Why Hazard Behavior Helps Identify a Severity Model
A severity model describes the distribution of a loss \(X\). Its hazard function gives the instantaneous rate of loss occurrence at size \(x\), conditional on the loss having exceeded \(x\). Hazard behavior can therefore provide a useful way to compare candidate severity families: a constant hazard suggests a different structure from a hazard that rises with size or falls toward zero.
The pattern is evidence, not a substitute for checking the full model. Several families can have hazards that increase or decrease, and a finite set of observed hazard values may not establish a family uniquely. Support and limiting behavior, together with the relevant parameter restrictions, provide important additional clues.
In this tutorial, the four families use these parameterizations: exponential rate \(\lambda>0\); gamma shape \(a>0\) and scale \(\theta>0\); Type I Pareto minimum \(x_m>0\) and shape \(\alpha>0\); and Weibull shape \(k>0\) and scale \(\eta>0\). This makes the parameter-to-hazard relationships explicit and avoids confusing a scale with a rate.
Hazard Signatures of the Four Families
| Family and support | Hazard | Hazard behavior and identifying features |
|---|---|---|
| Exponential, \(x\geq 0\) | \(h_X(x)=\lambda\) | Constant over the full support; rate \(\lambda\). |
| Gamma, \(x\geq 0\) | \(h_X(x)=f_X(x)/S_X(x)\) | Decreasing for \(0<a<1\), constant for \(a=1\), increasing for \(a>1\). For \(a>1\), it approaches \(1/\theta\) as \(x\) grows. |
| Type I Pareto, \(x\geq x_m\) | \(h_X(x)=\alpha/x\) | Decreases toward zero; the support begins at \(x_m\), and the hazard at that boundary is \(\alpha/x_m\). |
| Weibull, \(x\geq 0\) | \(h_X(x)=\dfrac{k}{\eta}(x/\eta)^{k-1}\) | Decreasing for \(0<k<1\), constant for \(k=1\), increasing for \(k>1\). |
The exponential distribution has density \(f_X(x)=\lambda e^{-\lambda x}\) and survival function \(S_X(x)=e^{-\lambda x}\). The gamma distribution has density \(f_X(x)=x^{a-1}e^{-x/\theta}/(\Gamma(a)\theta^a)\) for \(x>0\), with survival function given by the corresponding upper incomplete gamma ratio. A gamma model with \(a=1\) is exponential with rate \(1/\theta\).
For Type I Pareto, \(S_X(x)=(x_m/x)^\alpha\) on \(x\geq x_m\). For Weibull, \(S_X(x)=e^{-(x/\eta)^k}\). In particular, Weibull with \(k=1\) is exponential with rate \(1/\eta\). Thus an exact constant hazard over the full support is consistent with either an exponential label or a gamma or Weibull parameterization at its exponential special case: the distribution is the same.
Why the Constant-Hazard Result Holds
The relationship between hazard and survival explains why a constant hazard is especially informative when it is known over the entire support. For a continuous severity with \(S_X(0)=1\), the density satisfies \(f_X(x)=-S_X'(x)\). Therefore,
If \(h_X(x)=\lambda\) for every \(x\geq 0\), integrate from 0 to \(x\). Since \(\log S_X(0)=\log 1=0\), this gives
This is the exponential survival function with rate \(\lambda\). The conclusion depends on the hazard being exactly constant over the full support and on \(S_X(0)=1\). An approximately constant hazard, or one observed to be constant only over part of the support, does not establish an exponential model.
Determine where the severity can lie. A Type I Pareto model starts at \(x_m\), whereas the standard exponential, gamma, and Weibull models start at zero.
Decide whether it is exactly constant, increasing, decreasing, or only approximately described by one of those patterns.
For example, an increasing gamma hazard approaches a positive limit, while an increasing Weibull hazard with \(k>1\) grows without bound.
Distinguish an exact implication under a specified model from a plausible identification based on limited or qualitative evidence.
Worked Examples
Worked Example: An Exact Constant Hazard
A hypothetical portfolio’s positive severity \(X\), measured in thousands of dollars, is modeled on \(x\geq0\). Its hazard is specified to be exactly \(0.4\) per thousand dollars at every \(x\geq0\), and \(S_X(0)=1\). Identify the distribution and calculate \(P(X>5)\), its mean, and its variance. No interest or mortality assumption is relevant; this is a severity-only model.
Method. Apply the constant-hazard derivation over the full support. The hazard is \(h_X(x)=0.4\), so the rate is \(\lambda=0.4\) per thousand dollars, and \(X\) is exponential with rate \(0.4\).
