Fit a Power Curve for Tail Development

Why Fit a Power Curve?

At late development ages, selected link ratios often lie close to one, with their excess above one decreasing as age increases. A power curve offers one way to describe that pattern and extrapolate beyond the observed ages. The fitted curve can supply future age-to-age factors; multiplying those factors gives development over a specified range. An infinite tail requires an additional convergence check.

This method is a model-based extrapolation, not a guarantee that late development will follow the fitted pattern. Its results depend on the selected observations, the development-age scale, and whether the power-curve shape is credible for the loss experience being analyzed. The earlier material on diagnosing development patterns and selecting coherent factors is relevant when deciding whether the late-age observations are suitable for fitting.

Definition: Let $f_x$ be the age-to-age link ratio from development age $x$ to age $x+1$. A power-curve model for late-age link ratios is $$ f_x=1+a x^{-b}, $$ where $a>0$ and $b>0$. The model describes the excess $f_x-1$, not the entire link ratio. Development ages must use a consistent scale, such as years.

Fitting the Power Curve

Taking logarithms after subtracting one transforms the model into a straight-line relationship. Define $z_x=\ln(f_x-1)$ and $t_x=\ln x$. Then

z_x=\ln a-bt_x.

Fit a linear regression of $z_x$ on $t_x$. If its intercept is $\widehat{\alpha}$ and its slope is $\widehat{\beta}$, the power-curve estimates are

\widehat{a}=e^{\widehat{\alpha}}, \qquad \widehat{b}=-\widehat{\beta}, \qquad \widehat{f}_x=1+\widehat{a}x^{-\widehat{b}}.

This regression assumes that the power curve is a reasonable description of the late-age pattern and, when interpreting ordinary least squares as a statistical fit, that deviations from the straight line are appropriately modeled on the log-excess scale. The transformation requires $f_x>1$ for every fitted observation. A zero or negative excess cannot be logged; a ratio at or below one is not directly fit by this positive-excess model. Do not silently discard such observations or force them into the regression. Investigate the data and consider whether this model is suitable.

The selected development ages also matter. Use the ages corresponding to the observations, not row numbers or a reindexed sequence. Rescaling every age by the same positive constant changes the fitted $a$, although it does not change $b$ under an exact power curve. Mixing age units or mislabeling the ages can therefore distort the extrapolation.

Worked Example: Fitting a Power Curve by Regression

A hypothetical liability portfolio has selected late-age link ratios of 1.0400 at age 4, 1.0100 at age 8, and 1.0025 at age 16. Fit the power curve using these three points, and verify the fitted values.

Method and assumptions. Assume the late-age ratios follow $f_x=1+a x^{-b}$, with positive excesses. Regress $\ln(f_x-1)$ on $\ln x$. The three observed excesses are 0.0400, 0.0100, and 0.0025, so the transformation is defined.

The transformed points, rounded to six decimal places, are:

Age $x$	$f_x-1$	$\ln x$	$\ln(f_x-1)$
4	0.0400	1.386294	-3.218876
8	0.0100	2.079442	-4.605170
16	0.0025	2.772589	-5.991465

For each doubling of age, $\ln x$ increases by $\ln 2$, while $\ln(f_x-1)$ decreases by $\ln 4$. Thus the fitted slope is

\widehat{\beta} =\frac{-\ln 4}{\ln 2} =-2, \qquad \widehat{b}=2.

Using the age-4 point, the intercept is

\widehat{\alpha} =\ln(0.0400)-(-2)\ln 4 =\ln(0.0400)+2\ln 4 =\ln(0.64), \qquad \widehat{a}=0.64.

Therefore the fitted curve is $\widehat{f}_x=1+0.64x^{-2}$. At ages 4, 8, and 16, its excesses are, respectively, $0.64/16=0.0400$, $0.64/64=0.0100$, and $0.64/256=0.0025$. It reproduces all three observations, as expected for these exact power-curve data. The check confirms both the parameter estimates and the use of $f_x-1$ in the transformation.

