Why Density Diagnostics Need the Sampling Mechanism
A density estimate can look smooth and still describe the wrong population. If only losses below a reporting threshold are observed, the sample favors smaller losses. If follow-up ends before some losses are settled, those censored observations do not reveal their eventual values. In either case, a conventional kernel density estimate of the recorded numbers can mistake the observation process for the loss process.
The diagnostic task is not simply to choose a smoother bandwidth. It is to ask whether the estimate respects the support of the target distribution, whether its tail is supported by observed information, and whether apparent features persist when incomplete-data adjustments and reasonable smoothing choices are compared. The methods here are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.
Start With the Target and the Observable Support
Let \(X\) denote the loss of interest, with density \(f_X\), distribution function \(F_X\), and survival function \(S_X\). A kernel density estimate based on complete observations \(X_1,\ldots,X_n\) is
where \(K\) is a kernel with integral one and \(h>0\) is a bandwidth. This formula estimates the density of the sample's generating distribution when the observations are complete and comparably sampled. It does not itself correct for incomplete observation.
For right-truncated losses observed only when \(X\leq T\), the observed values have the conditional density
Thus, a naive estimate from the observed sample targets \(g_T\), not \(f_X\). If \(q=F_X(T)\) is known, then \(f_X(x)=qg_T(x)\) for \(x\leq T\). This does not determine the shape of the density above \(T\); the unobserved tail requires additional information or assumptions. A hard upper edge in the observed sample is therefore not evidence that the underlying loss density ends there.
With varying truncation thresholds, an inverse-probability weighted estimate can reduce the over-representation of observations that were easier to include. If observation \(i\) has known inclusion probability \(\pi_i>0\), set \(w_i=1/\pi_i\). A normalized weighted kernel estimate is
This estimate requires correctly specified inclusion probabilities and positive chances of observation over the target region. If a region has zero inclusion probability, weighting cannot recover its density from these data.
What a Censoring Adjustment Estimates
A right-censored value records that the loss exceeds a known censoring point, not that the loss equals that point. Under independent censoring, the Kaplan-Meier estimator uses the risk set and event counts to estimate the loss distribution. If \(t_j\) are distinct event times, \(d_j\) is the number of events at \(t_j\), and \(Y_j\) is the number at risk immediately before \(t_j\), its survival estimate is
The estimated event probability mass at \(t_j\) is \(\Delta\widehat F_X(t_j)=\widehat S_X(t_j-)d_j/Y_j\). Smoothing these event masses gives
Censoring times contribute to the risk sets but are not event masses. Also, if the largest follow-up time is censored, the estimated event masses may sum to less than one. The remaining estimated survival probability represents unresolved tail mass, not an observed event. Automatically renormalizing the event masses to sum to one would instead describe the event distribution conditional on occurrence by the observed endpoint.
Why the Adjustments Work
The right-truncation adjustment follows directly from conditional probability. For any small interval around \(x\leq T\), conditioning on inclusion divides the probability of that interval by the probability of inclusion:
Dividing by \(\Delta\) and taking the limit as \(\Delta\) approaches zero gives \(g_T(x)=f_X(x)/F_X(T)\). Multiplying by the known \(F_X(T)\) recovers the underlying density below \(T\). No corresponding calculation recovers the shape above \(T\), because those losses are never observed.
For censoring, the Kaplan-Meier product estimates the probability of surviving each event time among those still at risk. The decrease in estimated survival at an event time is exactly the estimated event probability there:
Smoothing these probability increments spreads estimated event mass across nearby loss values. Censoring times do not produce such a decrement, which is why treating them as exact losses changes the target being estimated.
Worked Example: Diagnosing a Right-Truncated Density
A hypothetical audit observes losses only when \(X\leq 10\). An external source establishes that \(F_X(10)=0.80\). The four observed losses are \(4,5,6,7\). Estimate the density at \(x=5\) using a uniform kernel with bandwidth \(h=2\), first naively and then with the fixed-threshold truncation adjustment. State what the comparison does and does not establish.
