Selection Bias From Truncated Loss Samples

Why Truncated Loss Samples Can Mislead

A loss database may contain only claims above a reporting threshold, only losses below an upper eligibility limit, or only losses lying within a specified range. In each case, some losses are absent altogether. The observed losses are therefore not generally a representative sample from the population severity distribution: the rule for inclusion changes which values can appear.

This tutorial focuses on the resulting selection bias. The central calculation is to describe the distribution of $X$ among included observations and compare its mean with the population mean $E[X]$. Interest and mortality assumptions do not apply to these severity-only examples; all monetary amounts are hypothetical.

Definition: Let $X$ be a loss with distribution function $F_X$, density $f_X$, and finite mean $E[X]$. If a loss is observed only when it belongs to a specified set $A$, the observed loss distribution is the distribution of $X$ conditional on inclusion: $X\mid\{X\in A\}$. For a continuous loss, its density is $$ f_{\mathrm{obs}}(x)=\frac{f_X(x)}{P(X\in A)},\qquad x\in A, $$ and is zero outside $A$, provided $P(X\in A)>0$.

The denominator is essential. It rescales the population density over the part of the loss range that can be observed so that the observed density integrates to one. For example, under left truncation at a fixed threshold $a$, only losses exceeding $a$ enter the data. The observed density is $f_X(x)/S_X(a)$ for $x>a$, where $S_X(a)=P(X>a)$. Under right truncation at $b$, only losses below $b$ enter, and the denominator is $F_X(b)$.

The same principle applies to interval truncation: if losses are observed only when $a<X<b$, the denominator is $F_X(b)-F_X(a)$. Endpoints make no difference for a continuous distribution. The mean in any of these cases is an expectation conditional on inclusion, not the unconditional population mean.

Deriving the Mean Shift

Let $I_A$ equal 1 when $X\in A$ and 0 otherwise. The observed mean follows directly from the definition of conditional expectation:

E[X\mid X\in A] =\frac{E[X I_A]}{P(X\in A)} =\frac{\int_A x f_X(x)\,dx}{P(X\in A)}.

Subtracting the population mean $\mu=E[X]$ gives a useful expression for the bias. Since $\operatorname{Cov}(X,I_A)=E[XI_A]-E[X]E[I_A]$ and $E[I_A]=P(X\in A)$,

E[X\mid X\in A]-E[X] =\frac{\operatorname{Cov}(X,I_A)}{P(X\in A)}.

This identity explains why the direction of the bias depends on the inclusion rule. For fixed left truncation, inclusion becomes more likely as the loss increases: $I_A=1(X>a)$ is a nondecreasing function of $X$. Its covariance with $X$ is nonnegative, so the observed mean is at least the population mean. For fixed right truncation, $I_A=1(X<b)$ is nonincreasing, so the observed mean is at most the population mean. These inequalities are strict when the selection rule separates loss values with positive probability and the loss is nondegenerate.

One justification for the covariance signs uses an independent copy $X'$ of $X$. For any nondecreasing function $g$,

\operatorname{Cov}(X,g(X)) =\frac12 E\big[(X-X')(g(X)-g(X'))\big]\geq 0,

because the two differences always have the same sign. For a nonincreasing $g$, they have opposite signs, so the covariance is nonpositive. Taking $g(X)=I_A$ gives the stated results for one-sided truncation. For interval truncation, inclusion is not monotone in $X$, so the direction of the mean shift cannot be inferred from this argument alone.

Worked Example: A Reporting Threshold for Exponential Losses

A hypothetical portfolio has individual losses $X\sim\operatorname{Exponential}(\text{mean } \$1{,}000)$, with density $f_X(x)=0.001e^{-0.001x}$ for $x\geq0$. A study records a loss only if it exceeds $500. Find the inclusion probability and the mean of recorded losses. No interest or mortality assumptions apply.

Method. Apply the conditional density for fixed left truncation. The population survival function is $S_X(x)=e^{-0.001x}$, so the probability a loss is recorded is

P(X>500)=S_X(500)=e^{-0.5}=0.6065.

The observed density, for $x>500$, is

f_{\mathrm{obs}}(x) =\frac{0.001e^{-0.001x}}{e^{-0.5}} =0.001e^{-0.001(x-500)}.

This is the density of $500 plus an exponential loss with mean $1,000. Therefore,

E[X\mid X>500]=500+1{,}000=\$1{,}500.

The recorded mean is $500 greater than the population mean of $1,000, as expected under left truncation. Check: the observed density integrates to one because its integral over $x>500$ is $e^{-0.5}/e^{-0.5}=1$. Common mistake avoided: using the population mean of $1,000 as the mean of the recorded claims ignores that no claim at or below $500 can enter the sample.

Worked Example: A Lower Eligibility Limit for Uniform Losses

A hypothetical warranty portfolio has losses $X\sim\operatorname{Uniform}(0,120)$, measured in thousands of dollars. A research file includes only losses above $30,000. Find the probability of inclusion, the population mean, and the observed mean. Interest and mortality assumptions do not apply.

The population density is $1/120$ on the interval from 0 to 120. The inclusion probability is

P(X>30)=\frac{120-30}{120}=\frac{90}{120}=0.75.

After division by this probability, the observed density is $(1/120)/0.75=1/90$ on $30<X<120$. Thus the observed loss is uniform over that interval. Its mean is the midpoint:

E[X]=\frac{0+120}{2}=60,\qquad E[X\mid X>30]=\frac{30+120}{2}=75.

The population mean is $60,000, while the observed mean is $75,000: a difference of $15,000. The retained fraction, 0.75, is consistent with the ratio of interval lengths. Common mistake avoided: applying the original mean to the truncated sample would miss the shift in the lower endpoint from 0 to 30.

