I'm getting my information from: http://sphweb.bumc.bu.edu/otlt/mph-modules/ep/ep713_analyticoverview/EP713_AnalyticOverview5.html It says that in a case-control study, you can not compute the probability of disease in each exposure group because you don't have the total amount of people in your population. But you just made a new population for the case control study - why not use this total as the population total? Of course it's a sample of the population you're really interested in so that introduces sampling bias and error but that is true for the odds ratio as well. So you'd have to say it's the approximation of the true relative risk.

Another argument I've read is that in a case-control study, you haven't really taken a cross-section of the population because you started with a fixed amount of people with the same outcome (=diseased), but the amount of individuals in this group (disregarding noise) is irrelevant: if this group is bigger, the amount of of exposed people grows with the same multiplier as the amount of unexposed people. So when calculating the relative risk (risk exposed group / risk in unexposed group) both the numerator and the denominator grow with the same amount, which then makes no difference in the value of the fraction.

Can anyone explain what I'm missing? All thoughts appreciated.


To obtain the relative risk you have to know the risk for each level of the exposure. If you sample people with each level of the exposure then you can estimate their risk of disease. You can then compare those risks which is what you are interested in. In a case control study you do not do that, you ascertain a number of cases in some way (usually not on the basis of a formal sampling strategy) and then find some controls who are similar to the cases (for some meaning of similar) except in their disease status. They are unlikely to be representative of the general population.

  • $\begingroup$ Yeah so a case control study doesn't have perfect sampling, now why does that invalidate calculating the relative risk? What is it about odds ratio that makes it perfect for imperfect sampling? $\endgroup$ – Plumpie Apr 30 '17 at 14:48
  • $\begingroup$ You cannot calculate the relative risk because you don't know the probability $P(Y|X=1)$ and $P(Y|X=0)$ (that you could only derive if you knew the real distribution either of $Y$ or of $X$). On the contrary, since the odds ratio does depend on the distribution of $Y|X$ $(P(Y=1|X=1)$ and $P(Y=1|X=0))$, but equivalently of $X|Y$ $(P(X=1|Y=1)$ and $P(X=1|Y=0))$, we can calculate it, since the distribution of $X|Y$ is precisely what we observe. $\endgroup$ – Federico Tedeschi Mar 20 '18 at 22:18

Inspired by Federico's comment, here is my take on formalizing the statistics:

Using the contingency table notation in Wikipedia:

                | Intervention (I)      | Control (C)
Events (E)      | IE                    | CE
Non-events (N)  | IN                    | CN

$\textrm{Odds Ratio}= \frac{P(I\cap E)/P(I\cap N)} {P(C\cap E)/P(C\cap N)}= \frac{P(I| E)/P(I| N)} {P(C| E)/P(C| N)}$


$\textrm{Relative Risk}= \frac{P(I\cap E)/P(I)} {P(C\cap E)/P(C)}= \frac{P(I| E)/P(I)} {P(C| E)/P(C)}$

The key difference being, for odds ratio, one needs to know the conditional probabilities $P(I| N), P(C| N)$, which are known in case control studies; whereas for relative risk, $P(I), P(C)$ are not known from a retrospective study design.

  • $\begingroup$ Cool! Is there also a way to describe it more intuitively? $\endgroup$ – Plumpie Mar 29 at 23:12

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