Weibull Hazard Shapes And Aging

Why Weibull Hazard Shape Matters

A severity model describes the size of a loss. For a continuous loss $X$, its hazard at size $x$ measures the density near $x$ relative to the probability that the loss is at least $x$. In actuarial applications, this is a way to describe how the distribution of losses behaves as the threshold gets larger. It is not, by itself, a claim that a loss grows over time.

For the Weibull severity model, one parameter controls a particularly useful feature: whether the hazard decreases, remains constant, or increases as loss size grows. This behavior can help distinguish possible models and provides a concise check on fitted parameters. The conclusions below assume an unshifted, continuous Weibull distribution on positive loss sizes.

Definition: Let $X$ have a Weibull distribution with shape parameter $\alpha>0$ and scale parameter $\theta>0$, where $x$ and $\theta$ are expressed in the same loss units. For $x\geq 0$, $$ S_X(x)=\exp\left[-\left(\frac{x}{\theta}\right)^\alpha\right], \qquad F_X(x)=1-S_X(x). $$ For $x>0$, the hazard is $h_X(x)=f_X(x)/S_X(x)$. It describes the density at size $x$ conditional on survival to $x$, in the sense that $P(x\leq X<x+\Delta x\mid X\geq x)$ is approximately $h_X(x)\Delta x$ for small positive $\Delta x$.

The scale $\theta$ has units of loss, while $\alpha$ is dimensionless. The hazard has units of inverse loss. For example, if losses are measured in dollars, $h_X(x)$ is measured per dollar. The formula requires a continuous, unshifted Weibull model; for a discrete or mixed severity, the density-over-survival hazard does not directly describe point masses. A shifted Weibull model also requires using the distance from its shift in place of $x$.

Deriving the Weibull Hazard

The hazard follows by differentiating the survival function to obtain the density, then dividing by survival. This derivation also shows exactly where the shape parameter enters.

Differentiate the survival function.
For $x>0$, $f_X(x)=-\frac{d}{dx}S_X(x)$. Applying the chain rule gives the Weibull density.

Divide density by survival.
The exponential factor appears in both $f_X(x)$ and $S_X(x)$, so it cancels in their ratio.

Differentiate the hazard.
The sign of $h_X'(x)$ determines whether the hazard rises or falls for positive $x$.

\begin{aligned} f_X(x) &=-\frac{d}{dx}\exp\left[-\left(\frac{x}{\theta}\right)^\alpha\right]\\ &=\frac{\alpha x^{\alpha-1}}{\theta^\alpha} \exp\left[-\left(\frac{x}{\theta}\right)^\alpha\right],\\[4pt] h_X(x) &=\frac{f_X(x)}{S_X(x)} =\frac{\alpha x^{\alpha-1}}{\theta^\alpha},\\[4pt] h_X'(x) &=\frac{\alpha(\alpha-1)}{\theta^\alpha}x^{\alpha-2}, \qquad x>0. \end{aligned}

Because $\theta^\alpha>0$, $x^{\alpha-2}>0$ for $x>0$, and $\alpha>0$, the sign of the derivative is the sign of $\alpha-1$. This gives the central classification.

Theorem: For an unshifted Weibull severity with $\alpha>0$, $h_X(x)$ decreases for $0<\alpha<1$, is constant for $\alpha=1$, and increases for $\alpha>1$, for positive $x$. The scale parameter changes the hazard level but not whether it rises or falls.

At the boundary $x=0$, the hazard has different limiting behavior depending on the shape: it tends to infinity when $0<\alpha<1$, equals $1/\theta$ when $\alpha=1$, and tends to zero when $\alpha>1$. The derivative test above is stated for $x>0$; it does not require the hazard to be finite at zero.

Shape parameter	Hazard behavior for positive loss sizes	Interpretation as the threshold grows
$0<\alpha<1$	Decreasing	The hazard at higher thresholds is lower
$\alpha=1$	Constant	The hazard does not change with the threshold
$\alpha>1$	Increasing	The hazard at higher thresholds is higher

Worked Applications

A hazard ratio is often easier to interpret than two separate hazard values. Dividing the hazards at two thresholds cancels the scale parameter, leaving a relation that depends only on the shape.