Common mistake avoided: taking the logarithm of $f_x$ instead of $f_x-1$ does not linearize this model. The required regression response is $\ln(f_x-1)$.

From Projected Links to Development

A projected link ratio applies to a specific transition: $f_x$ takes a cumulative amount at age $x$ to age $x+1$. If development is needed from age $k$ through age $m$, the cumulative development factor is the product of the links beginning at $k$ and ending at $m-1$. This product convention prevents an off-by-one error.

\widehat{F}_{k\text{ to }m} =\prod_{x=k}^{m-1}\widehat{f}_x =\prod_{x=k}^{m-1}\left(1+\widehat{a}x^{-\widehat{b}}\right).

Worked Example: Projecting Links Through Age 12

Use the fitted curve $\widehat{f}_x=1+0.64x^{-2}$ from the previous calculation. A hypothetical book of claims is evaluated at age 8. Calculate its projected development factor from age 8 through age 12.

Method and assumption. Apply the curve to the four transitions 8-to-9, 9-to-10, 10-to-11, and 11-to-12. The factor is the product of $\widehat{f}_8$ through $\widehat{f}_{11}$, not through $\widehat{f}_{12}$.

Calculate each factor:

\begin{aligned} \widehat{f}_8 &= 1+\frac{0.64}{8^2}=1.010000,\\ \widehat{f}_9 &= 1+\frac{0.64}{9^2}=1.007901,\\ \widehat{f}_{10} &= 1+\frac{0.64}{10^2}=1.006400,\\ \widehat{f}_{11} &= 1+\frac{0.64}{11^2}=1.005289. \end{aligned}

Multiplying the unrounded factors gives

\widehat{F}_{8\text{ to }12} =1.010000 \times 1.0079012346 \times 1.006400 \times 1.0052892562 \approx 1.029914.

Rounded to six decimal places, the projected factor is 1.029914. The projected cumulative amount at age 12 is therefore approximately 1.029914 times the amount at age 8. As a check, every link exceeds one, so the product must exceed one; the computed factor does.

Common mistake avoided: including $\widehat{f}_{12}$ would project the transition from age 12 to age 13, not development ending at age 12.

When Does an Infinite Tail Converge?

A curve can produce sensible finite-horizon factors even when it cannot produce a finite factor to ultimate. If the model is used for an infinite tail beginning at age $k$, its implied factor is the infinite product

F_{k\text{ to ultimate}} =\prod_{x=k}^{\infty}(1+a x^{-b}).

For $a>0$, this product is finite exactly when $b>1$. Here is why. Taking logarithms of the partial product gives a sum of $\ln(1+a x^{-b})$. For sufficiently large $x$, $a x^{-b}$ is small, and $\ln(1+a x^{-b})$ is comparable to $a x^{-b}$. The sum of $x^{-b}$ over integer ages converges when $b>1$ and diverges when $b\leq 1$, by the p-series criterion. Consequently, the logarithm of the product has a finite limit only when $b>1$; in that case the product has a finite limit as well.

This condition is about the infinite extrapolation, not whether any particular finite product can be calculated. A finite range contains finitely many factors and is finite whenever those factors are finite. If the estimated exponent is at or below one, report a finite-horizon projection only; do not label the divergent infinite product a finite tail-to-ultimate factor.

Worked Example: A Finite Projection Does Not Ensure a Finite Ultimate

A hypothetical selected curve is $f_x=1+0.20x^{-1}$. Calculate development from age 10 to age 12, then determine whether this curve implies a finite factor from age 10 to ultimate.

Method and assumptions. This is the power model with $a=0.20$ and $b=1$. The finite projection includes $f_{10}$ and $f_{11}$; an infinite tail requires the convergence test $b>1$.

