Why Rescale Event Times?
A point-process model describes not only how many events occur, but also when they occur. If its fitted rate changes over time, raw gaps between events can be hard to interpret: a gap of one time unit may be surprisingly long in a high-rate period and ordinary in a low-rate period. Cumulative intensity puts time on a scale where the model predicts a constant rate. On that scale, the gaps have a simple reference distribution that can reveal systematic disagreement between data and model.
Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \((N(t))_{t\ge0}\) be a nonhomogeneous Poisson process with state space \(S=\{0,1,2,\ldots\}\), natural filtration \(\mathcal{F}_t=\sigma(N(s):0\le s\le t)\), and deterministic intensity function \(\lambda(t)\). Assume \(\lambda\) is continuous and strictly positive. Its cumulative intensity is \(\Lambda(t)=\int_0^t\lambda(s)\,ds\). Write \(S_k\) for the time of the \(k\)th event, and put \(S_0=0\). We first consider a process on \(t\ge0\) for which \(\Lambda(t)\to\infty\); this ensures that every event number \(k\) occurs eventually.
The phrase “fully known” matters. It means that the intensity used for rescaling is the model's actual intensity, rather than an estimate chosen using the same event times being checked. Later we will separate the ideal reference result from practical diagnostics with fitted parameters.
Why the Rescaled Spacings Are Exponential
Proof. Fix a positive integer \(k\). For ordered event times \(0<t_1<\cdots<t_k\), the joint density of the first \(k\) event times is \(e^{-\Lambda(t_k)}\prod_{i=1}^k\lambda(t_i)\): the factors \(\lambda(t_i)\) give the event densities, and \(e^{-\Lambda(t_k)}\) is the probability of no further event before \(t_k\). Since \(\lambda\) is positive, \(u_i=\Lambda(t_i)\) is a strictly increasing change of variables, with \(du_i/dt_i=\lambda(t_i)\). Its Jacobian cancels the product of rates. Thus the joint density of \(0<U_1<\cdots<U_k\) is \(e^{-u_k}\). Set \(z_1=u_1\) and \(z_i=u_i-u_{i-1}\) for \(2\le i\le k\). This triangular change of variables has Jacobian \(1\), its domain is \(z_1,\ldots,z_k>0\), and \(u_k=\sum_{i=1}^k z_i\). The joint density becomes \(e^{-\sum_i z_i}=\prod_{i=1}^k e^{-z_i}\). It factors into unit-exponential densities, proving independence and the asserted marginal laws for every \(k\). \(\square\)
Proof. For \(0<r<1\), the event \(R_k\le r\) is equivalent to \(Z_k\le-\log(1-r)\). Therefore \(\mathbb{P}(R_k\le r)=1-e^{-[-\log(1-r)]}=r\). The endpoint probabilities are \(0\) at \(r=0\) and \(1\) at \(r=1\). Applying a separate deterministic transformation to each of the independent \(Z_k\) preserves independence, so the residuals are independent uniforms. \(\square\)
Worked Example: Rescaling an Increasing Rate
Suppose the true rate is \(\lambda(t)=2t\) for \(t\ge0\). For event times \(S_1=\tfrac12\), \(S_2=1\), and \(S_3=\tfrac32\), calculate the cumulative event times, spacings, and uniform residuals.
The cumulative intensity is \(\Lambda(t)=\int_0^t2s\,ds=t^2\). Thus \(U_0=0\), \(U_1=(\tfrac12)^2=\tfrac14\), \(U_2=1^2=1\), and \(U_3=(\tfrac32)^2=\tfrac94\). Subtracting successive cumulative event times gives
Transforming each spacing gives \(R_1=1-e^{-1/4}\), \(R_2=1-e^{-3/4}\), and \(R_3=1-e^{-5/4}\). These are three residual values to compare with a Uniform\((0,1)\) reference; the theorem does not say that every small sample must look evenly spread.
Check: Inverting the cumulative intensity gives \(\Lambda^{-1}(u)=\sqrt{u}\) for \(u\ge0\). Hence \(\sqrt{\tfrac14}=\tfrac12\), \(\sqrt{1}=1\), and \(\sqrt{\tfrac94}=\tfrac32\), recovering the event times. Also \(\tfrac14+\tfrac34+\tfrac54=\tfrac94=U_3\), as required for successive spacings.
