Why the Number Changes When the Clock Changes
A hazard is a rate defined relative to time. Its numerical value therefore depends on the unit used to measure time, even when the underlying event process has not changed. A hazard of \(0.002\) per day and a hazard of \(0.73\) per year can describe the same rate if one year is taken to contain 365 days. The values differ because “per day” and “per year” use different time units, not because the event process became faster.
This distinction matters when reading a model output, comparing reports, or calculating what a constant rate implies over an interval. Before interpreting a hazard, identify its time unit. Then express the rate and the elapsed interval in compatible units. The same unit-conversion rule applies to a cause-specific hazard and a subdistribution hazard: their risk-set definitions differ, but both are rates with respect to the chosen time coordinate.
For a cause-specific hazard, the relevant risk set consists of people who have not yet experienced any event that removes them from the event-free state. For a subdistribution hazard, the risk set is defined differently. Those distinctions affect the meaning of the rate, but not the arithmetic of changing the time unit. A hazard is not itself a probability: it describes a local rate, and an interval probability requires additional information and assumptions.
The Unit-Conversion Rule
Suppose \(t\) measures time in days and \(u\) measures the same elapsed time in years, using 365 days per year. Then \(u=t/365\), or equivalently \(t=365u\). If \(\lambda_{\text{day}}\) is a hazard per day, the hazard per year is
The conversion factor goes in this direction because one year contains 365 daily intervals. Conversely, \(\lambda_{\text{day}}=\lambda_{\text{year}}/365\). More generally, if one unit of the new time scale contains \(c\) units of the old scale, the numerical hazard per new unit equals \(c\) times the hazard per old unit.
One way to verify the rule is to use the definition of a hazard through a small time increment. If the hazard is approximately \(\lambda_{\text{day}}\) per day, the conditional event probability over a short increment \(dt\) is approximately \(\lambda_{\text{day}}dt\). Since \(dt=365\,du\), the same increment is approximately \(\lambda_{\text{day}}(365\,du)\). Therefore its rate per year is \(365\lambda_{\text{day}}\). The event process and the small-interval probability have not changed; only the numerical rate and the unit attached to it have changed.
A second check uses cumulative hazard. For a hazard function \(\lambda(t)\), cumulative hazard over an interval is the integral of the hazard with respect to time. It is dimensionless: the inverse-time unit in the hazard cancels the time unit in the integration. Under a time rescaling, the hazard gains the conversion factor while the time increment loses it, leaving the cumulative hazard unchanged.
This equality is a change of variable, with \(s=365v\) and \(ds=365\,dv\). It shows why a properly converted hazard gives the same cumulative hazard over the same physical duration. For a constant hazard \(\lambda\), the cumulative hazard over a duration \(\Delta\) is simply \(\lambda\Delta\), provided the rate and duration use matching units.
Worked Example: Converting a Daily Hazard to a Yearly Hazard
Suppose a hypothetical constant hazard for a specified endpoint is \(0.002\) per day. Convert it to a yearly rate using 365 days per year, and calculate the event probability over 30 days under a single-event constant-hazard model with no competing events.
Convert the rate. One year contains 365 days, so
Calculate the interval probability under the stated model. With a constant hazard and no competing events, the probability of the event by duration \(\Delta\) is \(1-\exp(-\lambda\Delta)\). Using days gives
The model-implied probability over 30 days is about 5.82%. The same calculation in years uses \(30/365\) years:
The products are equal because the rate and duration were converted together. The probability calculation relies on the single-event constant-hazard model; the hazard conversion itself does not require that model.
Rate Is Not Interval Probability
A hazard value cannot be read directly as the probability of an event over a specified interval. For a single-event process with a constant hazard \(\lambda\), the event probability over duration \(\Delta\) is \(1-\exp(-\lambda\Delta)\), not \(\lambda\) alone. When \(\lambda\Delta\) is small, \(1-\exp(-\lambda\Delta)\) is approximately \(\lambda\Delta\), but this is a short-interval approximation, not an identity.
For a cause-specific hazard in a competing-risks setting, the focal cause-specific hazard describes the instantaneous rate of that cause among those still event-free. The focal event’s probability over a longer interval also depends on the possibility of other causes removing people from that risk set. Thus, do not use \(1-\exp(-\lambda_k\Delta)\) as the focal cause’s interval probability merely because its cause-specific hazard is constant. The single-event formula in the example applies only under its stated no-competing-events assumption.
The unit-conversion result remains valid in either setting. If time is rescaled, the rate for the same hazard definition receives the corresponding inverse-time conversion. This does not change the risk set, turn one hazard definition into another, or supply missing information about other event causes.
Worked Example: Converting a Weekly Hazard to a Daily Hazard
In a hypothetical model, a constant hazard is \(0.021\) per week. Take one week to be exactly seven days. Convert the rate to a daily unit and calculate the model-implied event probability over 14 days under a single-event constant-hazard model with no competing events.
