Weibull Hazard Shapes

Why Weibull Hazard Shape Matters

A lifetime model describes not only how long an item may last, but also how its failure risk changes as it ages. A device whose failures are concentrated early in service may have a decreasing hazard; a component subject to wear may have an increasing hazard. The Weibull model represents these different patterns through one parameter, the shape parameter.

This tutorial uses the shape-scale parameterization: $k$ is the shape parameter and $\lambda$ is the scale parameter. The model applies to a continuous, nonnegative lifetime $T$, with $k>0$ and $\lambda>0$. The parameter $\lambda$ is measured in the same time units as $T$; $k$ is dimensionless. The examples are hypothetical, and no interest or mortality assumption is involved. The focus is the hazard function and its interpretation, not the model’s moments or a general treatment of its probabilities.

Definition: For a continuous lifetime $T$ with density $f(t)$ and survival function $S(t)=P(T>t)$, the hazard rate at age $t$ is the limiting conditional failure rate per unit time: $$ h(t)=\lim_{\Delta t\to 0^+} \frac{P(t<T\le t+\Delta t\mid T>t)}{\Delta t} =\frac{f(t)}{S(t)}, $$ where $S(t)>0$. It has units of inverse time. A hazard rate is not itself a probability.

Deriving the Weibull Hazard

Under the shape-scale parameterization, the Weibull distribution has survival function $S(t)=\exp[-(t/\lambda)^k]$ for $t\ge 0$. Its cumulative distribution function is $F(t)=1-S(t)$. Differentiating $F(t)$ gives its density for $t>0$.

S(t)=\exp\left[-\left(\frac{t}{\lambda}\right)^k\right], \qquad f(t)=\frac{k}{\lambda}\left(\frac{t}{\lambda}\right)^{k-1} \exp\left[-\left(\frac{t}{\lambda}\right)^k\right].

Substitution in $h(t)=f(t)/S(t)$ cancels the exponential factor. The resulting expression is the key calculation: the scale affects the level of the hazard, while the power of $t$ determines how the hazard changes over time.

\begin{aligned} h(t) &=\frac{\frac{k}{\lambda}\left(\frac{t}{\lambda}\right)^{k-1} \exp[-(t/\lambda)^k]} {\exp[-(t/\lambda)^k]}\\[4pt] &=\frac{k}{\lambda}\left(\frac{t}{\lambda}\right)^{k-1} =\frac{k t^{k-1}}{\lambda^k}, \qquad t>0. \end{aligned}

The cancellation also has a direct conditional-probability explanation. For small $\Delta t>0$, the probability of failure between $t$ and $t+\Delta t$, given survival to $t$, is approximately $f(t)\Delta t/S(t)$. Dividing by $\Delta t$ and taking the limit gives $f(t)/S(t)$, not $f(t)$ alone.

Theorem: For a Weibull lifetime with shape $k>0$ and scale $\lambda>0$, the hazard is $h(t)=\frac{k}{\lambda}(t/\lambda)^{k-1}$ for $t>0$. Its units are inverse time. The value of $k$ determines the hazard’s direction of change; $\lambda$ changes its scale.

How the Shape Parameter Controls the Hazard

For fixed $k$ and $\lambda$, the hazard is a positive constant multiplied by $t^{k-1}$. Its derivative makes the direction of change explicit:

h'(t)=\frac{k(k-1)t^{k-2}}{\lambda^k},\qquad t>0.

Since $k$, $t$, and $\lambda$ are positive, the sign of $h'(t)$ is the sign of $k-1$. Therefore:

Shape parameter	Hazard pattern	Interpretation
$0<k<1$	Decreasing	Hazard is higher at earlier ages and falls as age increases.
$k=1$	Constant	$h(t)=1/\lambda$ for all $t>0$.
$k>1$	Increasing	Hazard rises as age increases.

The boundary behavior is consistent with these patterns. If $0<k<1$, the hazard tends to infinity as $t$ approaches zero from above and then decreases. If $k=1$, the hazard is constant. If $k>1$, it tends to zero as $t$ approaches zero from above and increases thereafter. These are properties of the model, not claims that a real item must have exactly that behavior at age zero.

The scale parameter is not a shape parameter. For example, multiplying $\lambda$ by a positive factor changes the hazard level through $1/\lambda$ and $\lambda^k$, but does not change whether it rises or falls. Changing time units requires changing $\lambda$ to the same units; $k$ remains unchanged.