Finite Observation Windows Change the Reference
Data are usually observed only up to a fixed time \(T\). Put \(L=\Lambda(T)\). The transformed observation window is then \([0,L]\), not the entire positive half-line. The gaps \(Z_k\) between consecutive observed events are not, considered as a selected finite list, an unrestricted independent sample of exponential spacings: which gaps appear depends on how many events occurred before \(T\). The interval from the last observed event to \(T\) is censored, meaning it ends at the observation boundary rather than at an event. It must not be treated as a completed interarrival spacing.
There is a useful exact statement for a fixed window. Conditional on observing \(N(T)=n\), the transformed event locations \(U_i/L\), for \(1\le i\le n\), have the distribution of the order statistics of \(n\) independent Uniform\((0,1)\) variables. They are ordered locations, not an independent sequence. This is often a more appropriate reference for checking event locations when the observed count is fixed.
Proof. The joint density of the first \(n\) transformed event times together with no additional event by transformed time \(L\) is \(e^{-L}\) on \(0<u_1<\cdots<u_n<L\). The probability of exactly \(n\) events by \(T\) is \(e^{-L}L^n/n!\), since the count is Poisson with mean \(L\). Dividing the joint density by this probability gives the conditional density \(n!/L^n\). For \(n\) independent Uniform\((0,L)\) variables, each of the \(n!\) possible orderings has equal density \(1/L^n\); sorting them therefore gives density \(n!/L^n\) on the ordered region. Rescaling by \(L\) gives the claimed Uniform\((0,1)\) order statistics. When \(n=0\), there are no event locations and the statement is vacuous. \(\square\)
Worked Example: A Homogeneous Rate on a Fixed Window
Suppose \(\lambda(t)=2\), observe on \(0\le t\le2\), and suppose the event times are \(\tfrac14,\tfrac34,\tfrac54\). Find the transformed spacings and identify the window-end censoring.
Here \(\Lambda(t)=2t\), so \(L=\Lambda(2)=4\). The transformed event times are \(U_1=2(\tfrac14)=\tfrac12\), \(U_2=2(\tfrac34)=\tfrac32\), and \(U_3=2(\tfrac54)=\tfrac52\). Thus the completed spacings are \(Z_1=\tfrac12\), \(Z_2=\tfrac32-\tfrac12=1\), and \(Z_3=\tfrac52-\tfrac32=1\). The residuals are \(1-e^{-1/2}\), \(1-e^{-1}\), and \(1-e^{-1}\).
The transformed interval from the last event to the end of observation is \(L-U_3=4-\tfrac52=\tfrac32\). It is not another completed interarrival time. Conditional on the count being \(3\), the scaled event locations are \(\tfrac18,\tfrac38,\tfrac58\), which should be regarded as ordered uniform locations, not as three independent uniforms.
Check: The transformed gaps plus the final censored interval sum to \(\tfrac12+1+1+\tfrac32=4=L\). Dividing the event locations by \(L=4\) gives \((\tfrac18,\tfrac38,\tfrac58)\), all strictly between \(0\) and \(1\) and in increasing order.
Diagnostics When the Intensity Is Fitted
For a fitted model, replace \(\Lambda\) by the fitted cumulative intensity \(\widehat{\Lambda}(t)\), and calculate \(\widehat{Z}_k=\widehat{\Lambda}(S_k)-\widehat{\Lambda}(S_{k-1})\). These are useful diagnostics, but the theorem's exact reference law applies only when the rescaling intensity is the true, fixed model intensity. If parameters are estimated from the same data, the residuals are affected by that estimation. In particular, fitting can force one or more features of the transformed data, so treating the resulting residuals as an independent uniform sample can give misleading test thresholds.
Worked Example: Fitting a Rate That Increases Linearly
On the window \([0,2]\), consider \(\lambda_\theta(t)=\theta t\), with \(\theta>0\), and event times \(\tfrac12,1,\tfrac32\). Find the maximum-likelihood estimate and the fitted spacings.
