Why Compare Moment-Matched Approximations?
A moment approximation compresses information about an aggregate loss into a small set of summary measures. That makes it useful when an exact aggregate distribution is unavailable or inconvenient to evaluate. But two approximations fitted to the same moments need not assign similar probabilities to a particular threshold. Their support and tail shapes matter, especially when the question concerns an unusual loss or a probability near a boundary.
This tutorial compares three choices for an aggregate loss \(S\): a normal distribution fitted to the mean and variance, a translated gamma distribution fitted to the first three central moments, and a shifted lognormal distribution fitted to the same three moments when a feasible fit exists. These are approximations, not assertions that the actual loss follows one of the fitted distributions.
Three Approximations and Their Fits
The normal approximation uses only the first two moments: \(S\) is approximated by \(N(m,v)\), with \(v\) as the variance. A normal distribution is symmetric about its mean and has support over the entire real line. In particular, it assigns some probability to negative losses, even when the actual aggregate loss cannot be negative.
For a translated gamma approximation, represent the fitted variable as \(a+G\), where \(G\) has a gamma distribution with shape \(\alpha>0\) and scale \(\theta>0\). Thus \(G\) has mean \(\alpha\theta\), variance \(\alpha\theta^2\), and third central moment \(2\alpha\theta^3\). Matching the variance and third central moment gives the following parameters when \(\kappa_3>0\):
The fitted translated gamma has support \(S>a\). It has an exponential-type upper tail, unlike the normal’s symmetric tail. If the aggregate loss must be nonnegative, a nonnegative lower endpoint additionally requires \(a\geq 0\). A negative \(a\) is permissible as a mathematical translated gamma fit, but then the approximation assigns positive probability to negative values.
For a shifted lognormal fit, write \(S=a+Z\), where \(Z\) is lognormal: \(\ln Z\sim N(\mu,\sigma^2)\). Define \(w=e^{\sigma^2}-1\). The lognormal component has coefficient of variation squared \(w\), skewness \((w+3)\sqrt{w}\), and mean \(q=\sqrt{v/w}\) when its variance is matched to \(v\). Consequently, for \(\kappa_3>0\), solve
The shift \(a\) must be considered when interpreting support: the fitted variable is greater than \(a\). For a nonnegative lower endpoint, require \(a\geq0\). Unlike the translated gamma fit, finding \(w\) generally requires solving a nonlinear equation. A nondegenerate shifted lognormal fit has positive skewness, so it cannot match a zero or negative third central moment.
Worked Moment Match
Worked Example: Fitting All Three Models
A hypothetical portfolio has aggregate loss moments \(m=100\), \(v=400\), and \(\kappa_3=32{,}000\), measured in thousands of dollars and the corresponding powers of that unit. Fit the three approximations. Check the fitted moments and state each support.
Method. Fit the normal to \(m\) and \(v\); use the translated gamma equations for \(\kappa_3>0\); then solve the shifted lognormal skewness equation.
Normal. The fitted distribution is \(N(100,400)\), so its standard deviation is \(\sqrt{400}=20\). It matches the mean and variance. Its third central moment is zero, so it does not match \(32{,}000\).
Translated gamma. First calculate the scale: \( \theta=32{,}000/(2\cdot400)=40 \). Then calculate the shape: \( \alpha=4(400)^3/(32{,}000)^2=256{,}000{,}000/1{,}024{,}000{,}000=0.25 \). The gamma component has mean \(0.25\cdot40=10\), so the shift is \(a=100-10=90\). Its variance is \(0.25\cdot40^2=400\), and its third central moment is \(2\cdot0.25\cdot40^3=32{,}000\). The fitted aggregate is \(90+G\), with support greater than 90.
Shifted lognormal. The observed skewness is \( \gamma_1=32{,}000/400^{3/2}=32{,}000/8{,}000=4 \). Setting \(w=1\) gives \((w+3)\sqrt{w}=4\), so this solves the skewness equation. Thus \(q=\sqrt{400/1}=20\), \(a=100-20=80\), and \(\sigma^2=\ln 2\). Also, \( \mu=\ln20-\tfrac12\ln2=\ln(10\sqrt2) \). The lognormal component has mean 20 and variance \(20^2(1)=400\); its skewness is 4, so its third central moment is \(4(400^{3/2})=32{,}000\). The fitted aggregate is \(80+Z\), with support greater than 80.
Check. Each approximation has mean 100 and variance 400. The two skewed fits also have third central moment 32,000. The normal’s support extends below zero; the translated gamma and shifted lognormal have positive lower endpoints in this fit. A common mistake is to report the gamma shift as its support endpoint without checking the lognormal shift separately.
Why the Gamma Parameters Take This Form
The translated gamma formulas follow directly from the component moments, so they are not merely a parameter lookup. Translation changes the mean but not the variance or central moments of order two and higher. Therefore, matching variance and third central moment requires \(v=\alpha\theta^2\) and \(\kappa_3=2\alpha\theta^3\).
\(\kappa_3/v=(2\alpha\theta^3)/(\alpha\theta^2)=2\theta\), hence \(\theta=\kappa_3/(2v)\).
From \(v=\alpha\theta^2\), \(\alpha=v/\theta^2=4v^3/\kappa_3^2\).
The fitted mean is \(a+\alpha\theta\), so \(a=m-\alpha\theta=m-2v^2/\kappa_3\).
This derivation also shows why a positive third central moment is required for a standard translated gamma fit with positive scale. It does not establish that a gamma distribution is the true aggregate-loss model; it only identifies the parameters that match the stated moments.
Support and Probability Estimates
Worked Example: Comparing Probabilities Near the Lower Endpoint
Use the fitted models from the preceding example to estimate the probability that the aggregate loss is at most 85 thousand dollars. State the distribution and parameters used in each calculation.
