Incomplete Data

Study guide: Exam MAS-I

Incomplete data are observations whose values are only partly known — censored (known only to lie in a range) or truncated (never recorded at all outside a range). Insurance data are incomplete by design, because deductibles and policy limits decide which losses are seen and how exactly. Estimation must model this through the likelihood rather than treat what was recorded as a complete sample.

$L(\theta) = \prod_{i=1}^{n} \frac{f(x_i \mid \theta)^{\delta_i}\; S(x_i \mid \theta)^{1-\delta_i}}{S(d_i \mid \theta)}$

$\delta_i = 1$ if $x_i$ is an exact value and $0$ if it is right-censored at $x_i$ (the loss is known only to be at least $x_i$ ). $d_i$ is the left-truncation point, such as the deductible; with no truncation $S(d_i) = S(0) = 1$ . $S = 1 - F$ is the survival function.
Right censoring. A loss at or above the maximum covered loss is recorded only as "at least $u$ ", so it contributes $S(u)$ . Treating it as exactly $u$ biases severity estimates down.
Left truncation. Losses below a deductible $d$ are never reported, so each reported loss is drawn from $X \mid X > d$ and contributes $f(x)/S(d)$ . Ignoring this biases the average of the reported losses up, since it estimates $E[X \mid X > d]$ rather than $E[X]$ (only the larger losses are seen). A distribution fitted without the adjustment can err in either direction (Truncation).
Grouped (interval) data are censored on both sides: a value known only to lie in $(a, b]$ contributes $F(b) - F(a)$ .
Exponential shortcut. For an exponential model with any mix of truncation and right censoring, $\hat\theta = \dfrac{\sum_i (x_i - d_i)}{\text{number of exact observations}}$ , summing over all observations, censored ones included.
The same adjustments carry over to non-parametric estimates of the survival function, such as the Kaplan–Meier product-limit estimator. They matter again when fitting the severity distribution behind coverage modifications.

Fitting Ground-Up Severity From Payment Data A policy has an ordinary deductible of 500 and a maximum covered loss of 5,000. The insurer's records show five claims with ground-up losses of 1,000, 2,000 and 3,500, plus two claims paid at the limit (losses of at least 5,000). Assuming ground-up losses are exponential, find the MLE of $\theta$ .
Answer Every observation is truncated at $d = 500$ . The two limit claims are right-censored at 5,000:
 $\begin{align*} L(\theta) &= \frac{f(1000)\,f(2000)\,f(3500)\,S(5000)^2}{S(500)^5} \\ &= \theta^{-3}\,e^{-(500 + 1500 + 3000 + 4500 + 4500)/\theta} \\ &= \theta^{-3}\,e^{-14{,}000/\theta} \end{align*}$ 
Setting $\frac{d}{d\theta}\ln L = -\frac{3}{\theta} + \frac{14{,}000}{\theta^2} = 0$ :
 $\begin{align*} \hat\theta &= \frac{14{,}000}{3} \\ &= 4{,}666.67 \end{align*}$ 
Treating the two limit claims as exact losses of 5,000 would divide by 5 and give $2{,}800$ — understating mean severity by 40%.

MLE From Grouped Claim Sizes Of 100 claims, 60 were below 1,000, 30 were between 1,000 and 3,000, and 10 exceeded 3,000. Fit an exponential distribution by maximum likelihood.
Answer Let $p = e^{-1000/\theta}$ , so $S(1000) = p$ and $S(3000) = p^3$ . Each group contributes its interval probability:
 $\begin{align*} L &= (1-p)^{60}\,(p - p^3)^{30}\,(p^3)^{10} \\ &= p^{60}\,(1-p)^{90}\,(1+p)^{30} \end{align*}$ 
Differentiate $\ln L$ with respect to $p$ and clear denominators:
 $\begin{align*} \frac{60}{p} - \frac{90}{1-p} + \frac{30}{1+p} &= 0 \\ 3p^2 + p - 1 &= 0 \\ p &= \frac{-1 + \sqrt{13}}{6} \\ &= 0.43426 \end{align*}$ 
 $\begin{align*} \hat\theta &= \frac{-1000}{\ln 0.43426} \\ &= 1{,}198.9 \end{align*}$ 
The fitted mean claim size is about 1,199, using only the counts in each size band.