Why Parameterization Matters
A severity family can be written using different parameter conventions. The same distribution may therefore appear with different-looking parameters in a textbook, software output, or problem statement. A parameter called “scale” in one model does not necessarily play the same mathematical role as a parameter called “scale” in another.
The safest approach is to identify the distribution from its defining density or survival function, then translate the stated parameters into that form. This is especially important for gamma rate versus scale, lognormal parameters on the log scale, Pareto minimum versus shape, and Weibull scale conventions. A parameterization error can produce plausible-looking but materially incorrect means, variances, or tail probabilities.
Gamma: Rate and Scale Are Reciprocals
Let \(X\) have a gamma distribution with shape \(\alpha>0\). Two common conventions specify either a rate \(\beta>0\) or a scale \(\theta>0\). In this tutorial, the rate form is defined by the density below; the scale form uses the same distribution with \(\theta=1/\beta\).
The rate has inverse units to \(X\), while the scale has the same units as \(X\). The mean and variance are \(E[X]=\alpha/\beta=\alpha\theta\) and \(\operatorname{Var}(X)=\alpha/\beta^2=\alpha\theta^2\). These formulas are paired: use the rate formulas with \(\beta\), or the scale formulas with \(\theta\).
In \(e^{-\beta x}\), \(\beta\) is the rate. In \(e^{-x/\theta}\), \(\theta\) is the scale.
Use \(\theta=1/\beta\) or \(\beta=1/\theta\), without changing the shape \(\alpha\).
For example, a rate of \(0.5\) gives mean \(\alpha/0.5\), not \(\alpha(0.5)\).
Worked Example: Gamma Rate Versus Scale
A hypothetical claim amount \(X\), measured in thousands of dollars, has a gamma distribution with shape \(\alpha=2\) and rate \(\beta=0.5\) per thousand dollars. Find its equivalent scale, mean, variance, and \(P(X\leq 2)\).
Method and parameter conversion. The rate is \(0.5\), so the equivalent scale is \(\theta=1/\beta=1/0.5=2\) thousand dollars. The model can be written using either \((\alpha,\beta)=(2,0.5)\) or \((\alpha,\theta)=(2,2)\).
Mean. Using the rate form, \(E[X]=\alpha/\beta=2/0.5=4\) thousand dollars.
Variance. \(\operatorname{Var}(X)=\alpha/\beta^2=2/(0.5)^2=2/0.25=8\) thousand-dollars squared. Equivalently, the scale form gives \(\alpha\theta^2=2(2^2)=8\).
Probability. A gamma distribution with shape 2 has \(F_X(x)=1-e^{-\beta x}(1+\beta x)\). At \(x=2\), \(\beta x=0.5(2)=1\), so \(P(X\leq2)=1-e^{-1}(1+1)=1-2/e=0.2642\), to four decimal places.
Check and trap avoided. The mean of 4 is twice the scale and equals \(\alpha/\beta\). Using the rate as though it were a scale would incorrectly give \(2(0.5)=1\) thousand dollars. The probability is between 0 and 1, as required.
Lognormal: Parameters Belong to the Logarithm
A positive random variable \(X\) is lognormal when \(\ln X\) is normal. The convention used here is \(\ln X\sim N(\mu,\sigma^2)\), where \(\mu\) is the mean of \(\ln X\) and \(\sigma\) is the standard deviation of \(\ln X\). Thus, \(\mu\) and \(\sigma\) are not generally the mean and standard deviation of \(X\).
A useful justification for the mean formula follows from the normal moment-generating function. Since \(X=e^{\ln X}\), and \(Y=\ln X\) is normal with mean \(\mu\) and variance \(\sigma^2\), the normal exponential moment is \(E[e^Y]=e^{\mu+\sigma^2/2}\). This also explains why the arithmetic mean exceeds the median whenever \(\sigma>0\).
Worked Example: Lognormal Parameters on the Log Scale
A hypothetical claim amount \(X\), in dollars, satisfies \(\ln X\sim N(\ln 100,(\ln 1.5)^2)\). Find its median, arithmetic mean, and probability of exceeding $150.
