Why Feasibility Matters
A translated gamma approximation matches an aggregate loss distribution using its mean, variance, and third central moment. Moment matching gives candidate parameters, but those parameters do not automatically define a valid gamma distribution. In particular, the ordinary gamma distribution requires a positive scale, which places a restriction on the sign of the third central moment.
A second check is needed when calculating probabilities: the fitted gamma variable is the aggregate loss minus its location shift. Omitting that shift changes the threshold supplied to the gamma distribution and can produce a substantially incorrect probability. This tutorial develops the parameter checks and probability calculation, aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.
The scale parameter \(\theta\) is positive; it is not a rate parameter. A gamma distribution specified by a rate uses the reciprocal \(1/\theta\). Keeping the parameterization clear is essential both for fitting the model and for evaluating probabilities.
Deriving the Moment-Matching Parameters
The location shift changes the mean but not the variance or any higher central moment. Consequently, the variance and third central moment of \(S\) must match those of \(G\), while the mean of \(S\) must match \(c+E[G]\).
For a short justification of the gamma moment formulas, the cumulant-generating function of a gamma variable with shape \(\alpha\) and scale \(\theta\) is \(K_G(t)=-\alpha\ln(1-\theta t)\), for \(t<1/\theta\). Expanding the logarithm gives \(K_G(t)=\alpha\sum_{r=1}^{\infty}\theta^r t^r/r\). The coefficient of \(t^r/r!\) is the \(r\)th cumulant, so \(\kappa_r=(r-1)!\alpha\theta^r\). In particular, \(\kappa_2=\alpha\theta^2\) is the variance and \(\kappa_3=2\alpha\theta^3\) is the third central moment. Adding a constant \(c\) changes only the first cumulant, which justifies the three matching equations above.
Dividing the third-moment equation by the variance equation gives \(\kappa_3/\sigma^2=2\theta\). Substitution then determines the shape, and the mean equation determines the shift.
Here \(\sigma^4=(\sigma^2)^2\) and \(\sigma^6=(\sigma^2)^3\). These formulas require \(\sigma^2>0\) and \(\kappa_3\ne0\) to be evaluated. More importantly, an ordinary gamma fit requires \(\theta>0\). Since \(\sigma^2>0\), this is equivalent to \(\kappa_3>0\). A zero or negative third central moment therefore does not yield a valid ordinary translated gamma fit, even though the algebraic expression for \(\alpha\) may be positive.
An equivalent form uses the aggregate coefficient of skewness \(\gamma_1=\kappa_3/\sigma^3\), where \(\sigma\) is the positive standard deviation. When \(\gamma_1>0\), the fitted parameters are \(\alpha=4/\gamma_1^2\), \(\theta=\sigma\gamma_1/2\), and \(c=\mu-2\sigma/\gamma_1\). The third-central-moment form is often convenient when the aggregate moments have already been calculated directly.
Calculating Probabilities With the Shift
If \(S\) is approximated by \(c+G\), then \(S\le x\) is equivalent to \(G\le x-c\). The shift must therefore be subtracted from the loss threshold before evaluating the gamma probability. Define the regularized lower incomplete gamma function by
For a threshold \(x\), the fitted cumulative probability is \(0\) when \(x\le c\), because \(G\) is positive. When \(x>c\), the gamma threshold must also be divided by the scale \(\theta\).
For integer shape \(\alpha=n\), the gamma cumulative probability has the finite-sum form \(P(n,z)=1-e^{-z}\sum_{j=0}^{n-1}z^j/j!\). This is useful for hand calculations. For a noninteger shape, evaluate the regularized incomplete gamma function using an appropriate numerical method or calculator.
Worked Example: Fit Parameters and Calculate an Aggregate Tail
A hypothetical portfolio aggregate loss has mean \(20\), variance \(18\), and third central moment \(108\), with all monetary amounts expressed in consistent units. Fit the translated gamma parameters and estimate \(P(S>20)\).
Method and feasibility. Use the moment-matching formulas. The variance is positive and the third central moment is positive, so a valid ordinary gamma scale is possible.
Scale. \(\theta=108/(2\cdot18)=108/36=3\).
Shape. \(\alpha=4\cdot18^3/108^2=4\cdot5{,}832/11{,}664=2\).
Location. \(c=20-2\cdot18^2/108=20-648/108=20-6=14\).
Check the fitted moments. The fitted mean is \(14+2\cdot3=20\). Its variance is \(2\cdot3^2=18\), and its third central moment is \(2\cdot2\cdot3^3=108\). All three match the supplied moments.
Probability. The event \(S>20\) becomes \(G>20-14=6\). In scale units, \(z=6/3=2\). With shape \(2\), the gamma tail is \(e^{-z}(1+z)=e^{-2}(1+2)=3e^{-2}\). Thus \(P(S>20)\approx0.4060\), rounded to four decimal places.
The fitted shift is applied before dividing by the scale. Treating 20 itself as the gamma threshold would instead use \(20/3\), which is not the threshold corresponding to \(S>20\).
