Testing An Exponential Severity Restriction

Why Test the Exponential Restriction?

An exponential model is often a useful starting point for positive claim severities because it has a single parameter and a simple likelihood. A gamma model can represent a wider range of shapes. Comparing the two asks a precise question: does allowing the gamma shape to differ from one improve the fit enough to justify the additional parameter?

This is a nested-model comparison. The exponential model is not a separate, unrelated alternative: it is the gamma model with its shape fixed at one. The likelihood-ratio test measures the improvement in maximized likelihood when that restriction is removed. The test is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Suppose $X_1,\ldots,X_n$ are independent, identically distributed, positive, continuously measured severities, with no truncation or censoring. Under the null hypothesis, $X_i$ has an exponential distribution with mean $\theta$. Under the alternative, $X_i$ has a gamma distribution with shape $\alpha>0$ and scale $\theta>0$. Thus $H_0:\alpha=1$, with $\theta$ estimated, versus $H_1:\alpha>0,\ \theta>0$, with both parameters estimated. The exponential model is the gamma model restricted to $\alpha=1$.

Using the shape-scale parameterization, the gamma density is $f(x)=x^{\alpha-1}e^{-x/\theta}/\{\Gamma(\alpha)\theta^\alpha\}$ for $x>0$. At $\alpha=1$, this becomes $f(x)=e^{-x/\theta}/\theta$, the exponential density with mean $\theta$. State the parameterization before calculating: a gamma rate parameter is the reciprocal of its scale.

The Likelihood-Ratio Statistic

Let $\ell_0$ be the largest log-likelihood under the exponential restriction and $\ell_1$ the largest log-likelihood under the unrestricted gamma model. The generalized likelihood-ratio statistic is twice the difference, with the larger likelihood in the positive direction.

T=2(\ell_1-\ell_0).

For the usual chi-square calibration, the observations must satisfy the model assumptions, the likelihood must be correctly specified, and the null parameter must be an interior point of the alternative parameter space. Here $\alpha=1$ is interior to $\alpha>0$. The alternative has two estimated parameters, while the null has one; the difference is one. Under the regularity conditions for Wilks’ theorem, $T$ is approximately distributed as chi-square with one degree of freedom under $H_0$. A large $T$, and hence a small upper-tail p-value, is evidence against the exponential restriction.

Result: For an uncensored sample of positive severities, the exponential-versus-gamma likelihood-ratio test uses $T=2(\ell_1-\ell_0)$ and, under the null and standard regularity conditions, compares it with a $\chi^2_1$ reference distribution. The one degree of freedom comes from estimating the additional gamma shape parameter.

Deriving the Gamma Fit and the Statistic

Write $\bar{x}=n^{-1}\sum_{i=1}^n x_i$ for the sample mean and $D=\ln(\bar{x})-n^{-1}\sum_{i=1}^n\ln(x_i)$. The quantity $D$ is nonnegative by the arithmetic-mean/geometric-mean inequality. For gamma data, the log-likelihood is

\ell(\alpha,\theta) =(\alpha-1)\sum_{i=1}^n\ln(x_i) -\frac{\sum_{i=1}^n x_i}{\theta} -n\alpha\ln(\theta) -n\ln\Gamma(\alpha).

For fixed $\alpha$, differentiating with respect to $\theta$ and setting the derivative to zero gives $\widehat{\theta}=\bar{x}/\alpha$. Substituting this estimate into the likelihood and differentiating with respect to $\alpha$ gives the shape equation below. The function $\psi(\alpha)=d\ln\Gamma(\alpha)/d\alpha$ is the digamma function.

\ln(\widehat{\alpha})-\psi(\widehat{\alpha})=D, \qquad \widehat{\theta}=\frac{\bar{x}}{\widehat{\alpha}}.

Except in special cases, the shape equation is solved numerically. With $\widehat{\alpha}$ and $\widehat{\theta}$ substituted into the gamma likelihood, while the restricted exponential estimate is $\widehat{\theta}_0=\bar{x}$, subtraction gives a convenient expression for the test statistic:

T=2n\left[ -(\widehat{\alpha}-1)D +\widehat{\alpha}\ln(\widehat{\alpha}) -\ln\Gamma(\widehat{\alpha}) -\widehat{\alpha}+1 \right].

The expression depends on the data through $n$ and $D$; the sample mean sets the scale estimate but cancels from the likelihood difference. That cancellation also explains why changing the measurement unit for every severity does not change $T$.

Why the Likelihood Difference Has This Form

The shape estimate and the likelihood-ratio formula follow directly from the likelihood, rather than from a separate goodness-of-fit statistic. To see the key steps, differentiate the gamma log-likelihood and compare the two maximized fits.