The survival function is $$ S_X(x)=e^{-0.4x}. $$ Thus $$ P(X>5)=S_X(5)=e^{-0.4(5)}=e^{-2}=0.1353. $$ For an exponential distribution, \(E[X]=1/\lambda\) and \(\operatorname{Var}(X)=1/\lambda^2\). Therefore, $$ E[X]=\frac{1}{0.4}=2.5 \quad\text{and}\quad \operatorname{Var}(X)=\frac{1}{(0.4)^2}=6.25. $$ The mean is \(2.5\) thousand dollars, and the variance is \(6.25\) squared thousand dollars. The survival probability is to four decimals.
Check. At \(x=0\), \(S_X(0)=e^0=1\), and differentiating gives \(f_X(x)=0.4e^{-0.4x}\), so \(f_X(x)/S_X(x)=0.4\), as specified. A common mistake is to infer exponentiality from a hazard that only looks flat over a limited range; here the exact full-support condition is what justifies the conclusion.
Worked Example: An Increasing Weibull Hazard
A hypothetical equipment-loss severity has support \(x\geq0\), where \(x\) is measured in thousands of dollars. Its hazard is \(h_X(x)=x/2\) per thousand dollars. Identify the Weibull parameters and find \(P(X>2)\). No interest or mortality assumption applies.
Method. For a Weibull distribution with shape \(k\) and scale \(\eta\), $$ h_X(x)=\frac{k}{\eta}\left(\frac{x}{\eta}\right)^{k-1}. $$ The stated hazard is proportional to \(x\), so its power of \(x\) is 1. Hence \(k-1=1\), giving \(k=2\). The coefficient is \(k/\eta^k=2/\eta^2\), which must equal \(1/2\). Thus \(\eta^2=4\), and the positive scale is \(\eta=2\) thousand dollars.
The hazard increases with \(x\), as expected for \(k=2>1\). Using the Weibull survival function, $$ P(X>2)=e^{-(2/2)^2}=e^{-1}=0.3679. $$ The probability is rounded to four decimals.
Check. Substituting \(k=2\) and \(\eta=2\) into the hazard gives \(\frac{2}{2}(x/2)=x/2\), matching the stated function. The common mistake is to identify the scale from the coefficient as \(\eta=2\) without accounting for the power \(\eta^k\); matching the full coefficient confirms the parameter.
Worked Example: An Increasing Gamma Hazard That Levels Off
A hypothetical severity \(X\), in thousands of dollars, follows a gamma distribution with shape \(a=2\) and scale \(\theta=4\). Find its hazard at \(x=4\) and \(x=12\), and identify its limiting hazard. No interest or mortality assumption applies.
Method. For shape 2 and scale \(\theta\), the gamma density and survival function simplify to $$ f_X(x)=\frac{x}{\theta^2}e^{-x/\theta}, \qquad S_X(x)=e^{-x/\theta}\left(1+\frac{x}{\theta}\right). $$ Dividing gives $$ h_X(x)=\frac{x}{\theta(\theta+x)}. $$ With \(\theta=4\), this becomes \(h_X(x)=x/[4(4+x)]\).
At \(x=4\), $$ h_X(4)=\frac{4}{4(4+4)}=\frac{4}{32}=0.125. $$ At \(x=12\), $$ h_X(12)=\frac{12}{4(4+12)}=\frac{12}{64}=0.1875. $$ The hazard is increasing: differentiating \(x/[4(4+x)]\) gives \(1/(4+x)^2>0\). Its limit is $$ \lim_{x\to\infty}\frac{x}{4(4+x)}=\frac14=0.25. $$ All hazard values are per thousand dollars.
Check. The values rise from \(0.125\) to \(0.1875\) toward \(0.25\), consistent with the gamma shape \(a=2>1\). For comparison, a Weibull hazard with shape greater than 1 keeps increasing without bound. The common mistake is to treat every increasing hazard as Weibull; the finite positive limiting hazard here supports the gamma form.
Worked Example: A Decreasing Pareto Hazard
A hypothetical severity follows a Type I Pareto model with support \(x\geq100\) thousand dollars and hazard \(h_X(x)=3/x\). Identify its parameters and calculate \(P(X>300)\) and \(E[X]\). No interest or mortality assumption applies.
Method. The Type I Pareto hazard is \(\alpha/x\) on \(x\geq x_m\). Comparing the stated hazard with this form gives \(\alpha=3\); the support boundary gives \(x_m=100\) thousand dollars. The hazard decreases from \(3/100=0.03\) to \(3/300=0.01\) over the stated range.