\frac{h_X(x_2)}{h_X(x_1)} =\left(\frac{x_2}{x_1}\right)^{\alpha-1}, \qquad x_1,x_2>0.

Worked Example: Checking a Decreasing Weibull Hazard

A hypothetical portfolio models individual loss size $X$ with a Weibull distribution having $\alpha=0.5$ and $\theta=4$ dollars. Find the hazard at $x=1$ and $x=4$ dollars, calculate the derivative at $x=1$, and determine the hazard pattern. Use the Weibull hazard formula and derivative test.

Step 1: Substitute the parameters into the hazard. $$ h_X(x)=\frac{0.5x^{-0.5}}{4^{0.5}} =\frac{0.5}{2}x^{-0.5} =0.25x^{-0.5}. $$

Step 2: Evaluate at both loss sizes. $$ h_X(1)=0.25(1)^{-0.5}=0.25 \quad\text{per dollar}, $$ $$ h_X(4)=0.25(4)^{-0.5} =\frac{0.25}{2}=0.125 \quad\text{per dollar}. $$ The ratio is $0.125/0.25=0.5$, so the hazard at four dollars is half the hazard at one dollar.

Step 3: Differentiate and check the sign. $$ h_X'(x)=0.25(-0.5)x^{-1.5}=-0.125x^{-1.5}, \qquad h_X'(1)=-0.125. $$ The derivative is negative for all $x>0$, consistent with $\alpha=0.5<1$ and with the calculated hazard values.

Check and common mistake avoided. The ratio from the formula is $(4/1)^{0.5-1}=4^{-0.5}=0.5$, matching the direct calculation. When differentiating $0.25x^{-0.5}$, the coefficient is $0.25(-0.5)=-0.125$; dropping or changing the coefficient would give an incorrect derivative.

Worked Example: Testing Whether a Hazard Ratio Is Possible

A hypothetical severity analysis proposes an unshifted Weibull model and reports $h_X(12)/h_X(3)=1/4$. Determine the implied shape parameter and decide whether any valid Weibull model can have this ratio. Use the hazard-ratio equation and the requirement $\alpha>0$.

Step 1: Set up the ratio. The scale cancels, and $12/3=4$, so $$ \frac{h_X(12)}{h_X(3)} =\left(\frac{12}{3}\right)^{\alpha-1} =4^{\alpha-1}. $$

Step 2: Solve the stated equation. $$ 4^{\alpha-1}=\frac14=4^{-1} \quad\Longrightarrow\quad \alpha-1=-1 \quad\Longrightarrow\quad \alpha=0. $$

Step 3: Check admissibility. A Weibull shape parameter must satisfy $\alpha>0$. The algebraic solution $\alpha=0$ is therefore not a valid Weibull shape. No unshifted Weibull model with a positive shape parameter attains the reported ratio.

Check and common mistake avoided. For positive $\alpha$, $4^{\alpha-1}>4^{-1}=1/4$; the value $1/4$ is approached only as $\alpha$ tends to zero from above. Reporting $\alpha=0.5$ would be incorrect: it gives a ratio $4^{-0.5}=1/2$, not $1/4$. An algebraic solution must always be checked against the model's parameter restrictions.

Worked Example: Inferring an Increasing Hazard and Finding Its Level

A hypothetical claim-size model is Weibull. Its hazard at $18$ dollars is nine times its hazard at $2$ dollars, and $h_X(2)=1/36$ per dollar. Find the shape and scale parameters, then calculate the hazard at $10$ dollars. Use the ratio first, then the absolute hazard.

Step 1: Infer the shape from the ratio. $$ 9=\frac{h_X(18)}{h_X(2)} =\left(\frac{18}{2}\right)^{\alpha-1} =9^{\alpha-1}. $$ Since $9=9^1$, $\alpha-1=1$, so $\alpha=2$. This is an increasing-hazard Weibull model.

Step 2: Use the hazard at $2$ dollars to find $\theta$. $$ \frac{1}{36} =h_X(2) =\frac{2(2)^{2-1}}{\theta^2} =\frac{4}{\theta^2}. $$ Thus $\theta^2=144$, and because the scale is positive, $\theta=12$ dollars.