Check the parameterization.
Confirm that the model uses $S(t)=\exp[-(t/\lambda)^k]$, with shape $k$ and scale $\lambda$.

Write the hazard formula.
Use $h(t)=\frac{k}{\lambda}(t/\lambda)^{k-1}$, with $t$ and $\lambda$ in the same time units.

Read the shape.
Compare $k$ with 1 to identify decreasing, constant, or increasing hazard.

Interpret the result.
Report a rate per unit time and explain what it says about an item that has survived to age $t$.

Worked Lifetime and Reliability Examples

Worked Example: Conditional Reliability and Age-Specific Hazard

A hypothetical instrument has a Weibull lifetime $T$ measured in years, with $k=1.5$ and $\lambda=8$ years. Find (a) the probability that it lasts at least 10 years given that it has lasted 6 years, and (b) its hazard at ages 6 and 10 years. There is no interest or mortality assumption. Use the Weibull survival function and hazard formula.

Method. For part (a), use conditional survival, $P(T>10\mid T>6)=S(10)/S(6)$. For part (b), substitute each age into the Weibull hazard. The conditional-survival calculation avoids treating the probability of lasting from age 6 to 10 as the unconditional probability of lasting 10 years from the start.

(a) Conditional survival. The model survival function is $S(t)=\exp[-(t/8)^{1.5}]$, so

\begin{aligned} P(T>10\mid T>6) &=\frac{S(10)}{S(6)}\\ &=\exp\left[-\left(\frac{10}{8}\right)^{1.5} +\left(\frac{6}{8}\right)^{1.5}\right]. \end{aligned}

The two powers are $(10/8)^{1.5}=1.397542486$ and $(6/8)^{1.5}=0.649519053$. The exponent is therefore $-1.397542486+0.649519053=-0.748023433$. Thus the probability is $\exp(-0.748023433)=0.473301$, or 0.4733 to four decimal places.

(b) Hazards. The hazard is $h(t)=\frac{1.5}{8}(t/8)^{0.5}$, measured per year. At age 6,

h(6)=0.1875\sqrt{0.75} =0.1875(0.8660254) =0.1623798\text{ per year}.

At age 10,

h(10)=0.1875\sqrt{1.25} =0.1875(1.1180340) =0.2096314\text{ per year}.

To four decimal places, the hazards are 0.1624 per year at age 6 and 0.2096 per year at age 10. The increase is consistent with $k=1.5>1$. The final probability is between 0 and 1, and the later-age hazard is greater than the earlier-age hazard, as the model predicts.

Worked Example: Decreasing Hazard for a Hypothetical Sensor

A hypothetical sensor’s lifetime is modeled by a Weibull distribution with $k=0.6$ and $\lambda=12$ months. Calculate its hazard at age 1 month and at age 12 months, and interpret the comparison. No interest or mortality assumption applies. Use months throughout.

Method. Substitute each age into $h(t)=\frac{k}{\lambda}(t/\lambda)^{k-1}$. This calculation avoids assuming that a hazard is constant just because the same model is used at both ages.

Here $k/\lambda=0.6/12=0.05$ per month and $k-1=-0.4$. At age 1 month,

\begin{aligned} h(1) &=0.05\left(\frac{1}{12}\right)^{-0.4}\\ &=0.05(12^{0.4})\\ &=0.05(2.701920) =0.135096\text{ per month}. \end{aligned}

At age 12 months,

h(12)=0.05\left(\frac{12}{12}\right)^{-0.4} =0.05(1)=0.050000\text{ per month}.

To four decimal places, the hazards are 0.1351 per month at age 1 and 0.0500 per month at age 12. The decrease agrees with $0<k<1$. These are conditional rates for sensors that have survived to the stated ages; they are not unconditional probabilities of failure.

Worked Example: Increasing Hazard for a Hypothetical Pump

A hypothetical pump has a Weibull lifetime measured in operating hours, with $k=2$ and $\lambda=1{,}000$ hours. Calculate its hazard at 500, 1,000, and 1,500 operating hours. Interpret the pattern. No interest or mortality assumption applies.

