In meta-analysis: How do we convert hazard ratios in some studies to odds ratio? There are case control and cohort studies to be included and some of them report hazard ratios. The raw data is not reported in a way to calculate odds ratio.
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asked Dec 24, 2014 at 16:14
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1$\begingroup$ I cannot comment (rep), but hopefully this 'answer' is excused..... This paper makes a very convincing statement that HR = odds, note: not odds ratio (OR). Hazard Ratio in Clinical Trials ....and perhaps like the OP I would be delighted to see a derivation, particularly as the terms in HR are quite complicated (ratio in the limit of two rates derived from conditional probabilities), whereas the odds formula is not so complicated. $\endgroup$– Big Old DaveCommented Jun 8, 2016 at 13:39
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2$\begingroup$ @BigOldDave That reference may be convincing to some, but 'odds' are never really defined and then the HR is in turn defined in terms of odds. If the generally accepted definition for odds were implied (rather that some colloquial notion of risk) then the definition of HR is simply wrong. That's a very poor reference for this issue. Odds are not rates since they have no notion of interval of observation. Rate ratios and hazard ratios at least share a correction for time under observation. $\endgroup$– DWinCommented Jul 9, 2016 at 18:48
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1$\begingroup$ See also stats.stackexchange.com/questions/15897/… which backs up @DWin $\endgroup$– mdeweyCommented Jul 11, 2016 at 11:03
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$\begingroup$ @DWin the odds they speak of is a strange U-like statistic that takes p/(1-p) of the probability that a randomly selected case experiences an outcome faster than a randomly selected control. But that p is related to the RR by RR = p/(1-p) N0/N1 where N0 is the p-y exposure in the control, N1 for exposed. if the outcome is rare, N0, N1 are very large and the ratio tends to 1 leaving p $\approx$ RR/(1+RR) and the "odds" they speak of, p/(1-p) is just RR. $\endgroup$– AdamOCommented Oct 20, 2017 at 19:23
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4 Answers
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If there was an extremely low proportion of subjects with an event in all experiments (let's say <10%) and the hazard and odds ratios are vey close to 1, then hazard, odds and relative risk ratios will be relatively close to each other.
If that is not the case the fundamental differences between these measures will be more and more noticable. For a given trial duration, particular distribution for event occurence and a particular drop-out pattern, there is a correspondence of hazard ratio to odds ratio to relative risk ratio. If all your experiments in your meta-analysis are similar in these respects, it might be possible to convert them. Once you have experiments with different durations, different drop-out patterns or different event time distributions, a hazard ratio might be constant across experiments and is probably the better relative risk measure, but an odds or risk ratio will essentially never be (even if the hazard ratio is, while the same odds ratio would correspond to different hazard ratios across experiments).
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$$\text{RR} = (1 - e^{\text{HR}\ln(1-r)})/r$$ where HR is the hazard ratio and r is the rate for the reference group. If r is not reported, it is probably reported elsewhere (e.g. baseline death rate). See Shor et al. (2017) doi: 10.1016/j.socscimed.2017.05.049
Zhang and Yu (1998), doi: 10.1001/jama.280.19.1690 provide an approximate of RR based on the odds ratio (OR). $$\text{RR} = \frac{\text{OR}}{(1-P_o)+(P_o\cdot \text{OR})}$$ where Po is the incidence of the outcome of interest in the non-exposed group.
You can then calculate OR.
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2$\begingroup$ Note about Zhang, Yu: their calculation of the 95% CI is biased because it does not account for the prevalence of the outcome in the unexposed group, despite the term specifically appearing in their calculation. See Muller Maclehose doi: 10.1093/ije/dyu029.This is relevant to meta-analyses which utilize a forestplot to evaluate consistency. $\endgroup$– AdamOCommented Oct 20, 2017 at 19:10
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$\begingroup$ You're leaving out a lot of assumptions here, for a start the conversion from HR to other measures assumes no drop-outs, equal follow-up etc., which for some time-to-event studies (e.g. long event-driven cardiovascular outcome studies, most Oncology studies etc.) may very badly violated. $\endgroup$– BjörnCommented Apr 25, 2023 at 8:24
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Exploiting the assumption that hazard ratios are asymptotically similar to relative risks, you can use exploit the formula recommendeed by Grant et al, BMJ 2014:
RR = OR / (1 - p + (p * OR)
where RR is the relative risk, OR is the odds ratio, and p is the control event rate, which leads to the following:
OR = ((1 - p) * RR) / (1 - RR * p).
Thus, for instance, a RR of 2.0 with a p of 0.1 would lead to an OR of 2.25, whereas if p increases to 0.2 it would lead to an OR of 2.67.
answered Apr 29, 2016 at 14:07
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9$\begingroup$ But RR $\neq$ hazard ratio. $\endgroup$ Commented Apr 29, 2016 at 14:22
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1$\begingroup$ Unfortunately I am not aware of any formula for HR... But in meta-analysis it is not uncommon to have people use HR as they were RR $\endgroup$ Commented Apr 29, 2016 at 14:44
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See also this paper on the relation between OR, RR, and HR
Stare, J. & Maucort-Boulch, D. Odds ratio, hazard ratio and relative risk. Advances in Methodology and Statistics 13, 59–67 (2016).
Available in https://mz.mf.uni-lj.si/article/view/159/262
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1$\begingroup$ Welcome to CV. As links easily become outdated, please consider updating your answer with a full citation to the paper as well as a brief summary, so that your answer is helpful and informative even without downloading the paper. $\endgroup$– dipetkovCommented Apr 25, 2023 at 7:35