Comparing Gamma And Weibull Hazard Patterns

Why Compare Hazard Patterns?

A severity density describes where loss values are concentrated; a hazard describes the rate of loss occurrence at a particular value, conditional on the loss having exceeded that value. The distinction matters when comparing distributions: a density can rise and then fall even when its hazard increases throughout. Gamma and Weibull models make this contrast especially clear.

This tutorial uses the standard shape-scale parameterizations of the two distributions. It compares their hazards, shows how the density and survival functions determine those hazards, and identifies the parameter regimes. These results are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: For a continuous, positive severity $X$ with density $f_X(x)$ and survival function $S_X(x)=P(X>x)$, the hazard at $x>0$ is $$ h_X(x)=\frac{f_X(x)}{S_X(x)}. $$ It is a rate per unit of loss, not a probability. For a small increment $\Delta x>0$, $h_X(x)\Delta x$ approximates the conditional probability that $X$ lies in $(x,x+\Delta x]$, given $X>x$.

Gamma and Weibull Parameterizations

For a gamma severity with shape $\alpha>0$ and scale $\theta>0$, the density and survival function are

f_X(x)=\frac{x^{\alpha-1}e^{-x/\theta}}{\Gamma(\alpha)\theta^\alpha}, \qquad S_X(x)=\int_x^\infty f_X(t)\,dt, \qquad x>0.

Here $\Gamma(\alpha)$ is the gamma function. The scale $\theta$ sets the loss-unit scale; the corresponding rate parameter, if used, is $1/\theta$. The hazard depends on both parameters, but its monotonicity is determined by $\alpha$.

For a Weibull severity with shape $k>0$ and scale $\lambda>0$,

S_X(x)=e^{-(x/\lambda)^k}, \qquad f_X(x)=\frac{k}{\lambda}\left(\frac{x}{\lambda}\right)^{k-1}e^{-(x/\lambda)^k}, \qquad x>0.

The parameterizations are not interchangeable: $\theta$ is the gamma scale and $\lambda$ is the Weibull scale. In each model, the scale has the same units as $X$, while the hazard has inverse-loss units.

Theorem: For a gamma distribution, the hazard decreases when $0<\alpha<1$, is constant when $\alpha=1$, and increases when $\alpha>1$. For a Weibull distribution, it decreases when $0<k<1$, is constant when $k=1$, and increases when $k>1$. Neither standard family has a nonmonotone hazard.

How the Hazard Patterns Follow from the Formulas

For Weibull, divide the density by the survival function. The exponential terms cancel, leaving

h_X(x)=\frac{k}{\lambda}\left(\frac{x}{\lambda}\right)^{k-1}.

The power of $x$ is $k-1$. Equivalently, differentiating gives

h_X'(x)=\frac{k(k-1)}{\lambda^2}\left(\frac{x}{\lambda}\right)^{k-2}.

All factors other than $k-1$ are positive for $x>0$. Thus the sign of the derivative is negative, zero, or positive according as $k<1$, $k=1$, or $k>1$. The scale changes the hazard’s level but not its direction of change.

For gamma, the result can be justified directly from the density-to-survival ratio. Since $S_X'(x)=-f_X(x)$, differentiating $h_X=f_X/S_X$ gives

h_X'(x) =h_X(x)\left(\frac{f_X'(x)}{f_X(x)}+h_X(x)\right), \qquad \frac{f_X'(x)}{f_X(x)} =\frac{\alpha-1}{x}-\frac{1}{\theta}.

The expression in parentheses determines the sign of $h_X'$. For $\alpha>1$, set $c=1/\theta-(\alpha-1)/x$. If $c\leq 0$, then $f_X'(x)/f_X(x)=-c\geq0$, so $h_X'(x)>0$. If $c>0$, then for every $t\geq x$,

\frac{f_X'(t)}{f_X(t)} =\frac{\alpha-1}{t}-\frac{1}{\theta} \leq -c.

Integrating $f_X'(t)\leq-cf_X(t)$ from $x$ to infinity yields $-f_X(x)\leq-cS_X(x)$, or $h_X(x)\geq c$. Therefore $f_X'(x)/f_X(x)+h_X(x)=-c+h_X(x)\geq0$, proving that the hazard increases.