Step 3: Calculate the hazard at $10$ dollars. $$ h_X(10)=\frac{2(10)^{1}}{12^2} =\frac{20}{144} =\frac{5}{36} \approx 0.1389 \quad\text{per dollar}. $$ As a check, $h_X(18)=36/144=1/4$, while $h_X(2)=4/144=1/36$; their ratio is $(1/4)/(1/36)=9$, as specified.

Common mistake avoided. The ratio determines $\alpha$, not $\theta$. The scale must be found from an absolute hazard value after the shape is known. Also, the positive square root is required because a Weibull scale parameter is positive.

Interpreting Hazard Shape Correctly

A decreasing Weibull hazard means that, as the threshold $x$ rises, the density at that threshold relative to the remaining survival probability falls. An increasing hazard means this relative density rises. Neither statement says that individual losses increase or decrease over time, and neither alone determines the expected loss or the probability of exceeding a particular threshold.

The hazard is also not itself a probability: it has units of inverse loss and may exceed one. For a small increment $\Delta x$, $h_X(x)\Delta x$ is an approximation to the conditional probability of a loss in the narrow interval from $x$ to $x+\Delta x$, given $X\geq x$. A finite-interval probability requires more than simply reading off the hazard at one point.

The ratio equation is useful when two hazard levels are compared because the scale cancels. It does not make every proposed ratio feasible. In particular, when $x_2>x_1$, the ratio is greater than one if $\alpha>1$, equal to one if $\alpha=1$, and less than one if $0<\alpha<1$. A proposed ratio must agree with these restrictions and lead to a strictly positive shape.

Common Exam Traps

Confusing the scale with the shape. The positive scale $\theta$ affects the hazard's magnitude, while $\alpha$ determines its direction of change.
Ignoring parameter admissibility. Solving a hazard-ratio equation may produce a value outside the allowed range. A Weibull model requires $\alpha>0$, not merely a real algebraic solution.
Reversing the hazard ratio. If the question asks for $h_X(x_2)/h_X(x_1)$, the size ratio must be $x_2/x_1$ in the same order.
Using the derivative at zero. The derivative expression is for $x>0$. For $0<\alpha<1$, the hazard diverges as $x$ approaches zero, so evaluating the formula at zero is not valid.
Calling severity threshold movement aging in time. Here the argument is loss size. A decreasing hazard with loss size is not automatically a statement about a person's or policy's risk over calendar time.

Key takeaway: For an unshifted Weibull severity, $h_X(x)=\alpha x^{\alpha-1}/\theta^\alpha$, and the sign of $h_X'(x)$ is the sign of $\alpha-1$. Use hazard ratios to isolate the shape, then verify that the result satisfies $\alpha>0$.

Check Your Understanding

For each problem, use the Weibull hazard formula and check any inferred shape against its parameter restriction.

A hypothetical severity model has $\alpha=1.4$ and $\theta=5$ dollars. State whether its hazard increases, decreases, or stays constant as the loss threshold grows, and calculate $h_X(5)$. (Answer: Increasing; $h_X(5)=1.4/5=0.2800$ per dollar.)
A Weibull model satisfies $h_X(20)/h_X(5)=2$. Find its shape parameter and state the hazard pattern. (Answer: $\alpha=1.5$; increasing.)
A model has $\alpha=0.75$ and $\theta=2$ dollars. Find $h_X(16)/h_X(1)$ and classify the hazard. (Answer: $16^{-0.25}=0.5000$; decreasing.)
A hypothetical fit proposes $h_X(10)/h_X(2)=1/5$. Solve for the implied shape and decide whether it is admissible. (Answer: $\alpha=0$; inadmissible, so no positive-shape Weibull model attains this ratio.)
A Weibull severity has $\alpha=2$ and $\theta=5$ dollars. Calculate $h_X(2)$ and $h_X'(2)$, including their units, and state whether the hazard rises or falls. (Answer: $h_X(2)=0.1600$ per dollar; $h_X'(2)=0.0800$ per dollar squared; increasing.)

Shape parameter	Hazard behavior for positive loss sizes	Interpretation as the threshold grows
\(0<\alpha<1\)	Decreasing	The hazard at higher thresholds is lower
\(\alpha=1\)	Constant	The hazard does not change with the threshold
\(\alpha>1\)	Increasing	The hazard at higher thresholds is higher

What You'll Learn