Bandwidth Selection With Incomplete Loss Observations

Why Bandwidth Selection Changes With Incomplete Data

A kernel density estimate depends strongly on its bandwidth $h$: a small value can leave a jagged estimate, while a large value can smooth away important features. With incomplete loss observations, the choice has an additional complication. A loss may be more likely to appear in the data because of the truncation or censoring mechanism. The observations then contain unequal amounts of information about the underlying loss distribution.

A bandwidth rule applied as if every recorded loss were a complete, equally likely observation can therefore target the wrong distribution or overstate the amount of information available. The remedy is to incorporate observation probabilities into the estimate and into bandwidth assessment. In practice, a weighted normal-reference rule can provide a starting value, while weighted cross-validation can compare candidate bandwidths.

Definition: Let $X$ be a loss with target density $f_X$, and let $p(x)$ be the probability that a loss of size $x$ is observed under the sampling design. If an observed loss $x_i$ has observation probability $p_i=p(x_i)>0$, its inverse-probability weight is $q_i=1/p_i$. Larger weights compensate for losses that were less likely to be observed.

For example, with independent left truncation at a random threshold $L$, the inclusion probability at loss $x$ is $\alpha(x)=P(L\le x)$. With independent right censoring time $C$, an event at $x$ is observed when $C\ge x$, with probability $G(x-)=P(C\ge x)$. If both mechanisms operate independently, the probability that a loss at $x$ is included and observed as an event is $p(x)=\alpha(x)G(x-)$. These probabilities must be positive over the loss range for which the target density is to be estimated. If they are zero in a region, weighting cannot recover information about that region from these data alone.

Weighted Kernel Estimates and Effective Information

For exact observations with positive weights $q_1,\ldots,q_n$, a normalized weighted kernel density estimate is

\widehat f_h(x)=\frac{1}{W}\sum_{i=1}^n q_i K_h(x-X_i), \qquad W=\sum_{i=1}^n q_i, \qquad K_h(z)=\frac{1}{h}K\left(\frac{z}{h}\right).

Here $K$ is a kernel integrating to one and $h>0$ is the bandwidth. Normalizing by $W$ makes the estimate integrate to one. Inverse-probability weighting corrects unequal observation probabilities; it does not make the corrected data as informative as a complete sample of the same nominal size.

Definition: The effective sample size of positive weights $q_i$ is $$ n_{\mathrm{eff}}=\frac{\left(\sum_i q_i\right)^2}{\sum_i q_i^2}. $$ It equals $n$ when all $n$ weights are equal and is smaller when the weights are unequal. It is a diagnostic for the information lost through weighting, not a replacement for the actual sample size in every inferential calculation.

The formula has a simple interpretation. If all weights equal a constant $c$, then $n_{\mathrm{eff}}=(nc)^2/(nc^2)=n$. If a few observations carry much larger weights, the squared weights in the denominator rise quickly; the estimate is then driven by fewer effective observations. That is a reason to avoid interpreting an apparently detailed density curve too confidently in regions where only a small amount of weighted information is available.

Starting Values and Weighted Cross-Validation

For a smooth, approximately unimodal density and a Gaussian kernel, a normal-reference starting value is

h_{\mathrm{NR}}=1.06\,s_w\,n_{\mathrm{eff}}^{-1/5},

where $s_w$ is a weighted measure of spread, such as the square root of the weighted variance using weights $q_i/W$. This rule adapts the usual normal-reference scale to unequal observation weights. It is only a starting value: skewness, multiple modes, boundaries, and uncertain or highly variable weights can make the rule unsuitable as a final choice.

A more data-directed comparison uses weighted leave-one-out least-squares cross-validation. Define $W_{-i}=W-q_i$ and estimate the density at $X_i$ after omitting observation $i$:

\widehat f_{-i,h}(X_i) = \frac{1}{W_{-i}}\sum_{j\ne i}q_jK_h(X_i-X_j).

A weighted cross-validation score is

\operatorname{CV}(h) = \int \widehat f_h(x)^2\,dx - 2\sum_{i=1}^n\frac{q_i}{W}\widehat f_{-i,h}(X_i).