The cumulative intensity is \(\Lambda_\theta(t)=\theta t^2/2\). For \(n=3\) events, the likelihood as a function of \(\theta\) is \(L(\theta)=e^{-2\theta}\theta^3(\tfrac12)(1)(\tfrac32)\), because \(\Lambda_\theta(2)=2\theta\) and the event-rate factors are \(\theta t_i\). The log-likelihood, up to terms not depending on \(\theta\), is \(3\log\theta-2\theta\). Its derivative is \(3/\theta-2\), which is positive for \(\theta<\tfrac32\), zero at \(\tfrac32\), and negative for \(\theta>\tfrac32\). Hence \(\widehat{\theta}=\tfrac32\).
The fitted cumulative intensities at the three event times are \(\widehat{\Lambda}(\tfrac12)=\tfrac34(\tfrac14)=\tfrac3{16}\), \(\widehat{\Lambda}(1)=\tfrac34=\tfrac{12}{16}\), and \(\widehat{\Lambda}(\tfrac32)=\tfrac34(\tfrac94)=\tfrac{27}{16}\). Therefore
The corresponding fitted uniform residuals are \(1-e^{-3/16}\), \(1-e^{-9/16}\), and \(1-e^{-15/16}\). These values can be plotted against a uniform reference, but their finite-sample distribution is not guaranteed to be that of three independent uniforms: the same three event times were used to estimate \(\theta\).
Check: The fitted total cumulative intensity is \(\widehat{\Lambda}(2)=\tfrac32(4)/2=3\), equal to the observed count \(n=3\). The spacings sum to \((3+9+15)/16=27/16=\widehat{\Lambda}(\tfrac32)\). This checks the arithmetic; the count match is a consequence of the likelihood fit, not evidence by itself that the model is correct.
Useful checks include a plot of transformed event locations against their conditional uniform order-statistic reference, a uniform probability plot of \(1-e^{-\widehat{Z}_k}\), and an exponential probability plot of the completed spacings. Look for systematic curvature, long runs of unusually short or long gaps, or patterns associated with time or covariates. Also compare the observed event count with the model's predicted count distribution. These checks address different aspects of fit; no single plot proves a model is correct.
When a formal goodness-of-fit test is wanted after fitting, calibrate it to the fitting procedure. One practical approach is a parametric bootstrap: simulate data under the fitted model on the same observation window, refit the parameters to each simulated dataset, recompute the diagnostic statistic, and compare the observed statistic with that simulated reference distribution. Refitting is essential because the observed statistic was computed after fitting as well. If the count is conditioned on, the simulation and reference calculation must also respect that conditioning.
A Common Mistake and Its Correction
A common error is to transform event times using the instantaneous rate \(\lambda(S_k)\) rather than the cumulative intensity \(\Lambda(S_k)\). For the increasing-rate example with \(\lambda(t)=2t\) and \(S_1=\tfrac12,S_2=1,S_3=\tfrac32\), this incorrect calculation gives rate values \(1,2,3\). These are not cumulative event times, and their successive differences \(1,1,1\) are not the rescaled spacings. The correct cumulative event times are \(\tfrac14,1,\tfrac94\), giving spacings \(\tfrac14,\tfrac34,\tfrac54\). The rate describes intensity at an instant; integrating it over time is what converts elapsed time into cumulative expected event count.
Another pitfall is to regard every residual from a fitted, finite-window dataset as exactly independent Uniform\((0,1)\). The exact result is for spacings under the true intensity over the full process. A finite window censors its last gap, and estimating parameters changes the residual distribution. Keep these qualifications attached to the reference distribution whenever interpreting a plot or test.
Check Your Understanding
Use the distinction between cumulative event locations, completed spacings, and fitted residuals to answer the following.
- If \(\lambda(t)=3t^2\), what is \(\Lambda(t)\), and what transformation maps an event time \(S_k\) to cumulative time?
- For a process observed only until \(T\), why is the gap from the last event to \(T\) not a completed interarrival spacing?
- Under the exact model, how do you transform a unit-exponential spacing \(Z_k\) into a Uniform\((0,1)\) residual?
- Conditional on \(N(T)=n\), what is the distributional interpretation of the ordered transformed event locations divided by \(\Lambda(T)\)?
- Why should a bootstrap replicate the parameter-fitting step when calibrating a diagnostic test?