Percentile Matching

Study guide: Exam MAS-I

Percentile Matching estimates the parameters of a model by making its percentiles equal the corresponding percentiles of the data: with $p$ unknown parameters, choose $p$ percentile levels $g_1, \ldots, g_p$ and solve the $p$ equations below. It is the percentile analogue of the Method of Moments and works even when the model's moments do not exist.

$F(\hat\pi_{g_k} \mid \theta) = g_k, \quad k = 1, \ldots, p$

$\hat\pi_g = (1 - h)\,x_{(j)} + h\,x_{(j+1)}$

$j = \lfloor (n+1)g \rfloor, \qquad h = (n+1)g - j$

$\hat\pi_g$ is the smoothed empirical percentile: locate position $(n+1)g$ among the order statistics $x_{(1)} \le \cdots \le x_{(n)}$ and interpolate linearly between the neighbouring observations. For $n = 11$ and $g = 0.4$ , the position is $4.8$ , so $\hat\pi_{0.4} = 0.2\,x_{(4)} + 0.8\,x_{(5)}$ .
The exponential, Weibull and loglogistic give equations that can be solved by hand: for the two-parameter ones, dividing one equation by the other eliminates $\theta$ . A closed-form inverse CDF is not enough on its own — the two-parameter Pareto has $F^{-1}(g) = \theta\big((1 - g)^{-1/\alpha} - 1\big)$ , yet its two percentile equations have no analytic solution and need a numerical solver, as does the general case $F^{-1}(g \mid \theta) = \hat\pi_g$ .
Choices matter. Different percentile levels give different estimates, and the method uses only a few points of the sample. Maximum likelihood uses the data more efficiently and is generally preferred; percentile matching and moments give quick, explainable estimates and starting values for the MLE search.
Because percentiles always exist, the method can fit heavy-tailed severity models such as a Pareto with $\alpha \le 1$ , where the moment equations have no solution (see Tail Weight).

Exponential Fit to the Sample Median Ten claims are observed: $120, 180, 260, 330, 410, 560, 720, 980, 1{,}600, 2{,}400$ . Fit an exponential distribution by matching the smoothed empirical median.
Answer The position is $(10 + 1)(0.5) = 5.5$ , halfway between $x_{(5)} = 410$ and $x_{(6)} = 560$ , so $\hat\pi_{0.5} = 485$ . Match the exponential median:
 $\begin{align*} 1 - e^{-485/\theta} &= 0.5 \\ \hat\theta &= \frac{485}{\ln 2} \\ &= 699.7 \end{align*}$ 
The fitted mean severity is about $700, below the sample mean of $756 — the median match ignores how far out the two largest claims sit.

Weibull Fit to the Quartiles Eleven losses are observed: $54, 95, 140, 190, 230, 310, 380, 450, 620, 840, 1{,}300$ . Fit a Weibull, $F(x) = 1 - e^{-(x/\theta)^{\tau}}$ , by matching the 25th and 75th smoothed empirical percentiles.
Answer With $n + 1 = 12$ , the positions are $3$ and $9$ , so $\hat\pi_{0.25} = 140$ and $\hat\pi_{0.75} = 620$ . Each equation becomes $(x/\theta)^{\tau} = -\ln(1 - g)$ :
 $\begin{align*} (140/\theta)^{\tau} &= -\ln 0.75 \\ &= 0.28768 \\ (620/\theta)^{\tau} &= -\ln 0.25 \\ &= 1.38629 \end{align*}$ 
Dividing the second by the first eliminates $\theta$ :
 $\begin{align*} \left(\frac{620}{140}\right)^{\tau} &= 4.81884 \\ \hat\tau &= \frac{\ln 4.81884}{\ln 4.42857} \\ &= \frac{1.57253}{1.48808} \\ &= 1.0568 \\ \hat\theta &= \frac{620}{1.38629^{1/1.0568}} \\ &= 455.2 \end{align*}$ 
The fitted Weibull passes exactly through both quartiles; its shape $\hat\tau \approx 1.06$ is close to the exponential case $\tau = 1$ .