Choose the candidate bandwidth with the smallest score among those compared. The score estimates integrated squared error up to a term that does not depend on $h$, under suitable sampling and weighting assumptions. It is not an exact guarantee of good performance, particularly with few observations, estimated observation probabilities, or very large weights.

Why the Weights Enter the Criterion

The weighting principle follows from a direct expectation calculation. Suppose the observation mechanism selects a loss $x$ with probability $p(x)$. Among observed losses, the distribution is tilted in proportion to $p(x)f_X(x)$. Multiplying an observed contribution at $x$ by $1/p(x)$ removes that tilt:

p(x)f_X(x)\frac{1}{p(x)}=f_X(x).

After normalization, weighted observations can therefore represent the target loss distribution rather than the distribution among observed records alone. The same logic motivates weighting the cross-validation evaluation points: the term $q_i/W$ gives less-observed losses their appropriate representation in the target distribution. Leaving $i$ out prevents its own kernel contribution from making its fitted density at $X_i$ artificially large.

For right-censored data, a censored record does not supply an exact event loss at its censoring time. Under independent censoring, an observed event at $X_i$ can be weighted by $1/G(X_i-)$; with independent truncation as well, use $1/\{\alpha(X_i)G(X_i-)\}$ when both probabilities are known or estimated. Censored records inform the censoring and risk-set calculations, but should not be inserted as if their censoring times were exact losses. Interval-censored observations likewise do not become exact losses merely by assigning them interval midpoints; they require a method that respects their interval information.

Specify the target and observation mechanism.
Decide which underlying loss distribution is being estimated and define the probability that a loss at each value would be observed.

Construct appropriate weights.
Use inverse inclusion probabilities for truncated exact losses and inverse event-observation probabilities for exact events under censoring. State the assumptions used to obtain those probabilities.

Assess effective information.
Calculate $n_{\mathrm{eff}}$, and check whether a small number of records carry a large share of the total weight.

Compare plausible bandwidths.
Use a weighted rule of thumb as a scale check and weighted cross-validation when the data support it. Examine the score across a candidate range rather than trusting one isolated minimum.

Worked Example: A Weighted Normal-Reference Starting Value

A hypothetical loss study has four exact observed losses, in thousands of dollars: $10,20,30,40$. Their known observation probabilities are $1,0.5,1,0.5$, so their inverse-probability weights are $1,2,1,2$. Use the Gaussian normal-reference rule to calculate a starting bandwidth. Treat the weighted variance as the weighted population variance.

Method. First calculate the effective sample size, then calculate the weighted mean and spread. This accounts for unequal observation probabilities; using the raw count of four in the rule would ignore their effect.

The total weight is $W=1+2+1+2=6$, and the sum of squared weights is $1+4+1+4=10$. Therefore,

n_{\mathrm{eff}}=\frac{6^2}{10}=3.6.

The weighted mean is

\bar x_w=\frac{1(10)+2(20)+1(30)+2(40)}{6} =\frac{160}{6}=26.6667.

The weighted second moment is $\{1(10^2)+2(20^2)+1(30^2)+2(40^2)\}/6=5000/6=833.3333$. Thus the weighted variance and standard deviation are

s_w^2=833.3333-(26.6667)^2=122.2222, \qquad s_w=\sqrt{122.2222}=11.0554.

The normal-reference starting value is

h_{\mathrm{NR}} =1.06(11.0554)(3.6^{-1/5}) \approx 1.06(11.0554)(0.7736) \approx 9.06.

Thus a starting bandwidth is about $9.06$ thousand dollars. As a check, equal weights would have produced $n_{\mathrm{eff}}=4$; the unequal weights reduce effective information to $3.6$, making the starting bandwidth slightly larger. This is a scale suggestion, not proof that this bandwidth best fits the data.

