoring
Suppose a hypothetical single-event process remains under observation for at most 2 years. Under \(P\), its constant intensity is \(\lambda=0.20\) events per year; under the proposed \(Q\), it is \(\mu=0.50\) events per year. Both rates are positive and bounded, the initial state is the same, and at most one event can occur. Find the likelihood ratios for an event at 0.8 years and for no event by 2 years.
For an event at 0.8 years, the observed at-risk time is 0.8 years and there is one jump. Therefore \(L=\frac{0.50}{0.20}\exp\{-(0.50-0.20)(0.8)\}\). The arithmetic is \(2.5\exp(-0.24)=2.5(0.7866)=1.966\), to four significant figures. This particular path receives about 1.966 times its reference-law density under the proposed law.
For no event by 2 years, there is no jump factor, so \(L=\exp\{-(0.50-0.20)(2)\}=\exp(-0.60)=0.5488\). The event and event-free paths are reweighted in opposite directions, as expected when the proposed rate is higher.
The rates are finite and bounded, their ratio is \(2.5\), and the proposed rate is zero wherever the reference rate is zero (there are no such times here). Thus the stated sufficient conditions apply on the finite follow-up interval; the path calculations are part of a valid likelihood-ratio process.
Worked Example: Two Event Types and Unequal Path Counts
Consider a hypothetical recurrent-event process observed for 3 years, with two event types and no simultaneous jumps. Under \(P\), the constant intensities are \(\lambda_1=0.10\) and \(\lambda_2=0.20\) per year. Under \(Q\), they are \(\mu_1=0.20\) and \(\mu_2=0.10\) per year. Suppose a path has two type-1 events and one type-2 event, all before year 3. The initial state is common, and all rates are positive and bounded.
The jump factors are \(2\) for each type-1 event and \(0.10/0.20=0.5\) for the type-2 event. The integrated-rate difference is constant: \((\mu_1-\lambda_1)+(\mu_2-\lambda_2)=0.10-0.10=0\) per year. Hence
This path has twice its reference density under \(Q\). The example also shows why one must include every event type: the two intensity increases and decreases cancel in the cumulative-rate term, but the observed event counts still determine the jump contribution. Bounded rates and ratios, common initial law, and the absence of simultaneous jumps satisfy the sufficient conditions used here.
Worked Example: Checking Unit Expectation for Poisson Rates
Let \(N(t)\) be a hypothetical homogeneous Poisson process on \([0,1]\) year. Under \(P\), its intensity is \(\lambda=2\) events per year; under \(Q\), it is \(\mu=3\) events per year. The rates are positive and bounded, so the finite-horizon sufficient conditions hold. Verify directly that the proposed density has expectation one.
At time 1, the likelihood ratio is \(L_1=(3/2)^{N(1)}\exp\{-(3-2)(1)\}=(3/2)^{N(1)}e^{-1}\). Under \(P\), \(N(1)\) has a Poisson distribution with mean 2. Its probability generating function is \(E_P[z^{N(1)}]=\exp\{2(z-1)\}\). Substituting \(z=3/2\) gives
Thus \(L_1\) defines a probability measure. Under that measure, the event count over one year has the proposed Poisson rate of 3 events per year. This check illustrates why unit expectation matters: it ensures the reweighted probabilities sum to one.
Assumptions, Diagnostics, and Alternatives
Confirm that the filtration, event types, and initial-state distribution are shared, and that the assumed jump structure (including no simultaneous jumps here) fits the process being modelled.
Verify that the proposed law does not allow a type of jump at a time where the reference intensity is zero. Otherwise the proposed law cannot be represented by a density relative to the reference law.
Multiply the relative intensity at each observed jump and include the integrated-rate difference. Then establish \(E_P(L_\tau)=1\), using a sufficient condition or a direct argument where available.
Only after the density defines \(Q\) should the new compensated process be treated as a \(Q\)-local martingale with intensity \(\mu_j\).
A practical warning sign is a computed density with an expected value below one: it may be a strict local martingale rather than a probability density. A ratio that becomes unbounded, a process that can accumulate infinitely many jumps in finite time, or incompatible event support also calls for closer justification. These are mathematical checks on the change of measure, not merely numerical diagnostics of model fit.
The path-likelihood ratio is often the clearest approach when observed paths and intensities are explicit. The stochastic-exponential form is more useful for martingale arguments and predictable integrands. A score or log-likelihood calculation instead studies local changes in model parameters; it does not by itself establish a global probability measure for a proposed alternative intensity. In all approaches, the likelihood formula needs the same support and normalization scrutiny.
A common mistake is to see that \(L_t\) is a nonnegative local martingale and conclude immediately that \(E_P(L_t)=1\). Nonnegative local martingales are supermartingales, so only \(E_P(L_t)\leq L_0=1\) follows in general. The correction is to establish the true-martingale property on the time interval of interest, for example with suitable boundedness conditions, before using \(L_t\) as a density. A second error is to use \(\mu_j/\lambda_j\) at a point where \(\lambda_j=0\); the ratio is not defined there, and support compatibility must be checked rather than repaired by algebra.
Check Your Understanding
Answer each question, then compare your reasoning with the brief solutions below.
- Under constant rates \(\lambda=0.4\) and \(\mu=0.6\) events per year, what is the likelihood ratio for no event during 2 years?
- For the same rates, what is the likelihood ratio for exactly one event during 2 years, without specifying its time?
- Why is \(E_P(L_\tau)=1\) required to define \(Q(A)=E_P[L_\tau I(A)]\) as a probability measure?
- If \(\lambda_j=0\) on an interval but \(\mu_j>0\) there, what condition fails?
Solutions. (1) There is no jump factor, so \(L=\exp\{-(0.6-0.4)(2)\}=\exp(-0.4)=0.6703\). (2) For any one-event path, \(L=(0.6/0.4)\exp(-0.4)=1.5(0.6703)=1.005\); the event time does not affect the ratio because the rates are constant. (3) The total mass assigned by the proposed definition is \(Q(\Omega)=E_P(L_\tau)\), so it must equal one. (4) Support compatibility fails: the proposed law permits a jump that the reference law assigns zero intensity, so it cannot be represented by a density relative to \(P\).