Method. For the uniform kernel, \(K(u)=1/2\) when \(|u|\leq 1\), and zero otherwise. Thus \(K_h(x-X_i)=1/(2h)=1/4\) for observations within \(2\) of \(x=5\).
All four losses are within the window \([3,7]\). The naive estimate is therefore
This estimates the density among losses that passed the threshold. Multiply by \(F_X(10)=0.80\) to estimate the underlying density below the threshold:
The adjusted estimate is 0.2000 per loss unit at \(x=5\), compared with the naive conditional estimate 0.2500. This adjustment is valid below \(10\) under the stated truncation model and known inclusion probability. It does not estimate how the remaining 0.20 probability is distributed above \(10\). The common mistake is to label 0.2500 an estimate of the full-population density without accounting for selection.
Check. The adjusted estimate is smaller by the factor 0.80, as expected because the observed density is conditional on membership in the lower 80% of the distribution.
Worked Example: Comparing Naive and Censoring-Adjusted Smoothing
A hypothetical follow-up study records five losses: an event at time 1, censoring at time 2, events at times 3 and 4, and censoring at time 5. Estimate the density at \(x=3\) with a uniform kernel of bandwidth \(h=1\), using both the naive recorded-time estimate and the smoothed Kaplan-Meier event masses. Also identify the unresolved survival probability.
Naive estimate. The kernel window is \([2,4]\). The recorded times 2, 3, and 4 lie in it, so the naive estimate is
This calculation incorrectly treats the censoring time 2 as an event time.
Kaplan-Meier calculation. Initially \(Y_1=5\). The event at 1 has \(d_1=1\), giving survival \(1(1-1/5)=4/5\) and event mass \(1/5=0.2000\). The censoring at 2 reduces the risk set but creates no event mass. At time 3, \(Y_2=3\), \(d_2=1\), so the event mass is \((4/5)(1/3)=4/15=0.2667\), and survival becomes \(8/15\). At time 4, \(Y_3=2\), \(d_3=1\), so the event mass is \((8/15)(1/2)=4/15=0.2667\), and survival becomes \(4/15\). The censoring at 5 creates no further event mass.
At \(x=3\), the events at 3 and 4 are within one bandwidth; the event at 1 is outside the kernel window. Each included event mass is multiplied by \(1/(2h)=1/2\):
The estimated survival after time 4 is \(4/15=0.2667\), still unresolved at the last censoring time. The difference between 0.3000 and 0.2667 reflects, in part, the naive estimate's false event at time 2. The common mistake is to count every recorded endpoint as an observed loss.
Check. The event masses \(1/5+4/15+4/15=11/15\); the remaining \(4/15\) is survival, so the masses and residual survival sum to one.
Support, Tail, and Sensitivity Checks
A useful diagnosis compares estimates in several ways rather than treating a single curve as definitive. The following checks help separate features supported by loss information from features driven by incomplete observation.
- Support check. Confirm that the estimated density is being interpreted only where losses could be observed and the adjustment has positive support. A density that abruptly vanishes at a truncation boundary may reflect the observation rule, not the underlying distribution. Ordinary kernels can also place estimated mass outside a known physical support; this is a boundary-smoothing artifact, not evidence of negative or impossible losses.
- Tail check. Ask how much estimated probability lies beyond the largest informative loss. Right truncation supplies no direct observations above its threshold. With censoring, a residual Kaplan-Meier survival probability at the end of follow-up signals unresolved tail mass. A smooth-looking tail is not identified merely because a kernel extends past the last event.
- Sensitivity check. Compare plausible bandwidths and, where appropriate, naive and adjusted estimates on the same scale and region. A narrow spike that disappears under a modest bandwidth change may be sampling noise. A large difference between naive and adjusted curves suggests the inclusion mechanism matters. Neither result alone proves which curve is correct; model assumptions and data support still matter.
