Actuarial Notes

Benktander Method

Study guide: Exam 5

Benktander Method (Gunnar Benktander, 1976; also called the iterated BF or GB method) runs the Bornhuetter-Ferguson calculation a second time, using the BF ultimate in place of the a priori expectation. Mack (2000) shows the result is a credibility mixture of BF and the chain ladder, with the percentage developed as the credibility.

U_{\text{GB}} = C + \left(1 - \tfrac{1}{\text{CDF}}\right) U_{\text{BF}}

U_{\text{GB}} = \tfrac{1}{\text{CDF}} \, U_{\text{CL}} + \left(1 - \tfrac{1}{\text{CDF}}\right) U_{\text{BF}}

  • Notation. C is reported (or paid) losses to date, p = 1/\text{CDF} is the percentage developed, and q = 1 - p.
  • Credibility Z = p. The second identity is the useful one. Benktander weights chain ladder against BF with Z = p: the more of the year that has emerged, the more the answer leans on the data. The reserve mixes the same way, R_{\text{GB}} = q\,U_{\text{BF}} = p\,R_{\text{CL}} + q\,R_{\text{BF}}. That is Hovinen's (1981) reserve, the same method found independently.
  • Against the original a priori. Expressed against U_0, the weight on the chain ladder is p(2-p) = 1 - q^2:

U_{\text{GB}} = p(2 - p)\,U_{\text{CL}} + \left[1 - p(2-p)\right] U_{0}

  • Iterating further converges to the chain ladder. Mack proves it (his Theorem 1). Repeating R = qU then U = C + R from U_0 gives U^{(m)} = (1 - q^m)\,U_{\text{CL}} + q^m U_0. BF is m = 1, Benktander is m = 2, and the limit is the chain ladder.
  • Mean squared error. The comparison is Mack's (2000), not Benktander's. His model is E(C/U \mid U) = p and \text{Var}(C/U \mid U) = pq\,\beta^2(U), with \alpha^2(U) = U^2\beta^2(U). The a priori U_0 is independent of C, with E(U_0) = E(U). The optimal credibility on the chain ladder, c^*, depends on a single volatility measure t:

t = \frac{E[\alpha^2(U)]}{\text{Var}(U_0) + \text{Var}(U) - E[\alpha^2(U)]}

c^* = \frac{p}{p + t}

  • Which method wins. GB has a smaller mean squared error than BF if and only if t < 2 - p. It beats the chain ladder if and only if t > pq/(1+p). (Of the other two, BF beats the chain ladder if and only if p < t.) So GB is best at every maturity unless the payout pattern is extremely volatile or extremely stable. Mack's numerical examples give t \approx 0.3, well inside the GB region, and GB's standard error comes close to both the optimal c^* reserve's and an exact Bayesian model's.
  • The ordering is fixed by construction. Whichever of U_{\text{CL}} and U_{\text{BF}} is larger, U_{\text{GB}} lies between them. The method needs no data beyond what BF already requires.
  • The a priori must stay fixed. An a priori revised as the year develops is no longer independent of C. Mack likens it to choosing an unknown c, much less objective than Benktander's c = p.

Benktander Between BF and Chain Ladder An accident year has reported losses C = \$600, a priori expected losses U_0 = \$1{,}000, and \text{CDF} = 2.000 (so p = 50\%).

Compute the chain ladder, BF and Benktander ultimates.

Answer

\begin{align*}
U_{\text{CL}} &= \$600 \times 2.000 = \$1{,}200 \\[6pt]
U_{\text{BF}} &= \$600 + 0.50 \times \$1{,}000 \\
&= \$1{,}100 \\[6pt]
U_{\text{GB}} &= \$600 + 0.50 \times \$1{,}100 \\
&= \$1{,}150
\end{align*}

Checking against the credibility form:

0.50(\$1{,}200) + 0.50(\$1{,}100) = \$1{,}150 \;\checkmark

and against the a priori form, with p(2-p) = 0.5(1.5) = 0.75:

0.75(\$1{,}200) + 0.25(\$1{,}000) = \$1{,}150 \;\checkmark

Actual emergence is running ahead of the a priori (\$1{,}200 chain ladder against a \$1{,}000 expectation), so the ordering is U_0 < U_{\text{BF}} < U_{\text{GB}} < U_{\text{CL}}. Benktander moves 75\% of the way from the a priori to the chain ladder where BF moved only 50\%.

