Hazard Rate Ratio⚠ unverified
General Calculations / Reliability · Compute the ratio of Weibull hazard rates at two times
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| t1 | t1 | — | 1.0 | Reference time (denominator of the ratio), in time units |
| t2 | t2 | — | 1.0 | Comparison time (numerator of the ratio), in time units |
| shape | shape | — | 1.0 | Weibull shape parameter (beta), dimensionless |
| scale | scale | — | 1.0 | Weibull scale parameter (eta), in the same time units as ``t1`` and ``t2`` |
| Output | Symbol | Unit | Description |
|---|---|---|---|
| result | ratio | — | Ratio of the hazard rate at ``t2`` to the hazard rate at ``t1``, dimensionless. Returns positive infinity when the hazard rate at ``t1`` is zero |
The science & history
Understanding the Parameters
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The two times $t_1, t_2$ — the ages being compared; only their ratio $t_2/t_1$ matters. Doubling both times leaves the hazard ratio unchanged (for $t_2/t_1$ fixed) — a scale-free comparison.
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Shape parameter $\beta$ — the whole story of the ratio's behaviour:
- $\beta = 1$: ratio $= 1$ for any times — constant hazard, no aging (the exponential/memoryless case).
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$\beta > 1$: ratio $> 1$ for $t_2 > t_1$ — the item is aging; failure is more likely later. A wear-out item with $\beta = 3$ is $2^{2} = 4\times$ more hazardous at twice the age.
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$\beta < 1$: ratio $< 1$ for $t_2 > t_1$ — the item is improving (infant mortality burning off); failure is less likely later.
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Scale $\eta$ cancels — a key feature. Because both hazards carry the same $\beta/\eta$ and $\eta$-scaling, the ratio depends only on $\beta$ and $t_2/t_1$. So you can characterise aging without knowing the absolute timescale — useful when $\eta$ is uncertain but the shape is known. (The registry keeps $\eta$ as an input though it does not affect the result.)
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The output ratio — a dimensionless multiplier on risk. It quantifies aging directly, complementing the absolute hazard rate.
Derivation (Approaching a Proof)
Take the Weibull hazard rate at each time, $h(t) = \frac{\beta}{\eta}(t/\eta)^{\beta-1}$ (Weibull Failure Rate), and form the ratio. Every factor except the time-power is common to numerator and denominator:
$$\frac{h(t_2)}{h(t_1)} = \frac{\frac{\beta}{\eta}(t_2/\eta)^{\beta-1}}{\frac{\beta}{\eta}(t_1/\eta)^{\beta-1}} = \frac{(t_2/\eta)^{\beta-1}}{(t_1/\eta)^{\beta-1}} = \left(\frac{t_2}{t_1}\right)^{\beta-1}. \qquad\blacksquare$$
The prefactor $\beta/\eta$ and the $\eta$ inside each power cancel, leaving a pure function of $t_2/t_1$ and $\beta - 1$. This is the mathematical statement that the Weibull hazard is a power law: its ratio over any two times is set by the exponent alone, the signature of scale-invariance in the failure process.
Dimensional check. $t_2/t_1$ is dimensionless, so its power $(\cdot)^{\beta-1}$ is dimensionless — a pure ratio. $\checkmark$
History and Development
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Aging in one number. The hazard rate ratio distils the aging behaviour of a Weibull item into a single scale-free quantity, making it easy to communicate "twice as old, four times as likely to fail" to non-specialists.
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Proportional hazards. The idea that hazards can be compared as ratios is the foundation of Cox proportional-hazards models in survival analysis and reliability, where covariates (stress, temperature, design) multiply a baseline hazard — the ratio being the effect size.
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Maintenance timing. Comparing hazards across ages informs age-based replacement: when the hazard ratio relative to a new part climbs past a threshold, preventive replacement becomes worthwhile — a direct application of the $\beta > 1$ increasing-hazard regime.
Related Concepts: Weibull Failure Rate, Weibull Reliability, Weibull Distribution PDF, Bathtub Curve, Exponential Reliability, B10 Life
Notes: Registry calculator hazard-rate-ratio (unverified). $h(t_2)/h(t_1) = (t_2/t_1)^{\beta-1}$ — correct
as shipped; independent of $\eta$ (which the registry keeps as an unused input). $\beta=1$ ⇒ ratio $1$ (no
aging); $\beta>1$ ⇒ aging; $\beta<1$ ⇒ improving. Scale-free measure of aging. Returns $+\infty$ if $h(t_1)=0$.