Worked Example: Adjusting for Right Censoring

In a hypothetical portfolio, exact event losses of $12,18,30$ thousand dollars are observed. The estimated probabilities that censoring would allow events at those loss values to be seen are respectively $0.8,0.6,0.4$. Two other records are right-censored at $22$ and $35$ thousand dollars. Treat the stated event-observation probabilities as known for this calculation. Find the event weights, their effective sample size, and a Gaussian normal-reference starting bandwidth using the weighted population standard deviation.

Method. Apply inverse censoring probabilities to the exact events. The censored records are not exact event losses, so neither censoring time is added as a kernel data point.

The event weights are

q_1=\frac{1}{0.8}=1.25,\qquad q_2=\frac{1}{0.6}=\frac{5}{3},\qquad q_3=\frac{1}{0.4}=2.5.

Their sum is $1.25+1.6667+2.5=5.4167=65/12$. The sum of squared weights is \[ 1.25^2+(5/3)^2+2.5^2 =25/16+25/9+25/4 =1525/144. \] Hence

n_{\mathrm{eff}} =\frac{(65/12)^2}{1525/144} =\frac{4225}{1525} \approx 2.7705.

The weighted mean is $\{1.25(12)+(5/3)(18)+2.5(30)\}/(65/12)=120/(65/12)=22.1538$. The weighted second moment is $\{1.25(12^2)+(5/3)(18^2)+2.5(30^2)\}/(65/12)=2970/(65/12)=548.3077$. Therefore,

s_w=\sqrt{548.3077-(22.1538)^2}\approx 7.5838, \qquad h_{\mathrm{NR}}=1.06(7.5838)(2.7705^{-1/5})\approx 6.55.

The starting bandwidth is approximately $6.55$ thousand dollars. It is based on only three exact event losses and an effective sample size of about $2.77$; it should therefore be treated cautiously. The censoring times contribute information about which events were observable, but treating them as exact losses would change the target data and give a misleading bandwidth.

Worked Example: Comparing Bandwidths by Weighted Cross-Validation

A hypothetical incomplete-data study has exact losses $100,130,160,190$ thousand dollars, with inverse-observation weights $1,2,1,2$. Compare uniform-kernel weighted cross-validation at $h=20$ and $h=35$, where bandwidths are in thousands of dollars. Use $K_h(z)=1/(2h)$ when $|z|\le h$, and zero otherwise. Select the bandwidth with the smaller score.

Method. Calculate the integrated squared estimate and the weighted leave-one-out evaluation term separately for each candidate. This avoids choosing a bandwidth from the plotted curve alone.

For both candidates $W=6$, and $\sum q_i^2=10$. At $h=20$, each kernel has interval length $40$. The four diagonal overlap contributions to the integral are $40(10)=400$. Adjacent observations are $30$ apart, so their kernel intervals overlap by $40-30=10$; the sum of adjacent weight products is $1(2)+2(1)+1(2)=6$. Including both orders gives $2(10)(6)=120$. Thus

\int \widehat f_{20}(x)^2\,dx =\frac{400+120}{6^2(40^2)} =\frac{520}{57600} \approx 0.009028.

No other observed loss is within $20$ of a loss being evaluated, so every leave-one-out density at its omitted loss is zero. The weighted evaluation term is zero, and $\operatorname{CV}(20)\approx0.009028$.

At $h=35$, each kernel interval has length $70$. The diagonal overlap contribution is $70(10)=700$. Adjacent intervals overlap by $70-30=40$, contributing $2(40)(6)=480$. The two pairs separated by $60$ overlap by $70-60=10$, with weight products $1(1)$ and $2(2)$; including both orders gives $2(10)(1+4)=100$. Therefore,

\int \widehat f_{35}(x)^2\,dx =\frac{700+480+100}{6^2(70^2)} =\frac{1280}{176400} \approx 0.007256.

The leave-one-out densities at the four losses are $2/(5\cdot70)=0.005714$, $2/(4\cdot70)=0.007143$, $4/(5\cdot70)=0.011429$, and $1/(4\cdot70)=0.003571$. Weighting these by $q_i/W$, the evaluation term is

\sum_i\frac{q_i}{W}\widehat f_{-i,35}(X_i) =\frac{0.005714+2(0.007143)+0.011429+2(0.003571)}{6} \approx0.006429.