The two links are

f_{10}=1+\frac{0.20}{10}=1.020000, \qquad f_{11}=1+\frac{0.20}{11}=1.0181818\ldots.

Thus the finite factor, rounded to six decimal places, is

F_{10\text{ to }12} =1.020000\times1.0181818\ldots =1.038545\ldots \approx1.038545.

The finite factor is greater than one, consistent with both links exceeding one. But $b=1$ does not satisfy $b>1$. The sum of the late-age excesses behaves like $0.20\sum_{x=10}^{\infty}1/x$, which diverges. Therefore this curve does not imply a finite age-10-to-ultimate factor, even though the age-10-to-12 factor is finite and calculable.

Common mistake avoided: extrapolating a finite number of links successfully is not evidence that the infinite product converges. Check the fitted exponent before using an infinite tail.

Interpretation, Checks, and Common Traps

The power curve is useful when the excess of late-age link ratios appears to decline according to a power of age. It supplies a transparent relationship between age and projected excess, but it does not validate the selected data or eliminate judgment about the tail. Examine the observed and fitted ratios together, and consider whether a small number of late-age observations supports extrapolation over the intended range.

A frequent exam trap is to regress $\ln f_x$ on $\ln x$, as if $f_x=ax^{-b}$. That is a different model: the power curve here is $f_x=1+ax^{-b}$. For instance, with $f_4=1.08$ and $f_8=1.02$, the incorrect raw-ratio calculation gives

\frac{\ln(1.02/1.08)}{\ln(8/4)} \approx -0.0825.

That slope does not estimate the exponent of the specified excess model. The correct excesses are $0.08$ and $0.02$, so

\frac{\ln(0.02/0.08)}{\ln(8/4)} =\frac{\ln(0.25)}{\ln 2} =-2.

Thus the two points imply $b=2$ under an exact power curve, with $a=0.08(4^2)=1.28$. The fitted age-16 link is $1+1.28/16^2=1.005$. The correction is to subtract one before taking logarithms—not to interpret the raw-ratio slope as the power exponent.

Before reporting a tail estimate, check that the fitted observations have positive excesses, that age indices and units are consistent, and that the projected range uses the correct link ages. For an infinite tail, also verify $b>1$. State whether the result is a finite-horizon factor or an infinite factor to ultimate; those are not interchangeable.

Key takeaway: Fit $\ln(f_x-1)=\ln a-b\ln x$, then use $1+\widehat{a}x^{-\widehat{b}}$ for each projected link. Multiply the links over the required ages, and use the condition $\widehat{b}>1$ before treating an infinite product as a finite tail-to-ultimate factor.

Check Your Understanding

For each problem, identify the model or product convention needed before calculating.

A hypothetical portfolio has ratios 1.0600 at age 3 and 1.0150 at age 6. Assuming an exact power curve, find $a$ and $b$, then calculate the fitted age-12 link ratio. (Answer: $a=0.54$, $b=2$, and $f_{12}=1.00375$.)
For $f_x=1+0.30x^{-2}$, calculate the factor from age 5 through age 8, using the transitions at ages 5, 6, and 7. Round to six decimals. (Answer: $1.026681$.)
A fitted power curve has $a=0.50$ and $b=1.2$. Calculate the age-10 link ratio to four decimal places and state whether the infinite tail converges. (Answer: $f_{10}\approx1.0315$; the infinite tail converges because $b>1$.)
An analyst regresses $\ln f_x$ on $\ln x$ for $f_4=1.08$ and $f_8=1.02$. Give the correct power-curve exponent using the excesses, and identify the mistake in the analyst’s transformation. (Answer: $b=2$; the transformation must use $\ln(f_x-1)$, not $\ln f_x$.)
For $f_x=1+0.80x^{-2}$, calculate the age-16 link ratio and decide whether an infinite tail beginning at age 16 is finite. (Answer: $f_{16}=1.003125$; the infinite tail is finite because $b=2>1$.)

What You'll Learn