# How to explain Hazard Ratio in layperson's terms

In the context of a Cox regression, I recently heard someone say that "the hazard ratio of 0.70 between the two treatment groups indicates that, during the whole followup time, of 100 persons in the control group who have an event only 70 have an event in the intervention group." Is this a valid simplification of an HR or are there other non-technical interpretations?

• That isn't sensible English. Looks like some kind of typo. Commented Jun 11 at 11:52
• @PeterFlom The quoted sentence reads as cromulent English to me. (I would prefer the word "entire followup period" to "whole followup time" but the quoted sentence is understable with "whole followup time"). Commented Jun 12 at 16:28
• @Alexis Thanks for teaching me a new word! Commented Jun 12 at 16:36
• The interpreter may rather be interested in number-needed-to-treat (NNT). Commented Jun 12 at 17:04
• Sorry, English is not my mother tongue. I indeed meant what is discussed above (entire followup period). Commented Jun 14 at 10:43

A simple explanation of the hazard ratio is:

At any given instant, people in group A are XXX as likely to have the event as people in group B.

"Over the whole of the study period" is not right, because each person can have a different study period, due to either censoring (which could be left or right) or the event occurring.

• The obvious example being a study (like cancer treatment) that ran for so long that everyone dies in both groups. Event ratio would obviously be 1:1. Commented Jun 11 at 12:28
• +1 The quote in the question works sort of Ok as a simplified explanation, but as you point out obscures Cox PH's many assumptions. Commented Jun 12 at 16:50
• The problem with this interpretation is the confusion between incidence and cumulative incidence. "Have the event" is ambiguous, whereas "the event occurs" is more to the meaning. Commented Jun 12 at 16:55
• Thanks, that does make sense to me. If the hazard between the groups was proportional, then the HR applies at each timepoint t (See Frank‘s answer below). So why wouldnt this definition apply to the entire followup time? Does the noncollapsibility of HRs play a role in this as the risk set is reduced with ongoing time? Commented Jun 14 at 10:40

We need to start with a premise. Suppose that either the HR is constant (so that what’s below applies to any t) or that we are speaking at the hazard ratio at a specific time t if the HR is not constant. First think of discrete time. The group B to group A HR is the B:A ratio of the probabilities of failing at discrete time t, given the subject had not failed before t. Now consider continuous time. The B:A HR is the ratio of two limits (hazard rates) as the time interval shrinks to zero. The hazard rate is the limit as $$\delta$$ approaches zero of the ratio of the probability of failing within $$(t, t+\delta)$$ given not failing before $$t$$, to $$\delta$$.

• Not to be rude, but you seem to not given much notice to the "layman's terms" part of the question. Commented Jun 15 at 21:30
• The only thing I can think of expanding is this: A hazard function expresses the level of instantaneous risk and this varies over time. These instantaneous risk are called ‘force of mortality’ in demography and represent the instantaneous chance of suffering the event at a given time given the event has not occurred before that time. Commented Jun 16 at 12:37

What you really need to explain what hazard means. It goes like that: suppose someone tells you that there is a 50% probability that you are going to die during the next 20 years. This bad news but it could be much worse: what if I tell you that there is a 50% probability that you are going to die during the next 3 months? The risk* is the same, the forecast periods are not, so you have to take the time into account. A natural way of doing this is to divide the probability by the time, and this is the hazard. It is analogous to speed, which is distance divided by time. a So in the first case the hazard is 0.5/20=0.025, and in the second case the hazard is 0.5/0.25=2.