Worked Example: Fit From Skewness and Evaluate a Cumulative Probability
A hypothetical aggregate loss has mean \(50\), standard deviation \(10\), and coefficient of skewness \(1\). Use the translated gamma approximation to estimate \(P(S\le40)\).
Method and feasibility. The positive skewness permits an ordinary gamma fit. First obtain the third central moment: \(\kappa_3=\gamma_1\sigma^3=1\cdot10^3=1{,}000\).
Shape. \(\alpha=4/\gamma_1^2=4\).
Scale. \(\theta=\sigma\gamma_1/2=10/2=5\).
Location. \(c=\mu-\alpha\theta=50-4\cdot5=30\). The fitted mean is \(30+20=50\), its variance is \(4\cdot5^2=100\), and its third central moment is \(2\cdot4\cdot5^3=1{,}000\), consistent with the stated standard deviation and skewness.
Probability. The event \(S\le40\) becomes \(G\le40-30=10\), so \(z=10/5=2\). For shape \(4\), \(P(4,2)=1-e^{-2}(1+2+2^2/2!+2^3/3!)=1-(19/3)e^{-2}\approx0.1429\). The estimate, rounded to four decimal places, is \(P(S\le40)\approx0.1429\).
The calculation checks both transformations: subtract the location \(30\), then divide the resulting threshold by the scale \(5\).
Worked Example: Reject an Infeasible Moment Match
A hypothetical aggregate model produces a mean of \(30\), variance \(9\), and third central moment \(-18\). Determine whether an ordinary translated gamma approximation can match these moments.
Method and feasibility. Check the scale first, since an ordinary gamma distribution requires a positive scale. The candidate value is \(\theta=-18/(2\cdot9)=-1\), which is invalid.
The algebraic shape value would be \(\alpha=4\cdot9^3/(-18)^2=2{,}916/324=9\), and the algebraic shift would be \(c=30-2\cdot9^2/(-18)=30+9=39\). Neither value repairs the negative scale. In fact, a gamma variable with positive shape and scale has positive third central moment \(2\alpha\theta^3\), so it cannot match \(-18\).
Therefore, no valid ordinary translated gamma fit exists for these moments. Do not use the resulting algebraic values to calculate gamma probabilities; a different approximation method would be needed.
Practical Checks and Common Traps
A reliable calculation separates parameter feasibility from probability evaluation. First verify that the supplied moments are interpreted correctly: the input is the third central moment, not the third raw moment. If skewness is supplied instead, convert using \(\kappa_3=\gamma_1\sigma^3\). Then check the sign of \(\kappa_3\) before using the formulas.
Confirm that \(\sigma^2>0\), and identify whether the input is a third central moment or a coefficient of skewness.
An ordinary translated gamma requires \(\kappa_3>0\). If \(\kappa_3\le0\), stop: the fit is not feasible.
Find \(\theta\), \(\alpha\), and \(c\), then substitute them into the fitted mean, variance, and third central moment.
For a probability at threshold \(x\), use the gamma threshold \((x-c)/\theta\) when \(x>c\). If \(x\le c\), the fitted cumulative probability is zero.
A frequent exam trap is to calculate the gamma probability at \(x/\theta\) instead of \((x-c)/\theta\). The location shift is part of the model, not an optional adjustment. Another is to conclude that a fit is feasible merely because the computed shape is positive: when \(\kappa_3<0\), the shape formula squares the third moment, but the scale remains negative. Finally, a negative fitted location does not make the gamma parameters invalid. It does mean the approximation assigns probability to negative aggregate losses, so note that limitation when interpreting results.
Check Your Understanding
For each problem, check feasibility before calculating a probability.
- A hypothetical aggregate loss has mean \(18\), variance \(8\), and third central moment \(32\). Find the fitted shape, scale, and location. (Answer: \(\alpha=2,\ \theta=2,\ c=14\))
- A hypothetical aggregate loss has mean \(25\), variance \(16\), and third central moment \(128\). Fit the translated gamma and estimate \(P(S\le25)\), rounding the probability to four decimal places. (Answer: \(\alpha=1,\ \theta=4,\ c=21;\ P(S\le25)\approx0.6321\))
- A hypothetical aggregate loss has mean \(50\), variance \(36\), and third central moment \(216\). Fit the parameters and estimate \(P(S\le41)\), rounding the probability to four decimal places. (Answer: \(\alpha=4,\ \theta=3,\ c=38;\ P(S\le41)=1-(8/3)e^{-1}\approx0.0190\))
- A hypothetical aggregate loss has variance \(16\) and third central moment \(-64\). Is an ordinary translated gamma fit feasible? State the scale that the matching formula would produce. (Answer: No; \(\theta=-64/(2\cdot16)=-2\), which is not a valid gamma scale.)
- A hypothetical aggregate loss has mean \(12\), variance \(8\), and third central moment \(32\). Fit the parameters and estimate \(P(S\le7)\). (Answer: \(\alpha=2,\ \theta=2,\ c=8;\ P(S\le7)=0\), since \(7\le c\).)