For $0<\alpha<1$, set $C=1/\theta+(1-\alpha)/x$. For $t\geq x$, $1/\theta+(1-\alpha)/t\leq C$, so $f_X'(t)\geq-Cf_X(t)$. Integrating gives $-f_X(x)\geq-CS_X(x)$, hence $h_X(x)\leq C$. Since $f_X'(x)/f_X(x)=-C$, the derivative formula gives $h_X'(x)\leq0$. At $\alpha=1$, the gamma density is exponential with scale $\theta$, so its hazard is $1/\theta$, a constant.

Density Shape Is Not Hazard Shape

The parameter regimes also affect the density, but the density’s mode does not by itself classify the hazard. A gamma density has an interior mode at $x=\theta(\alpha-1)$ when $\alpha>1$; for $0<\alpha\leq1$, it decreases from the boundary at zero. A Weibull density has an interior mode at $x=\lambda((k-1)/k)^{1/k}$ when $k>1$; for $0<k\leq1$, it decreases from the boundary. Thus both families can have a density that first rises and then falls, while their hazards remain monotone.

The endpoint behavior reinforces the difference. A gamma hazard with $\alpha>1$ starts at zero and approaches $1/\theta$; with $0<\alpha<1$, it starts unbounded and decreases toward $1/\theta$. A Weibull hazard with $k>1$ starts at zero and increases without bound, while with $0<k<1$ it starts unbounded and decreases toward zero. These descriptions concern the hazard, not the density.

Worked Example: Correcting a Gamma Hazard Calculation

A hypothetical gamma severity has shape $\alpha=3$ and scale $\theta=2$. Its survival function is $S_X(x)=e^{-x/2}(1+x/2+x^2/8)$. Calculate its hazards at $x=2$ and $x=6$, then classify the pattern.

Method. Use $h_X(x)=f_X(x)/S_X(x)$. The gamma density is $f_X(x)=x^2e^{-x/2}/16$, since $\Gamma(3)=2$ and $\theta^3=8$, giving denominator $2\cdot8=16$.

At $x=2$, the density is $4e^{-1}/16=e^{-1}/4$, and the survival is $e^{-1}(1+1+1/2)=2.5e^{-1}$. Therefore

h_X(2)=\frac{e^{-1}/4}{2.5e^{-1}} =\frac{0.25}{2.5} =0.1000.

At $x=6$, the density is $36e^{-3}/16=2.25e^{-3}$, and the survival is $e^{-3}(1+3+36/8)=8.5e^{-3}$. Therefore

h_X(6)=\frac{2.25e^{-3}}{8.5e^{-3}} =\frac{2.25}{8.5} =\frac{9}{34} \approx0.2647.

The hazards are in inverse loss units. Since $\alpha=3>1$, the theorem also confirms that the gamma hazard is increasing; the computed values, $0.1000<0.2647$, are consistent. The common mistake is to use the survival polynomial but miscalculate the density coefficient, which here would understate each hazard by a factor of four.

Worked Example: Comparing Weibull Shapes

Two hypothetical Weibull severities have the same scale $\lambda=4$. Model A has shape $k=0.5$, and Model B has shape $k=2$. For each model, calculate the hazards at $x=1$ and $x=4$, and identify which hazard is increasing.

Method. From the density and survival ratio, $h_X(x)=(k/4)(x/4)^{k-1}$. For Model A, this becomes $0.125(x/4)^{-0.5}$. Thus

h_A(1)=0.125(0.25)^{-0.5}=0.125(2)=0.2500, \qquad h_A(4)=0.125(1)^{-0.5}=0.1250.

Model A’s hazard decreases, as expected because $k=0.5<1$. For Model B, the hazard is $0.5(x/4)$, so

h_B(1)=0.5(0.25)=0.1250, \qquad h_B(4)=0.5(1)=0.5000.

Model B’s hazard increases, as expected because $k=2>1$. At $x=1$, Model A has the larger hazard; at $x=4$, Model B has the larger hazard. These pointwise comparisons do not change the monotonicity classification. The trap is to infer the direction of a hazard from the scale alone: both models have the same scale, but their shapes give opposite patterns.

Worked Example: A Hump-Shaped Density with an Increasing Hazard

Compare a hypothetical gamma severity with $\alpha=2,\theta=3$ and a hypothetical Weibull severity with $k=2,\lambda=3$. Calculate both hazards at $x=1$ and $x=3$, and compare their densities and survivals at $x=3$.

