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I am trying to convert the Hazard Ratio from this paper (Risk of Gastrointestinal Adverse Events Associated With Glucagon-Like Peptide-1 Receptor Agonists for Weight Loss) to determine the prevalence of gastroparesis in a given population.

The absolute risk of an individual developing gastroparesis is 50 per 100,000 individuals in the US according to the National Institute of Diabetes and Digestive Kidney Diseases.

The adjusted HR from the paper is 3.67 which indicates that an individual is 3.67x more likely to contract gastroparesis relative to bupropion-naltrexone which is not a GLP-1 agonist.

Is it reasonable to multiply 50 (absolute risk) by 3.67 and apply that to a population? So for 100,000 individuals using semgalutide we would expect 183.5 to be diagnosed with gastroparesis?

David Andrews
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2 Answers2

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Hazard and risk are not directly related.

An extreme example, consider the risk of dying before 130 years old, and the hazard of dying due to smoking. If you smoke then you experience a higher hazard to die, but the risk of dying is eventually the same 100% for anybody.

The risk is an integral of the hazard $h(t)$

$$\text{Risk dying before time $t$} = P(T<t) = 1-e^{-\int_0^t h(t) dt } $$

See also this waiting time problem: What distribution to use to model time before a train arrives?

Approximation

If the risk is small, then we can use a Taylor series expansion like $e^x \approx 1-x$, and you get

$$\text{Risk dying} \approx \int_0^t h(t) dt $$

and hazard ratio's transfer (approximately) to risk ratio's.

Your further computations with risk may work if not too many people are using using semgalutide. The risk of 50 per 100000 is probably for both users and non-users. Only if the users groups is very small, then you can use the 50 per 100000 of the entire population as estimate for the non-users population.

Otherwise you need to incorporate the relative fraction of users and non-users into the computation. A related question to this is How to account for the difference between the sexes in a mortality rate

Sextus Empiricus
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The attempt is incorrect. Intuitively, if the prevalence were not 50/100,000 but 50%, it would exceed 100% if we multiply it by the hazard ratio 3.67.

The prevalence P is (# cases) / (# population) which does not have a time scope (i.e. age) in it and comes from a cross section. However, studies with hazard ratio model event occurrence with time, h(t) = f(t)/S(t), where f(t) is the derivative of event probability F(t) = Pr{event happens before t} at time t and S(t) = 1 - F(t) is the survival probability at time t. It must come from longitudinal data. Without reference to time t, the model does not stand. The probability or proportion that can be directly derived from a hazard ratio (HR) is a stochastic superiority: HR = Pr{S(t, X1) < S(t, X0)} / Pr{S(t, X1) >= S(t, X0)}. See https://en.wikipedia.org/wiki/Hazard_ratio#The_hazard_ratio_and_survival.

I would say that P is related to F(t), but I can find no source of mathematical expression relating these two, so I derive on my in own in the following. Comments and corrections are welcome. Assuming that the survival study has external validity so that it can be applied to the entire population, I believe that P = mean F(t) over t where t is age as distributed in the population, that is, event probability by age weighted according to the population age distribution. For example, if we estimate F(t) = 1 - S(t) = t/100 from the sample, so that the probability of event is 1% by age 1 and 99% by age 99. If we also find that age is uniformly distributed [0, 100] in the population (note that this information does not necessarily come from the study sample, and the age distribution in the study sample can be different from that in the population), then prevalence in the population cross section must be 50%. Therefore, to relate hazard ratio (or any survival analysis quantity) to prevalence, we must know the distribution of t (often the age) in a cross section so we can integrate it out.

DrJerryTAO
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