Worked Example: An Upper Eligibility Limit for Exponential Losses

A hypothetical equipment-loss model has $X\sim\operatorname{Exponential}(\text{mean } \$1{,}000)$. A study excludes every loss of $1,000 or more and records only losses below that amount. Calculate the inclusion probability and the observed mean. Interest and mortality assumptions do not apply.

The inclusion probability is

P(X<1{,}000)=1-e^{-1}=0.6321.

The conditional mean is the first moment over the retained range divided by its probability. With $\lambda=1/1{,}000$, integration by parts gives

\int_0^b x\lambda e^{-\lambda x}\,dx =\frac{1}{\lambda}\left[1-(1+\lambda b)e^{-\lambda b}\right].

For $b=1{,}000$, $\lambda b=1$, so the numerator is $1{,}000[1-2e^{-1}]$. Dividing by the inclusion probability gives

E[X\mid X<1{,}000] =1{,}000\frac{1-2e^{-1}}{1-e^{-1}} =1{,}000\frac{e-2}{e-1} =\$418.02.

The mean is rounded to the nearest cent; probabilities are given to four decimal places. It is below the population mean of $1,000, as right truncation predicts. Check: the unconditioned first moment below $1,000$ is approximately $264.24, and $264.24/0.6321$ is approximately $418.02. Common mistake avoided: multiplying the population mean by the inclusion probability gives an unconditioned partial moment, not the mean conditional on being included. The partial moment must be divided by the inclusion probability.

Worked Example: Truncation to a Middle Range

A hypothetical liability study has losses $X\sim\operatorname{Uniform}(0,200)$, in thousands of dollars. Only losses between $40,000 and $140,000 are recorded. Find the inclusion probability and compare the observed and population means. Interest and mortality assumptions do not apply.

The population density is $1/200$, so the probability of falling in the observed interval is

P(40<X<140)=\frac{140-40}{200}=0.5.

The conditional density is $(1/200)/0.5=1/100$ between 40 and 140. Hence the observed distribution is uniform on that interval, giving

E[X\mid 40<X<140]=\frac{40+140}{2}=90,\qquad E[X]=\frac{0+200}{2}=100.

The observed mean is $90,000, which is $10,000 below the population mean of $100,000, even though the sample is truncated at both ends. Check: the conditional density integrates to $100(1/100)=1$. Common mistake avoided: assuming that interval truncation must increase the mean because it removes low losses, or must decrease it because it removes high losses. Here the interval removes more distance from the high end than from the low end, so the mean falls.

What Changes When Inclusion Probabilities Vary

Fixed cutoffs are a useful starting point, but not every truncated-data mechanism has the same threshold for every claim. If a loss $x$ has probability $\pi(x)$ of being observed, then the density among observed losses is proportional to $f_X(x)\pi(x)$. In particular, when $E[\pi(X)]>0$,

f_{\mathrm{obs}}(x)=\frac{f_X(x)\pi(x)}{E[\pi(X)]}, \qquad E[X\mid \text{observed}] =\frac{E[X\pi(X)]}{E[\pi(X)]}.

These expressions follow by weighting each possible loss by its chance of appearing and normalizing the total weight. They also show why the direction of bias depends on the selection mechanism: if $\pi(x)$ tends to be larger for larger losses, the observed mean tends to rise. This is a description of selection, not an estimation procedure for recovering the original distribution.

Truncation should also be distinguished from censoring. With truncation, a loss outside the observation region does not appear as a record. With censoring, a record remains but reports limited information about its loss, such as that it exceeds a threshold. The likelihood contributions for these mechanisms differ, so treating an absent loss as if it were a censored value changes the data model.

Common Exam Traps

Forgetting to condition. The observed mean is a partial first moment divided by the inclusion probability. The partial moment alone is not a conditional mean.
Using the wrong tail probability. Under left truncation at $a$, use $P(X>a)=S_X(a)$. Under right truncation at $b$, use $P(X<b)=F_X(b)$.
Assuming all truncation raises the mean. Left truncation at a fixed lower threshold raises the mean, and right truncation at a fixed upper threshold lowers it. Double truncation has no universal direction.
Calling unrecorded losses censored. If there is no record at all for a loss outside the selection region, it is truncated under that sampling scheme. A censored record conveys a bound or interval about the loss.
Mixing units. If a model is stated in thousands of dollars, the calculated mean is in thousands until converted. State the unit before comparing means.

Key takeaway: Truncation changes the distribution of observed losses by conditioning on inclusion. Calculate the inclusion probability, normalize the density on the observable region, and only then calculate the observed mean. For fixed one-sided thresholds, the direction of selection bias follows from whether inclusion favors larger or smaller losses.

Check Your Understanding

For each hypothetical severity model, determine the requested inclusion probability or observed mean. Interest and mortality assumptions do not apply.

A loss is exponential with mean $600 and is recorded only above $300. Find the inclusion probability and the mean of recorded losses. (Answer: probability $0.6065$; observed mean $900.)
A loss is uniform from 0 to 160, in thousands of dollars, and only values above 40 are recorded. Find the population and observed means. (Answer: population mean $80,000; observed mean $100,000.)
A loss is exponential with mean $500 and only values below $500 are recorded. Find the observed mean, rounded to the nearest cent. (Answer: $209.01.)
A loss is uniform from 0 to 240, in thousands of dollars, and only values between 60 and 180 are recorded. Find the inclusion probability and compare the observed mean with the population mean. (Answer: probability $0.5000$; both means are $120,000.)
A nonnegative loss is uniform from 0 to 200, in thousands of dollars. A loss of amount $x$ is independently selected with probability $x/200$. Find the mean among selected losses. (Answer: $133,333.33.)

What You'll Learn