Normal method. Under \(N(100,400)\), the standard deviation is 20. The standardized value is \(z=(85-100)/20=-0.75\), so \(P(S\leq85)\approx\Phi(-0.75)=0.2266\).
Translated gamma method. The fitted variable is \(90+G\), where \(G\) has shape \(0.25\) and scale 40. Since \(85<90\), the event is outside its support, and \(P(S\leq85)=0\).
Shifted lognormal method. The fitted variable is \(80+Z\), with \(\ln Z\sim N(\ln(10\sqrt2),\ln2)\). The threshold for \(Z\) is \(85-80=5\). Its standardized log value is \[ \frac{\ln5-\ln(10\sqrt2)}{\sqrt{\ln2}} =\frac{-1.0397}{0.8326}\approx-1.2488. \] Therefore \(P(S\leq85)\approx\Phi(-1.2488)=0.1058\).
Check. The three estimates are 0.2266, 0, and 0.1058. The large differences are possible because 85 lies below both fitted skewed models’ lower endpoints but above the normal model’s center-minus-one-standard-deviation point. Do not treat agreement on moments as a guarantee of agreement at a threshold.
Worked Example: Comparing Upper-Tail Probabilities
For the same hypothetical aggregate and fitted distributions, estimate \(P(S>120)\). Use the appropriate tail calculation for each model and compare the results.
Normal method. Under \(N(100,400)\), \(z=(120-100)/20=1\), giving \(P(S>120)\approx1-\Phi(1)=0.1587\).
Shifted lognormal method. The event is \(Z>40\), with \(Z\) having log-mean \(\mu=\ln(10\sqrt2)\) and log-variance \(\sigma^2=\ln2\). Thus \[ z=\frac{\ln40-\ln(10\sqrt2)}{\sqrt{\ln2}} =\frac{1.0397}{0.8326}\approx1.2488. \] The estimated probability is \(1-\Phi(1.2488)\approx0.1058\).
Translated gamma method. Here \(S=90+G\), with gamma shape \(0.25\) and scale 40. The event is \(G>30\), so its standardized gamma argument is \(30/40=0.75\). In regularized upper-incomplete-gamma notation, the probability is \(Q(0.25,0.75)\approx0.1000\). For example, the corresponding lower-tail series is \[ P(0.25,0.75)= \frac{e^{-0.75}(0.75)^{0.25}}{\Gamma(1.25)} \left(1+\frac{0.75}{1.25} +\frac{0.75^2}{1.25\cdot2.25} +\frac{0.75^3}{1.25\cdot2.25\cdot3.25}+\cdots\right) \approx0.9000. \] Taking the complement gives approximately \(0.1000\).
Check. The three estimated upper-tail probabilities are 0.1587, 0.1000, and 0.1058 for the normal, translated gamma, and shifted lognormal fits, respectively. These values are approximation outputs, not known true probabilities. In this example the normal estimate is largest, but that ordering should not be assumed at every threshold.
Interpreting the Differences
The approximations differ for structural reasons. A normal distribution is symmetric and has an unbounded lower tail; it may assign probability to impossible negative aggregate losses. A translated gamma distribution has a finite lower endpoint and an exponential-type upper tail. A shifted lognormal also has a finite lower endpoint, but its upper tail is heavier than an exponential tail. For sufficiently large thresholds, the lognormal tail can therefore remain material when an exponential-tail approximation has become much smaller. At moderate thresholds, however, the ordering depends on the fitted parameters and threshold; tail descriptions do not replace calculation.
Moment matching is most informative when the moments are reliable and the threshold of interest is not excessively far into a tail. A normal fit is straightforward when the distribution is not strongly skewed and negative-loss probability is negligible for the application. A translated gamma or shifted lognormal fit can reflect positive skewness, but the fitted lower endpoint and tail assumptions should be checked against the intended use. Neither a good moment match nor a plausible support proves that the approximation is accurate in the tail.
Common Mistakes
- Using the variance as the normal standard deviation. For \(N(m,v)\), the standard deviation is \(\sqrt v\), not \(v\).
- Calling the gamma component mean the shift. The gamma component mean is \(\alpha\theta\); the translation is \(a=m-\alpha\theta\).
- Ignoring feasibility. A fitted shift can be negative, and a shifted lognormal fit requires a positive solution to its skewness equation.
- Assuming matched moments imply matched tails. The models can produce meaningfully different threshold probabilities, as the examples show.
- Comparing tail labels without specifying a threshold. “Heavier tail” describes asymptotic behavior; it does not establish the probability ordering at every finite loss level.
Check Your Understanding
For each problem, identify the relevant moment or support calculation before comparing or interpreting the approximations.
- An aggregate has mean 250 and variance 900. Give its normal moment approximation and standard deviation. (Answer: \(N(250,900)\), standard deviation 30.)
- An aggregate has \(m=80\), \(v=100\), and \(\kappa_3=500\). Find the translated gamma shape, scale, and shift. (Answer: \(\alpha=16\), \(\theta=2.5\), \(a=40\).)
- A shifted lognormal fit has \(m=60\), \(v=100\), and skewness 4. Find \(w\), the shift, and the log-variance. (Answer: \(w=1\), \(a=50\), \(\sigma^2=\ln2\).)
- Using the translated gamma fit in question 2, what is the probability assigned to an aggregate loss at most 39? (Answer: 0, because the fitted support is greater than 40.)
- A normal approximation is \(N(100,400)\), with loss amounts in thousands of dollars. Estimate the probability it assigns to a negative aggregate loss. Give four decimals. (Answer: \(\Phi(-5)\approx0.0000\).)