Consequently, $\operatorname{CV}(35)\approx0.007256-2(0.006429)=-0.005601$. Since $-0.005601<0.009028$, weighted cross-validation selects $h=35$ from these two candidates. The check is that the criterion includes both the integrated square and twice the weighted leave-one-out evaluation; comparing only the integrated squares would not be cross-validation.

Local Sparsity, Assumptions, and Exam Traps

A single global effective sample size can conceal local information gaps. For a region around $x$, one useful diagnostic is a local effective count based on nearby weighted observations. For example, with nonnegative proximity weights $a_j(x)$, define

n_{\mathrm{eff}}(x) = \frac{\left\{\sum_j q_j a_j(x)\right\}^2} {\sum_j\{q_j a_j(x)\}^2}.

A small value signals that only a few weighted records meaningfully support the estimate near $x$. A locally wider bandwidth may stabilize such a region, but it also increases smoothing bias. This diagnostic should prompt scrutiny, not an automatic local adjustment: adaptive bandwidth rules need their own validation, and cannot recover a region where observation probability is zero.

The weighting and bandwidth methods rely on the observation probabilities being correctly specified or estimated, and on the assumptions behind them. In particular, interpreting $G(x-)$ as an event-observation probability requires independent censoring, and using $\alpha(x)$ for truncation requires an appropriate independent truncation model. When probabilities are estimated, their uncertainty can make weights unstable. Inspect large weights and sensitivity to plausible probability estimates; a small cross-validation score does not remove this uncertainty.

Using the raw sample size. Substituting the number of records for $n_{\mathrm{eff}}$ in a weighted rule ignores the loss of information from unequal weights.
Using the observed distribution as the target. A density estimate that omits observation-probability correction can describe the selected records rather than the underlying losses.
Treating censoring times as events. A right-censoring time is not the exact loss. Use an approach that accounts for censoring rather than placing a full event kernel there.
Ignoring zero or tiny observation probabilities. Inverse weights are undefined at zero and can become highly unstable near zero. Weighting does not create information that was never observed.
Trusting one bandwidth score. With sparse data, weighted cross-validation may be variable or have a boundary minimum over the candidates. Compare a sensible range and report the limitation.

The central decision is not simply whether a curve looks smooth. A defensible bandwidth accounts for how observations entered the data, how much effective information remains after weighting, and whether the local regions being interpreted are actually supported by observations.

Check Your Understanding

Work each problem using the stated observation probabilities and assumptions.

A study has three exact losses with inverse-observation weights $1,1,2$. Calculate $n_{\mathrm{eff}}$. (Answer: $16/6\approx2.6667$)
Two exact events occur at $15$ and $25$ thousand dollars. Their independent censoring observation probabilities are $0.75$ and $0.50$. Give their inverse-censoring weights and calculate their effective sample size. (Answer: weights $4/3$ and $2$; $n_{\mathrm{eff}}=100/34\approx2.9412$)
For weights $1,2,1$, suppose the weighted normal-reference spread is $5$ thousand dollars and use $n_{\mathrm{eff}}$ in $h_{\mathrm{NR}}=1.06s_wn_{\mathrm{eff}}^{-1/5}$. Calculate the starting bandwidth. (Answer: $n_{\mathrm{eff}}=16/6\approx2.6667$, $h_{\mathrm{NR}}\approx4.40$ thousand dollars)
A loss is known only to lie between $40$ and $60$ thousand dollars because it is interval-censored. Should it be entered as an exact kernel observation at $50$ thousand dollars in the weighted leave-one-out calculation? State the reason. (Answer: No; the midpoint is not an observed exact loss, so a method accounting for the interval information is required.)
Two candidate bandwidths have weighted cross-validation scores $0.012$ and $-0.004$. Which is preferred by the criterion, and what important limitation remains if only three effective observations support the estimate? (Answer: Choose the candidate with score $-0.004$; the small effective sample means the selected bandwidth and resulting density may be unstable.)

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