In a Cox proportional hazards model of churn, what does a hazard ratio of 1.8 on an annual-plan flag mean?
answer
- a ratio of rates, not probabilities
- assumed identical at every time point
- the exponentiated coefficient
- the baseline hazard cancels out
- survival transforms as S1 = S0^HR
basics
~10 sAnnual-plan customers cancel at 1.8 times the instantaneous rate of the reference group, at every time point the model considers. A hazard ratio compares rates among those still subscribed; it is not a probability.
solid answer
~50 sA Cox model writes the hazard as `h(t | x) = h0(t) * exp(b'x)`, so exponentiating the coefficient on the annual-plan flag gives the hazard ratio. A value of 1.8 says that at any given time, among customers who are still subscribed, annual-plan customers cancel at 1.8 times the rate of the otherwise-identical reference customers — the instantaneous churn rate is 80% higher. Because the baseline hazard cancels out of that ratio, 1.8 carries no information about the absolute level or shape of churn over time. It is also not a statement that 80% more customers churn by day 90: risk ratios over a fixed window are pulled toward 1 relative to the hazard ratio. And it is not a claim that typical lifetime is 1.8 times shorter, since survival transforms as `S1(t) = S0(t)^1.8`.
go deeper
Be ready to say in one sentence that the exponentiated coefficient is a hazard ratio, and that 1.8 means an 80% higher instantaneous event rate, not an 80% chance of the event.
Expect to explain where the ratio comes from algebraically, including why the baseline hazard cancels and why the model assumes the ratio is the same at every time point.
Show that you know when the number misleads in a real report: risk over a window is compressed toward 1, absolute churn is not recoverable from the coefficient, and an averaged ratio hides a fading effect.
Own how these numbers get communicated. Decide whether stakeholders should see hazard ratios at all or absolute risk differences over a business-relevant horizon, and make the reference group and adjustment set part of the reporting standard.
## The model behind the number A Cox proportional hazards model writes the hazard for a subject with covariates `x` as `h(t | x) = h0(t) * exp(b1*x1 + b2*x2 + ...)` The hazard `h(t)` is the instantaneous event rate at time `t` among subjects who have not yet had the event. In a churn model it is the rate at which customers who are still subscribed at time `t` cancel right then. It is a rate per unit time, so it can exceed 1 and is not a probability. `h0(t)` is the baseline hazard — the hazard for a subject whose covariates are all zero — and it is left as an arbitrary, unspecified function of time. ## Where 1.8 comes from Fitting returns coefficients on the log-hazard scale. If the coefficient on the annual-plan flag is `b = 0.588`, then `exp(0.588) = 1.8`. Compare two customers identical on every other covariate, one on the annual plan (`x = 1`) and one on the reference plan (`x = 0`): `h(t | x=1) / h(t | x=0) = [h0(t) * exp(0.588)] / [h0(t) * exp(0)] = 1.8` Two things cancel. The baseline hazard `h0(t)` cancels, which is why the ratio does not depend on how churn rises or falls over the lifecycle. And `t` disappears entirely, which is the proportional-hazards assumption: the ratio is taken to be the same at day 3 and at day 300. ## What the number does say - It is a comparison between two covariate profiles that differ by one unit in the annual-plan flag, holding the other covariates in the model fixed. - It is a multiplicative statement about rates: 80% higher instantaneous churn rate for the annual-plan group. - It is conditional on still being subscribed. Every hazard comparison is made within the set of customers who have survived to that instant. - For a continuous covariate, the exponentiated coefficient is the ratio per one-unit increase, and a ten-unit increase multiplies the hazard by `1.8^10` — the model is log-linear in the covariate unless you deliberately add splines or categories. ## What the number does not say **It is not the ratio of event probabilities over a window.** Under proportional hazards the survival functions are related by `S1(t) = S0(t)^HR`. Suppose the reference group's 90-day churn probability is 10%, so `S0(90) = 0.90`. Then `S1(90) = 0.90^1.8 = 0.827`, a churn probability of 17.3%, and the 90-day risk ratio is 1.73 — close to 1.8 only because the baseline risk is small. If the reference group's 90-day churn were 50%, then `S1(90) = 0.5^1.8 = 0.287`, a churn probability of 71.3%, and the risk ratio is only 1.43. The higher the baseline risk, the more a risk ratio is compressed toward 1 while the hazard ratio stays at 1.8. **It is not a median-lifetime statement.** How much the median shifts depends on the shape of the baseline hazard. There is one clean special case: if the baseline hazard is constant over time (exponential survival), then the median is `ln(2)/h`, so multiplying the hazard by 1.8 divides the median lifetime by exactly 1.8. For any other baseline shape — churn that spikes in the first week and then flattens, for instance — the median ratio is something else, and you have to compute it from the fitted survival curves rather than read it off the coefficient. **It is not an absolute risk.** Nothing in the coefficient tells you whether either group churns at 2% a year or 40% a year. Absolute predictions require a separate estimate of the baseline hazard. ## Reporting it honestly Uncertainty is symmetric on the log scale, not the hazard-ratio scale, so build the interval on `b` and exponentiate the endpoints. An interval such as 1.8 (1.25 to 2.59) is the usual presentation; if the interval spans 1, the data do not distinguish the two groups. Always name the reference level and say which covariates are held fixed, because a hazard ratio without its adjustment set is uninterpretable. Finally, the single number is only meaningful if proportionality actually holds. If the annual-plan effect is strong in the first month and gone afterwards, 1.8 is an average of a large early ratio and a near-1 later one, and the average depends on how long you watched.
- Why is a hazard ratio of 1.8 not the same as being 1.8 times as likely to churn within 90 days?Because probabilities are bounded and hazards are not. Under proportional hazards `S1(t) = S0(t)^1.8`. If the reference 90-day churn is 10%, the other group's is `1 - 0.9^1.8 = 17.3%`, a risk ratio of 1.73. If the reference churn is 50%, the other group's is 71.3%, a risk ratio of only 1.43. The higher the baseline risk, the more the risk ratio is squeezed toward 1.
- Does a hazard ratio of 1.8 mean typical subscription length is 1.8 times shorter?Only in one special case. If the baseline hazard is constant over time, the median lifetime is `ln(2)/h`, so a hazard ratio of 1.8 divides the median by exactly 1.8. With any other baseline shape the median ratio differs, and you must read it off the fitted survival curves rather than infer it from the coefficient.
- How would you report uncertainty around a hazard ratio of 1.8?Build the interval on the log-hazard scale, where the estimate is roughly symmetric, then exponentiate both endpoints. That gives an asymmetric interval such as 1.25 to 2.59 around 1.8. If the interval covers 1, the data do not separate the groups. Always state the reference level and the covariates held fixed.
It is like saying one road has 1.8 times the accident rate per mile driven of another. That tells you nothing about how many miles either road carries, or how long a given driver lasts before a crash.
saying these in an interview costs you the question
- Calls a hazard ratio a probability of the event
- Reads 1.8 as 80% of customers churning
- Claims the same fixed 1.8x risk over any follow-up window
- Says median lifetime must be 1.8 times shorter
- Forgets the ratio is assumed constant over time
- Quotes a hazard ratio without naming the reference group