Identify the log-scale parameters. Here \(\mu=\ln100\) and \(\sigma=\ln1.5\approx0.4055\). These are parameters of \(\ln X\), not the mean and standard deviation of the dollar amount.
Median. The median is \(e^\mu=e^{\ln100}=\$100.00\).
Arithmetic mean. First, \((\ln1.5)^2/2\approx0.0822\). Therefore \(E[X]=e^{\ln100+(\ln1.5)^2/2}=100e^{0.0822}\approx\$108.56\), rounded to the nearest cent.
Exceedance probability. Standardizing the logarithm gives \[ P(X>150)=P\left(\frac{\ln X-\ln100}{\ln1.5}> \frac{\ln150-\ln100}{\ln1.5}\right). \] The standardized threshold is \(\ln(150/100)/\ln1.5=1\), so \(P(X>150)=P(Z>1)=0.1587\), to four decimal places.
Check and trap avoided. The median is $100.00, while the mean is larger, $108.56, consistent with the right-skewed lognormal distribution. Treating \(\mu=\ln100\) as the dollar mean would confuse a log-scale parameter with an original-scale moment.
Pareto: Minimum and Shape Have Different Roles
For a Pareto Type I severity, use \(\theta>0\) for the minimum and \(\alpha>0\) for the shape. The support begins at \(\theta\), and the survival function is a power of the ratio \(\theta/x\). Some sources use different letters, so the survival function is the most reliable way to confirm what a stated parameter means.
The minimum \(\theta\) sets the lower endpoint and has the same units as \(X\). The shape \(\alpha\) controls tail decay and is dimensionless. The mean exists only for \(\alpha>1\), in which case \(E[X]=\alpha\theta/(\alpha-1)\). The variance exists only for \(\alpha>2\), and is \(\alpha\theta^2/[(\alpha-1)^2(\alpha-2)]\). These moment formulas should not be used when their stated conditions fail.
Worked Example: Pareto Minimum Versus Shape
A hypothetical severity \(X\), in thousands of dollars, has a Pareto Type I distribution with minimum \(\theta=2\) and shape \(\alpha=3\). Find \(P(X>8)\), the mean, and the variance.
Tail probability. Since \(8\geq2\), use the Pareto survival function: \(P(X>8)=(2/8)^3=(1/4)^3=1/64=0.0156\), to four decimal places.
Mean. The shape is greater than 1, so the mean exists: \(E[X]=3(2)/(3-1)=6/2=3\) thousand dollars.
Variance. The shape is greater than 2, so the variance exists: \[ \operatorname{Var}(X)= \frac{3(2^2)}{(3-1)^2(3-2)} =\frac{12}{4(1)}=3 \] thousand-dollars squared.
Check and trap avoided. The support begins at 2, and the mean exceeds that minimum. The shape is the exponent in the tail probability; exchanging the two parameters would give the wrong tail and moment calculations.
Weibull: Scale Versus Power Coefficient
A common Weibull convention uses shape \(k>0\) and scale \(\lambda>0\), with survival function \(S_X(x)=\exp[-(x/\lambda)^k]\) for \(x\geq0\). The scale \(\lambda\) has the same units as \(X\). Another convention writes the survival function as \(\exp(-c x^k)\), where \(c>0\) is a power coefficient. Equating the exponents shows that \(c=\lambda^{-k}\), or \(\lambda=c^{-1/k}\).
The scale convention gives the mean \(E[X]=\lambda\Gamma(1+1/k)\) and the \(p\)-quantile \(x_p=\lambda[-\ln(1-p)]^{1/k}\). The power coefficient \(c\) is not the scale: it has units of \(X^{-k}\). Confusing these forms is particularly risky when the exponent \(k\) is not 1.