Gamma hazards. Its density is $f_X(x)=xe^{-x/3}/9$, and its survival is $S_X(x)=e^{-x/3}(1+x/3)$. Dividing gives $h_X(x)=x/[3(x+3)]$. Hence

h_X(1)=\frac{1}{12}\approx0.08333, \qquad h_X(3)=\frac{3}{18}=\frac{1}{6}\approx0.1667.

Weibull hazards. Here $h_X(x)=(2/3)(x/3)$, so

h_X(1)=\frac{2}{9}\approx0.2222, \qquad h_X(3)=\frac{2}{3}\approx0.6667.

Both hazards increase, though their levels differ. At $x=3$, the gamma survival is $2e^{-1}\approx0.7358$, and its density is $e^{-1}/3\approx0.1226$; their ratio is $1/6$. The Weibull survival is $e^{-1}\approx0.3679$, and its density is $2e^{-1}/3\approx0.2453$; their ratio is $2/3$. The gamma density has its mode at $3$, while the Weibull density has its mode at $3/\sqrt{2}\approx2.121$. The density shapes and levels differ, but neither has a nonmonotone hazard. The check is that each reported hazard equals its density divided by its survival.

Comparison and Exam Traps

Family	Decreasing hazard	Constant hazard	Increasing hazard
Gamma, shape $\alpha$	$0<\alpha<1$	$\alpha=1$, hazard $1/\theta$	$\alpha>1$
Weibull, shape $k$	$0<k<1$	$k=1$, hazard $1/\lambda$	$k>1$

Several checks help prevent errors in a comparison:

Use the survival in the denominator. The hazard is $f_X/S_X$, not $f_X$, and not $f_X/F_X$.
Keep scale and rate distinct. A gamma scale $\theta$ corresponds to exponential hazard $1/\theta$, not $\theta$. A Weibull scale $\lambda$ gives exponential hazard $1/\lambda$ when $k=1$.
Do not mistake a density mode for a hazard turning point. A unimodal density does not imply a nonmonotone hazard.
Do not claim a nonmonotone hazard for either standard family. Each has decreasing, constant, or increasing hazard according to its shape parameter; none has a change from decreasing to increasing or vice versa.

These classifications are useful for comparing conditional loss behavior: an increasing hazard means the local rate of moving beyond the current loss level rises with that level; a decreasing hazard means it falls. The classification alone does not establish which distribution has larger probabilities at every threshold. That requires comparing the survival functions at the thresholds of interest.

Key takeaway: Compute a hazard as density divided by survival, then classify it using the shape parameter: $\alpha$ for gamma and $k$ for Weibull. A density may be hump-shaped even though the hazard is monotone, and neither standard family has a nonmonotone hazard.

Check Your Understanding

For each problem, classify the hazard and calculate the requested values. Hazards are reported per loss unit.

A hypothetical gamma severity has shape $\alpha=1$ and scale $\theta=5$. Find its hazards at $x=2$ and $x=8$, and state whether they increase, decrease, or remain constant. (Answer: $0.2000$ at both points; constant.)
A hypothetical Weibull severity has shape $k=3$ and scale $\lambda=2$. Calculate its hazards at $x=1$ and $x=2$, and classify the pattern. (Answer: $0.3750$ and $1.500$; increasing.)
A hypothetical gamma severity has shape $\alpha=2$ and scale $\theta=4$. Its survival is $e^{-x/4}(1+x/4)$. Find the hazards at $x=4$ and $x=12$. (Answer: $0.1250$ and $0.1875$; increasing.)
A hypothetical Weibull severity has shape $k=0.75$ and scale $\lambda=9$. Calculate the hazards at $x=1$ and $x=9$, and classify the pattern. (Answer: $\sqrt{3}/12\approx0.1443$ and $1/12\approx0.08333$; decreasing.)
A hypothetical gamma severity has shape $\alpha=3$ and scale $\theta=2$, with survival $e^{-x/2}(1+x/2+x^2/8)$. Calculate the hazards at $x=2$ and $x=6$, showing the density-to-survival ratio. (Answer: $0.1000$ and $9/34\approx0.2647$; increasing.)

Family	Decreasing hazard	Constant hazard	Increasing hazard
Gamma, shape \(\alpha\)	\(0<\alpha<1\)	\(\alpha=1\), hazard \(1/\theta\)	\(\alpha>1\)
Weibull, shape \(k\)	\(0<k<1\)	\(k=1\), hazard \(1/\lambda\)	\(k>1\)

What You'll Learn