Truncation

Study guide: Exam MAS-I

Truncation occurs when observations below (or above) a threshold are not recorded at all — unlike Censoring, which records that an unobserved value exists. In insurance, left-truncation arises with deductibles: only losses exceeding the deductible $d$ are reported.

Left-truncated likelihood contribution (only losses $X > d$ are observed): $L_i(\theta) = \frac{f(x \mid \theta)}{S(d \mid \theta)}$

$= \frac{f(x \mid \theta)}{P(X > d \mid \theta)}$

$x > d$

Under left-truncation at $d$ , the effective distribution is the conditional distribution $X \mid X > d$
Failing to adjust for truncation biases the estimates. The average of the reported claims estimates $E[X \mid X > d]$ , not $E[X]$ , so it is biased upward (only the larger claims are seen). A distribution fitted without the adjustment can err in either direction: a lognormal fitted naively to truncated data overestimates $\mu$ , yet can under-predict the high quantiles
Right-truncation: only values below some upper limit are recorded (less common in insurance)
Truncation differs from censoring: with truncation, you do not even know an observation exists; with censoring, you know there is an observation but not its exact value

MLE With Left-Truncated Exponential Data Claims above a deductible $d = 1$ are reported. Two reported claims are $x_1 = 2$ and $x_2 = 3$ . Assume $X \sim \text{Exp}(\theta)$ . Write and maximize the likelihood.

Answer $S(1|\theta) = e^{-1/\theta}$ , so each claim contributes $f(x|\theta)/S(1|\theta) = \frac{1}{\theta}e^{-x/\theta}/e^{-1/\theta} = \frac{1}{\theta}e^{-(x-1)/\theta}$ . $L(\theta) = \frac{1}{\theta}e^{-(2-1)/\theta}\cdot\frac{1}{\theta}e^{-(3-1)/\theta} = \frac{1}{\theta^2}e^{-3/\theta}$ Solving $\ell'(\theta) = -2/\theta + 3/\theta^2 = 0$ gives $\hat{\theta} = 3/2 = 1.5$ . This equals $\bar{x} - d = 2.5 - 1 = 1.5$ , the sample mean of excess losses.