Worked Example: Converting a Weibull Power Coefficient
A hypothetical loss severity \(X\), measured in thousands of dollars, has survival function \(S_X(x)=e^{-0.01x^2}\) for \(x\geq0\). Identify its Weibull shape and scale, then calculate the median and mean.
Match the survival form. Comparing \(e^{-0.01x^2}\) with \(e^{-cx^k}\) gives \(k=2\) and \(c=0.01\). The scale is \(\lambda=c^{-1/k}=(0.01)^{-1/2}=10\) thousand dollars. Indeed, \((x/10)^2=0.01x^2\).
Median. Set \(S_X(x_{0.5})=0.5\): \(e^{-0.01x_{0.5}^2}=0.5\), so \(0.01x_{0.5}^2=\ln2\) and \(x_{0.5}=10\sqrt{\ln2}\approx8.33\) thousand dollars.
Mean. For shape 2, \(E[X]=10\Gamma(1+1/2)=10\Gamma(1.5)\). Using \(\Gamma(1.5)=\sqrt{\pi}/2\), the mean is \(10\sqrt{\pi}/2\approx8.86\) thousand dollars.
Check and trap avoided. Substituting the scale back gives \(e^{-(x/10)^2}=e^{-0.01x^2}\), as required. Treating \(c=0.01\) as the scale would invert the model’s scale and give an incorrect result.
Parameterization Checks for Exam Calculations
The same habit applies across all four families: verify the defining expression before applying a remembered formula. Units provide a fast independent check. A gamma rate has inverse-severity units, a gamma scale has severity units, a lognormal parameter is on the logarithmic scale, a Pareto minimum has severity units, and a Weibull power coefficient has units \(X^{-k}\).
| Family | Parameter to identify | Reliable check |
|---|---|---|
| Gamma | Rate \(\beta\) or scale \(\theta\) | Confirm \(\theta=1/\beta\) and use matching moments |
| Lognormal | Parameters of \(\ln X\) | Median \(e^\mu\); mean \(e^{\mu+\sigma^2/2}\) |
| Pareto Type I | Minimum \(\theta\) and shape \(\alpha\) | Support starts at \(\theta\); tail is \((\theta/x)^\alpha\) |
| Weibull | Scale \(\lambda\) or coefficient \(c\) | Match exponents using \(c=\lambda^{-k}\) |
A common exam trap is to identify a familiar family correctly but then use formulas from a different convention. For example, if the gamma density contains \(e^{-x/4}\), the scale is 4, not the rate. Reading that 4 as a rate would make the calculated mean too small by a factor of \(4^2\) relative to the correct mean for a fixed shape. Similarly, a lognormal \(N(\mu,\sigma^2)\) statement applies to \(\ln X\) when it defines a lognormal model; it does not make \(X\) normally distributed.
These checks matter because parameter names and symbols vary, while the underlying distribution does not. A reliable solution states the defining form, records the parameter roles, converts where needed, and then calculates. This is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.
Check Your Understanding
For each problem, identify the parameter convention before calculating.
- A gamma severity has shape 4 and rate 2 per unit. Find the equivalent scale, mean, and variance. (Answer: scale \(0.5\) units; mean \(2\) units; variance \(1\) unit squared.)
- A lognormal amount satisfies \(\ln X\sim N(\ln20,0.4^2)\). Find its median and mean, rounding money to the nearest cent. (Answer: median \(\$20.00\); mean \(\$21.67\).)
- A Pareto Type I severity has minimum 5 and shape 4. Find \(P(X>10)\) and \(E[X]\). (Answer: \(0.0625\); \(20/3\), or about \(6.67\) in the severity’s units.)
- A Weibull survival function is \(S_X(x)=e^{-0.008x^3}\). Find the shape and scale, then the median. (Answer: shape \(3\), scale \(5\) units, median \(5(\ln2)^{1/3}\approx4.42\) units.)
- A gamma model is given in scale form with shape 3 and scale 4. State the equivalent rate and calculate the mean and variance. (Answer: rate \(0.25\) per unit; mean \(12\) units; variance \(48\) units squared.)