- Weight check. Examine the range of inverse inclusion weights. Very large weights mean that a few observations can dominate the adjusted estimate. One summary is the effective sample size, \((\sum_i w_i)^2/\sum_i w_i^2\). A much smaller effective size than the recorded sample size warns that the weighted curve may be unstable.
- Mass check. Check the estimated probability mass as well as pointwise density height. A censoring-adjusted curve built from event increments may integrate to less than one when follow-up ends with positive estimated survival. Treating that deficit as a numerical error and forcing the curve to integrate to one can conceal the unknown tail.
Worked Example: Detecting Weight Sensitivity
A hypothetical study has observed losses \(2,4,6,8\), with known inclusion probabilities \(1,0.80,0.50,0.25\), respectively. Use a uniform kernel and bandwidth \(h=2\) to compare naive and normalized inverse-probability weighted estimates at \(x=8\). Calculate the effective sample size and explain what the diagnostics indicate.
Method. The inverse-probability weights are \(1,1.25,2,4\), with total \(8.25\). At \(x=8\), the window is \([6,10]\), so only losses 6 and 8 contribute. The naive estimate is
For the weighted estimate, the contributing weight total is \(2+4=6\), and the kernel height for each included observation is \(1/(2h)=1/4\):
The effective sample size is
The weighted estimate at 8 is higher because the observation with loss 8 receives a large weight. But the effective sample size is about 3.02, below the four recorded observations, so the apparent tail feature is sensitive to a small number of heavily weighted records. The appropriate conclusion is not that the weighted estimate is automatically more reliable: its validity depends on the supplied inclusion probabilities, and its precision is limited by weight concentration.
Check. The weighted estimate uses the two included observations' combined weight \(6\), divided by the total weight \(8.25\), and then multiplied by kernel height \(0.25\).
Common Diagnostic Traps
A common error is to treat truncation and censoring as interchangeable. Truncation means a loss is absent from the sample unless it satisfies an inclusion condition; censoring means the record is present but its exact loss is not known. The corresponding corrections are different. Conditional-density or inclusion-probability adjustments address truncation, while event-risk methods address censored outcomes.
Another trap is to overinterpret visual smoothness. Smoothing can make a density curve continuous, but it cannot create information in a region with no chance of observation or resolve a tail that remains censored. Report the target population, the observation mechanism, the adjustment, the bandwidth, and any unresolved tail mass together. When naive and adjusted estimates differ materially, investigate why before selecting one for actuarial use.
Check Your Understanding
For each problem, identify the relevant adjustment and interpret the final estimate in context.
- A hypothetical sample is right-truncated at \(T=12\). The known probability of inclusion is \(F_X(12)=0.75\), and the naive conditional density estimate at \(x=8\) is 0.32. Find the adjusted estimate at 8 and state whether the calculation identifies the density above 12. (Answer: 0.2400 below 12; no, the tail shape above 12 is not identified.)
- In a right-censored sample, the Kaplan-Meier event masses are 0.20 at time 2 and 0.30 at time 4, with residual survival 0.50 after follow-up. Using a uniform kernel with \(h=1\), find the smoothed density at \(x=4\). (Answer: 0.1500; the residual survival remains unresolved.)
- A weighted estimate uses weights \(1,2,3\). Calculate its effective sample size. (Answer: \(36/14=2.5714\).)
- A right-truncated sample has known inclusion probability 0.60 and a naive conditional density estimate of 0.45 at a point below the threshold. Find the adjusted density there and state the remaining limitation. (Answer: 0.2700; the density shape above the truncation threshold remains unidentified.)
- A study records an event at time 1, censoring at time 2, and an event at time 3, with three observations initially at risk. Under independent censoring, calculate the Kaplan-Meier event masses and residual survival after time 3. (Answer: event masses \(1/3\) at time 1 and \(1/3\) at time 3; residual survival \(1/3\).)