Estimate the scale for a fixed shape.
The scale score is $\partial\ell/\partial\theta=\sum x_i/\theta^2-n\alpha/\theta$. Setting it to zero yields $\widehat{\theta}=\bar{x}/\alpha$.

Estimate the shape.
Substitute $\bar{x}/\alpha$ for $\theta$. The resulting shape score, divided by $n$, is $n^{-1}\sum\ln(x_i)-\ln(\bar{x})-\psi(\alpha)+\ln(\alpha)=0$. Rearranging gives $\ln(\alpha)-\psi(\alpha)=D$.

Subtract the restricted maximum.
Under $H_0$, the exponential maximum occurs at $\widehat{\theta}_0=\bar{x}$. Substitute this and the unrestricted gamma estimates into their log-likelihoods. Their difference is the bracketed expression in the formula for $T$, multiplied by $n$.

Double the improvement.
By definition, the likelihood-ratio statistic is twice the unrestricted-minus-restricted maximized log-likelihood, so the resulting difference is multiplied by two.

The difference cannot be negative when both maximizations use the same data and the null model is nested inside the alternative: the unrestricted fit can always choose $\alpha=1$ and reproduce the restricted fit. A negative computed difference therefore signals an error in the likelihoods, parameterization, or subtraction order.

Worked Tests

Worked Example: A Significant Improvement Over the Exponential Fit

A hypothetical portfolio has $n=20$ independent, uncensored positive claim severities measured in dollars. The sample summaries are $\bar{x}=\$100$ and $D=0.270362845$. Test the exponential restriction against the two-parameter gamma model at the 5% significance level. Use the supplied value $D=\ln(2)-\psi(2)$.

Step 1: State the hypotheses and estimates. The null is $\alpha=1$; the alternative estimates $\alpha$ and $\theta$. Since the supplied $D$ equals $\ln(2)-\psi(2)$, the gamma shape equation gives $\widehat{\alpha}=2$. Thus $\widehat{\theta}=\bar{x}/\widehat{\alpha}=100/2=\$50$. Under the null, the exponential mean estimate is $\widehat{\theta}_0=\$100$.

Step 2: Calculate the log-likelihood improvement per observation. Since $\Gamma(2)=1$, $\ln\Gamma(2)=0$. The bracketed term in the statistic is $ -(2-1)(0.270362845)+2\ln(2)-0-2+1$. Using $2\ln(2)=1.386294361$, this is $-0.270362845+1.386294361-1=0.115931516$.

Step 3: Calculate the test statistic and p-value. $T=2(20)(0.115931516)=4.63726064$, or $4.6373$ to four decimal places. With one degree of freedom, the upper-tail p-value is approximately $0.0313$.

Conclusion and check. Because $0.0313<0.05$, reject the exponential restriction at the 5% level in favor of the gamma alternative. The unrestricted fit has a higher maximized likelihood, as required; its log-likelihood improvement is $4.63726064/2=2.31863032$. This conclusion is evidence against the restriction, not proof that the gamma model is an adequate description in every respect. The calculation uses the gamma shape likelihood equation and likelihood-ratio test, and avoids treating a goodness-of-fit statistic as the likelihood-ratio statistic.

Worked Example: A Larger Shape Estimate Does Not Guarantee Rejection

For a different hypothetical sample of $n=30$ independent, uncensored severities, suppose $\bar{x}=\$240$ and $D=0.368975134$. The value is $\ln(1.5)-\psi(1.5)$. Test the same restriction at the 5% level. Use $\ln\Gamma(1.5)=-0.120782238$ and $\psi(1.5)=0.036489974$.

Step 1: Fit both models. The shape equation gives $\widehat{\alpha}=1.5$, and $\widehat{\theta}=240/1.5=\$160$. The exponential estimate remains $\widehat{\theta}_0=\$240$.

Step 2: Evaluate the likelihood improvement per observation. The bracketed term is $-(1.5-1)(0.368975134)+1.5\ln(1.5)-\ln\Gamma(1.5)-1.5+1$. The first term is $-0.184487567$. Also, $1.5\ln(1.5)=0.608197662$, and $-\ln\Gamma(1.5)=0.120782238$. The remaining terms contribute $-0.5$. The total is $-0.184487567+0.608197662+0.120782238-0.5=0.044492333$.

Step 3: Test and interpret. The statistic is $T=2(30)(0.044492333)=2.66953998$, or $2.6695$ to four decimal places. Its $\chi^2_1$ upper-tail p-value is approximately $0.1023$. Because $0.1023>0.05$, do not reject the exponential restriction at the 5% level. The gamma estimate has a shape different from one, but the likelihood improvement is not large enough relative to its sampling variability to produce rejection. The likelihood difference is positive, which is an arithmetic check on the calculation.