Convert the rate. There are seven days in one week, so the daily rate is the weekly rate divided by seven:
Calculate the interval probability. Fourteen days equal two weeks. In weekly units, the cumulative hazard is \(0.021(2)=0.042\). In daily units it is \(0.00300(14)=0.042\). Therefore,
The probability under the stated model is about 4.11% over 14 days. Both unit systems produce the same dimensionless cumulative hazard and probability. The weekly number \(0.021\) is not a 2.1% weekly probability; the model-implied probability over one week is \(1-\exp(-0.021)\approx0.02078\), or about 2.08%.
Checking and Comparing Hazard Units
A reliable conversion has three parts: identify the original time unit, identify the target time unit, and apply the conversion factor in the correct direction. A useful dimensional check is to write the units alongside the arithmetic. Multiplying a rate per day by 365 days per year cancels “per day” and leaves “per year.” If the result still has the wrong unit, the conversion factor was inverted or the units were not tracked consistently.
Two hazard values are directly comparable numerically only when they use the same time unit and the same hazard definition. If two hazards are both converted using the same time-scale factor, their ratio is unchanged. For example, converting both a numerator and denominator from per day to per year multiplies each by 365, so the common factor cancels. This invariance does not make rates with different risk-set definitions interchangeable.
Be precise about what “year” means. A 365-day year is a stated conversion convention; calendar years can differ in length, and months do not all contain the same number of days. A model that records elapsed time in days should not be converted using an unstated approximation when exact calendar timing matters. If a report uses person-years, the time denominator also needs to be understood: the hazard is still a rate per unit of elapsed person-time, not a probability for one individual over a year.
A unit conversion is also different from changing the time scale itself. Re-expressing days as years preserves the ordering and distances of elapsed times by a constant factor. Replacing “time since enrollment” with “attained age,” for example, changes the time coordinate and may change the hazard pattern being described; it is not merely a unit conversion.
Worked Example: Checking an Annual Rate Against a Daily Rate
Two hypothetical reports describe the same constant hazard, but one gives \(1.095\) per year and the other gives \(0.00300\) per day. Use 365 days per year to check whether the numbers are consistent. Then find the model-implied event probability over 20 days under a single-event constant-hazard model with no competing events.
Convert the annual rate.
The reports are consistent to the stated precision. For a 20-day interval, the cumulative hazard is \(0.00300(20)=0.0600\). Equivalently, the annual calculation uses \(20/365\) years:
The model-implied probability is about 5.82%. This result is conditional on the single-event constant-hazard model; the agreement between the two rates follows from unit conversion alone.
Common Mistakes and Diagnostics
- Multiplying in the wrong direction. Since a year contains 365 days, a rate per day becomes 365 times as large numerically when expressed per year. To convert per year to per day, divide by 365.
- Changing the rate but not the interval. A probability calculation requires a rate and duration in matching units. Check that \(\lambda\Delta\) is dimensionless before applying an exponential formula.
- Calling a rate a probability. A hazard of \(0.73\) per year is not automatically a 73% probability during a year. Under a single-event constant-hazard model, the one-year probability is \(1-\exp(-0.73)\), approximately 51.8%.
- Using the single-event formula with competing events. A cause-specific hazard alone does not determine the focal cause’s interval probability when other causes may occur. Do not omit those causes from the interpretation.
- Treating a changing hazard as constant. If the hazard varies over time, multiplying one selected hazard value by the whole interval may not give the cumulative hazard. The rate must be integrated over that interval, or a justified constant-hazard approximation must be stated.
A practical diagnostic is to calculate the rate-duration product in both unit systems. If the two products differ for the same physical interval, first check the conversion factor, then the interval conversion, and then any rounding. If the arithmetic agrees but the probability interpretation seems wrong, check whether the single-event assumptions hold and whether the hazard definition matches the question.
Check Your Understanding
For each problem, state the units and show the calculation. Solutions are included below.
- A constant hazard is \(0.004\) per day. Convert it to a rate per year using 365 days per year.
- A constant hazard is \(0.14\) per week. Convert it to a rate per day, using seven days per week.
- Under a single-event constant-hazard model with no competing events, what is the event probability over 10 days if the hazard is \(0.002\) per day?
- Explain why a constant cause-specific hazard by itself does not generally give the focal cause’s probability over a long interval when competing events are present.
Solutions. (1) \(0.004(365)=1.46\) per year. (2) \(0.14/7=0.0200\) per day. (3) The cumulative hazard is \(0.002(10)=0.0200\), so the probability is \(1-\exp(-0.0200)\approx0.01980\), or about 1.98%, under the stated model. (4) Other causes can remove people from the event-free risk set before the focal event occurs. The cause-specific hazard describes the focal rate among those still event-free; the focal event probability over the interval therefore depends on the competing events as well as the focal hazard.