# Calculating risk difference and number needed to treat from hazard ratio in meta-analysis

If you have primary data, there exist several ways to calculate a risk difference (RD) or number needed to treat (NNT) from time-to-event data (see e.g. following pmid: 29955580, 31626655). However, I am performing a meta-analysis yielding a summary-HR, based on Cox regression models which were adjusted for different covariables. I have no primary data.

What I do have, on the other hand, is the number of participants at risk (crude or in person years, depending on study design) and the number of events.

My question is: Is is possible to calculate a risk difference based on my summary HR in this case?

• As a side issue, note that risk difference and NNT are functions of the entire set of covariate values. Though commonly believed, there is no such thing is "the NNT" or "the risk difference". – Frank Harrell Oct 21 '20 at 11:33
• I cannot fully follow your comment - the HR is also a function of the entire set of covariables. My problem here was that I would greatly appreciate a measure of absolute risk, since the relative effect and the incidence are small. I would like to clarify, that the effect of exposure on the outcome is small. I know that the NNT is probably not the right measure for small effects, but a RD would be great. In the end what matters to me is to illustrate what is clinically relevant and I think an absolute measure is mor suitable in my case. – ehi Oct 22 '20 at 8:45
• Absolute risk difference is a function of particular values of all the covariates, unlike the HR. – Frank Harrell Oct 22 '20 at 11:30

As a simple thought-experiment, consider a treatment that is protective against death and the proportional hazards assumption is valid. If you run the analysis, you will have one single hazard ratio describing the whole time. For RD, you would observe something like RD < 0 for $$t=10$$. However, as $$t \rightarrow \infty$$ the RD will go to zero (eventually everyone dies).