Method. Substitute each age into the hazard formula. Because $k=2$, the hazard simplifies to $h(t)=\frac{2}{1{,}000}(t/1{,}000)=2t/1{,}000{,}000$ per operating hour. This simplification makes the age dependence explicit and avoids confusing the hazard with a cumulative failure probability.

\begin{aligned} h(500)&=\frac{2(500)}{1{,}000{,}000} =0.001\text{ per operating hour},\\ h(1{,}000)&=\frac{2(1{,}000)}{1{,}000{,}000} =0.002\text{ per operating hour},\\ h(1{,}500)&=\frac{2(1{,}500)}{1{,}000{,}000} =0.003\text{ per operating hour}. \end{aligned}

The hazard increases from 0.001 to 0.002 to 0.003 per operating hour, as required by $k=2>1$. Each number describes the instantaneous failure rate for pumps that remain in service at that age. It does not mean that exactly 0.1%, 0.2%, or 0.3% of all pumps fail over an entire hour regardless of their age.

Common Mistakes and Interpretation

A hazard is a conditional rate among items that have survived to age $t$. It is useful in reliability analysis because it describes how failure propensity changes with age, but it should not be read as a probability without specifying a time interval and the relevant conditional model.

Using the density as the hazard. The hazard is $f(t)/S(t)$, not $f(t)$. In the pump model above, at age 500 hours the hazard is 0.001 per hour. The density there is $f(500)=h(500)S(500)=0.001\exp(-0.25)$, which is approximately 0.0007788 per hour. Calling that density the hazard would incorrectly omit the conditioning on survival to 500 hours.
Reversing the shape pattern. Since $h(t)$ is proportional to $t^{k-1}$, $k<1$ gives decreasing hazard and $k>1$ gives increasing hazard. The comparison is with 1, not with 0.
Mixing time units. If $t$ is measured in months, $\lambda$ must also be in months and the hazard is per month. A rate per month is not numerically a rate per year.
Using a different Weibull convention without converting. Some sources use a rate parameter rather than a scale parameter. The formula here assumes $S(t)=\exp[-(t/\lambda)^k]$; using a different parameterization without conversion can change the calculation.
Calling the hazard a probability. A value such as 0.2 per year is a rate. A probability over a stated interval depends on the model and the time interval; it is not obtained by treating the rate as a probability for any interval length.

Key takeaway: Under the Weibull shape-scale parameterization, $h(t)=\frac{k}{\lambda}(t/\lambda)^{k-1}$. The shape parameter determines the pattern: decreasing for $0<k<1$, constant for $k=1$, and increasing for $k>1$. Always retain the conditioning and the inverse-time units when interpreting a hazard. This treatment is aligned with the SOA VEE Mathematical Statistics syllabus.

Check Your Understanding

For each hypothetical lifetime model below, use the stated Weibull shape-scale parameterization. Give rates to four decimal places where needed; no interest or mortality assumption applies.

A component lifetime has $k=0.5$ and $\lambda=4$ days. Find the hazard at ages 1 day and 4 days, and state its pattern. (Answer: $h(1)=0.2500$ per day, $h(4)=0.1250$ per day; decreasing.)
A device has $k=1.8$ and $\lambda=6$ years. Calculate the hazard at age 3 years and at age 6 years, and state which is larger. (Answer: $h(3)=0.1723$ per year, $h(6)=0.3000$ per year; the age-6 hazard is larger.)
A test unit has $k=1.25$ and $\lambda=20$ operating hours. Calculate the hazards at 20 and 40 operating hours and identify the hazard pattern. (Answer: $h(20)=0.0625$ per operating hour, $h(40)=0.0743$ per operating hour; increasing.)
A hypothetical lifetime has $k=2$ and $\lambda=10$ years. Find the probability that an item survives from age 5 to age 10, conditional on having survived to age 5. (Answer: $\exp(-0.75)=0.4724$.)
A component has $k=3$ and $\lambda=2$ months. Find the hazards at ages 1 and 2 months, and compare them. (Answer: $h(1)=0.3750$ per month and $h(2)=1.5000$ per month; the age-2 hazard is four times the age-1 hazard.)

Shape parameter	Hazard pattern	Interpretation
\(0<k<1\)	Decreasing	Hazard is higher at earlier ages and falls as age increases.
\(k=1\)	Constant	\(h(t)=1/\lambda\) for all \(t>0\).
\(k>1\)	Increasing	Hazard rises as age increases.

What You'll Learn