Worked Example: Converting Units and Avoiding the Wrong Degrees of Freedom

A hypothetical sample contains $n=8$ independent, uncensored severities with mean $\$500$ and $D=0.270362845$, again corresponding to $\widehat{\alpha}=2$. An analyst records the same losses in thousands of dollars and reports a statistic of $1.8549$. Check the statistic, find its p-value, and identify the correct degrees of freedom.

Step 1: Check the unit conversion. In thousands of dollars, the sample mean is $0.5$, and every log severity decreases by $\ln(1000)$. Therefore $\ln(\bar{x})$ and the average log severity each decrease by $\ln(1000)$; their difference $D$ is unchanged. The fitted shape remains $2$, and the scale is $0.5/2=0.25$ thousand dollars, or $\$250$.

Step 2: Recalculate the statistic. The per-observation likelihood improvement is $0.115931516$, as in the $\widehat{\alpha}=2$ calculation above. Therefore $T=2(8)(0.115931516)=1.854904256$, which rounds to $1.8549$.

Step 3: Use the correct reference distribution. The unrestricted gamma model adds one free parameter, the shape, to the exponential model. Thus the reference distribution is $\chi^2_1$, not $\chi^2_2$. The upper-tail p-value is approximately $0.1729$, so the restriction is not rejected at 5%.

Common mistake and correction. Using two degrees of freedom just because the gamma model has two parameters would give the wrong comparison: the null also estimates one parameter, its mean. The degrees of freedom are the difference in parameter counts, $2-1=1$. Using $\chi^2_2$ here would give an upper-tail probability near $0.3956$, not the correct $0.1729$. Both the test statistic and the degrees of freedom have now been checked against the nested models.

Interpretation, Assumptions, and Common Traps

The null hypothesis is a restriction on the gamma shape, not a claim that the observed sample must look exactly exponential. A small p-value says that the observed likelihood improvement would be unusual under the restricted model, using the stated reference distribution. A large p-value means the data do not provide sufficient evidence to reject the restriction; it does not establish that the exponential model is true.

The chi-square result is an approximation, particularly relevant when the sample is not large. It relies on a correctly specified likelihood and the regularity conditions for Wilks’ theorem. The calculations above assume independent, uncensored observations from a continuous severity distribution. With truncation or censoring, the likelihood must reflect the observation mechanism; the displayed likelihood and estimates should not be applied unchanged. If the null parameter were on a boundary, the usual chi-square calibration could also fail.

Reversing the likelihood difference: use unrestricted minus restricted log-likelihood. Reversing the order can produce an impossible negative statistic.
Counting all alternative parameters: the degrees of freedom are the difference between model dimensions, not the number of parameters in the larger model.
Confusing scale and rate: with the shape-scale convention here, $\theta$ is the mean when $\alpha=1$, and the exponential rate is $1/\theta$.
Overstating a non-significant result: failure to reject is not proof of an exponential severity distribution.

Key takeaway: For the exponential-versus-gamma severity test, estimate the gamma shape and scale, compare the maximized likelihoods using $T=2(\ell_1-\ell_0)$, and use one degree of freedom because the gamma model adds one shape parameter. Interpret the p-value in light of the model assumptions and the approximate chi-square calibration.

Check Your Understanding

For each question, use the exponential restriction $\alpha=1$, the gamma shape-scale parameterization, and the $\chi^2_1$ approximation when appropriate.

A sample has $n=10$, and the gamma shape estimate is $2$, with $D=\ln(2)-\psi(2)=0.270362845$. Find the likelihood-ratio statistic. (Answer: $T=2.3186$.)
For $n=25$ and the same fitted shape and $D$ as in Question 1, calculate the statistic and decide whether to reject at 5%. (Answer: $T=5.7966$, p-value approximately $0.0161$; reject at 5%.)
A sample has $n=12$, $\widehat{\alpha}=1.5$, and $D=0.368975134$. Use the likelihood-ratio formula to calculate $T$ and the approximate p-value. (Answer: $T\approx1.0678$, p-value approximately $0.3015$.)
A data set is converted from dollars to thousands of dollars, with every severity divided by $1000$. What happens to $D$ and to the likelihood-ratio statistic, assuming the same observations and models? (Answer: Neither $D$ nor the statistic changes; both the log sample mean and average log severity shift by $-\ln(1000)$.)
The unrestricted gamma maximized log-likelihood is $-45.8$, and the restricted exponential maximized log-likelihood is $-48.0$. State the statistic, degrees of freedom, and 5% decision. (Answer: $T=2[-45.8-(-48.0)]=4.4$, with 1 degree of freedom; the p-value is approximately $0.0360$, so reject at 5%.)

What You'll Learn