For \(x\geq x_m\), the survival function is \(S_X(x)=(x_m/x)^\alpha\). Therefore, $$ P(X>300)=\left(\frac{100}{300}\right)^3 =\left(\frac13\right)^3 =\frac1{27}=0.0370. $$ For a Type I Pareto distribution, the mean is finite when \(\alpha>1\), and then \(E[X]=\alpha x_m/(\alpha-1)\). Here \(\alpha=3>1\), so $$ E[X]=\frac{3(100)}{3-1}=150 $$ thousand dollars. The probability is rounded to four decimals.
Check. The survival at the support boundary is \((100/100)^3=1\), and its hazard is \(\alpha/x=3/x\), as specified. A common mistake is to use \(x_m+x\) in the Pareto hazard denominator; for Type I Pareto the hazard is \(\alpha/x\), with \(x\geq x_m\).
Interpreting Similar Hazard Shapes
A decreasing hazard does not, by itself, identify one of these families. A gamma model with \(0<a<1\), a Weibull model with \(0<k<1\), and a Type I Pareto model all have decreasing hazards. Their endpoint behavior helps distinguish their standard forms:
- For gamma with \(0<a<1\), the hazard is unbounded near zero and approaches \(1/\theta\) as \(x\) grows.
- For Weibull with \(0<k<1\), the hazard is unbounded near zero and decreases toward zero.
- For Type I Pareto, the support begins at \(x_m>0\), and the hazard decreases toward zero as \(x\) tends to infinity.
Likewise, increasing hazard is shared by gamma with \(a>1\) and Weibull with \(k>1\). A gamma hazard approaches \(1/\theta\), whereas the Weibull hazard grows without bound. These limiting distinctions are theoretical properties of the specified families; a finite data range may not reveal them clearly. If only a qualitative observation is available, describe the compatible models rather than claiming a unique identification.
An exact constant hazard over the full support \(x\geq0\), with \(S_X(0)=1\), implies an exponential distribution. It is also represented by gamma shape \(a=1\) and scale \(\theta=1/\lambda\), or by Weibull shape \(k=1\) and scale \(\eta=1/\lambda\). These are different family parameterizations of the same distribution, not competing hazard shapes.
Common Exam Trap
Suppose a problem says that a fitted hazard is “approximately flat for observed severities between 10 and 40.” The incorrect calculation is to set \(h_X(x)=\lambda\) for all \(x\geq0\), integrate, and conclude \(S_X(x)=e^{-\lambda x}\). That calculation extends the given condition beyond the observed interval. The conclusion does not follow unless the hazard is specified to be exactly \(\lambda\) on the full support and the boundary condition \(S_X(0)=1\) holds.
The correction is to state that an approximately flat hazard over a restricted range is consistent with an exponential model, but does not prove it. Check the support, the hazard outside the observed range if known, and whether another model can produce similar behavior there. This distinction separates an exact mathematical implication from a model-selection judgment.
Check Your Understanding
For each problem, identify the family or explain what the stated hazard does and does not establish. Probabilities are rounded to four decimals where needed.
- A severity on \(x\geq0\) has exact hazard \(0.2\) per unit over its full support and \(S_X(0)=1\). Identify its distribution and rate, then calculate \(P(X>10)\). (Answer: Exponential with rate \(0.2\); \(P(X>10)=e^{-2}=0.1353\).)
- A Weibull severity has \(k=1.5\) and \(\eta=3\). Write its hazard and calculate \(P(X>6)\). (Answer: \(h_X(x)=0.5\sqrt{x/3}\); \(P(X>6)=e^{-2\sqrt{2}}\approx0.0591\).)
- A Type I Pareto severity has support \(x\geq250\) and hazard \(4/x\). Find its shape, minimum, survival probability above 1,000, and mean. (Answer: \(\alpha=4\), \(x_m=250\), \(P(X>1000)=1/256=0.0039\), and \(E[X]=333.33\).)
- A gamma severity has shape 2 and scale 5. Use \(h_X(x)=x/[\theta(\theta+x)]\) to find the hazard at \(x=5\), its limiting hazard, and \(P(X>5)\). (Answer: \(h_X(5)=0.1\), limiting hazard \(0.2\), and \(P(X>5)=2/e=0.7358\).)
- A severity is specified as Type I Pareto with support \(x\geq80\) and hazard \(5/x\). State its hazard at the support boundary and at \(x=400\), then say whether it decreases toward zero. (Answer: \(h_X(80)=0.0625\), \(h_X(400)=0.0125\), and yes, it decreases toward zero.)