How the Weighting Shifts With Maturity A line's reported percentages are 30\% at 12 months, 60\% at 24, and 85\% at 36. For each maturity, give the weight Benktander places on the chain ladder relative to the original a priori.

Answer The chain ladder weight is p(2-p); BF's is p:

Maturity p BF weight on CL GB weight on CL
12 mo 0.30 0.300 0.510
24 mo 0.60 0.600 0.840
36 mo 0.85 0.850 0.978

Benktander always gives the data more weight than BF and less than the chain ladder, and the extra weight over BF is largest in the middle. At 12 months it moves from 30\% to 51\% reliance on emergence; at 36 months it is already 98\% chain ladder.

Whether that extra weight is right is not a question of maturity alone. In Mack (2000) it depends on the line's volatility t: GB beats BF whenever t < 2 - p, and beats the chain ladder whenever t > pq/(1+p), at any of these ages.

Optimal Credibility and the Mean Squared Errors For one accident year, all as percentages of premium: the a priori ultimate is U_0 = 75\%, the payout pattern gives p = 0.40, and paid losses are C = 36\%. The actuary assesses \text{Var}(U) = (25\%)^2, \text{Var}(U_0) = (10\%)^2 and \text{Var}(C/U \mid U) = 0.08^2, the last not depending on U.

Using Mack's (2000) model, find t and c^*, the BF, chain ladder, Benktander and optimal reserves, and the standard error of each.

Answer With q = 0.60 and E(U) = E(U_0) = 75\%:

\begin{align*}
\beta^2 &= \frac{0.08^2}{0.40 \times 0.60} \\
&= 0.026667 \\
E[\alpha^2(U)] &= \big(0.25^2 + 0.75^2\big)\beta^2 \\
&= 0.625 \times 0.026667 \\
&= 0.016667 \\
t &= \frac{0.016667}{0.01 + 0.0625 - 0.016667} \\
&= 0.2985 \\
c^* &= \frac{0.40}{0.40 + 0.2985} \\
&= 0.5726
\end{align*}

Reserves:

\begin{align*}
R_{\text{BF}} &= 0.60 \times 75\% \\
&= 45.0\% \\
R_{\text{CL}} &= 0.60 \times 36\%/0.40 \\
&= 54.0\% \\
R_{\text{GB}} &= 0.40(54.0\%) + 0.60(45.0\%) \\
&= 48.6\% \\
R_{c^*} &= 0.5726(54.0\%) + 0.4274(45.0\%) \\
&= 50.2\%
\end{align*}

Mean squared errors, using \text{mse}(R_c) = E[\alpha^2(U)]\left(\frac{c^2}{p} + \frac{1}{q} + \frac{(1-c)^2}{t}\right)q^2, which gives BF at c = 0 and the chain ladder at c = 1:

Reserve c mse Standard error
BF 0 0.03010 17.3\%
Chain ladder 1 0.02500 15.8\%
Benktander 0.40 0.01964 14.0\%
Optimal 0.5726 0.01859 13.6\%

The conditions agree. p = 0.40 > t, so the chain ladder beats BF. t = 0.30 < 2 - p = 1.6, so GB beats BF, and t > pq/(1+p) = 0.171, so GB beats the chain ladder.

Benktander gets within 0.4 points of the optimal standard error without estimating any variance at all. These are unconditional errors with the payout pattern taken as known; uncertainty in p would add to all four.