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It may help to think of both cohort and case-control studies being limited, just in opposite ways. A case-control study cannot give you risk of disease, but a cohort study cannot give you the probability of exposure (it's not really a "risk" of exposure, because studies usually require the exposure to occur before the outcome). Only a cross-sectional study can give you both the probability of exposure and the probability of outcome (though, as Alexis mentioned below in a comment, these are prevalences not risks in a cross sectional study. The math is the same.).

It may help to think of both cohort and case-control studies being limited, just in opposite ways. A case-control study cannot give you risk of disease, but a typical cohort study where the numbers exposed and not exposed are fixed (e.g. 1000 treated vs 1000 not treated) cannot give you the probability of exposure (it's not really a "risk" of exposure, because studies usually require the exposure to occur before the outcome). A cross-sectional study can give you both the probability of exposure and the probability of outcome (though, as Alexis mentioned below in a comment, these are prevalences not risks in a cross sectional study. The math is the same.).

It may help to think of both cohort and case-control studies being limited, just in opposite ways. A case-control study cannot give you prevalence (risk) of disease, but a cohort study cannot give you prevalence of exposure. Only a cross-sectional study could give you both.

This is because prevalence (risk) is calculated from probabilities, and those probabilities in your study need to match what's in the actual population in order for your study to give good estimates. The moment you fix these probabilities to something other than reality, that limits what can be calculated from your study. For example, we force the probability of disease to 50% when we do a 1:1 case-control study. We force the probability of exposure to 50% when we do a 1:1 cohort study. These rarely match the true proportion in the population.

Here's an example. The true prevalences per group are shown in the table below from a cross-sectional study.

In a cohort study, you are fixing the number in each exposure group, say 1200 in each group. Thus, your study can't estimate the probability of being exposed anymore: you forced it to be 50%. Notice how the risk of disease is unchanged (far column), but the prevalences of exposure are all different now.

In a case-control study, you are fixing the number of cases and controls, say 1200 in each group. Thus, your study can't estimate the probability of disease (risk). Notice how the prevalence of exposure is the correct, but the risk of disease is not.

The simplest "risk" to calculate here is the prevalence of exposure given disease status: $P({Exposure}^+|{Disease}^+)\approx \frac{w}{y}$. Compare to the cohort study, where the simplest risk was the risk of disease given exposure: $P({Disease}^+|{Exposure}^+)\approx \frac{a}{b}$. The previous answers have explained why $\frac{w}{y}\neq \frac{a}{b}$.

[Back into the statistics, unlike $b$, $w+x$ is not constant. Therefore, the distribution in the top row is not $Bin(w+x,Risk_{{Exposure}^+})$. Since we cannot measure directly the risk, we'd have to use Bayes Theorem:$$Risk_{{Exposure}^+}=P({Disease}^+|{Exposure}^+)=\dfrac{P({Exposure}^+|{Disease}^+)P({Disease}^+)}{P({Exposure}^+)}.$$ We can estimate $P({Exposure}^+|{Disease}^+)\approx \frac{w}{y}$, because the binomial distribution holds. However, we cannot estimate $P({Disease}^+)\not\approx \frac{y}{y+z}$, because we chose $y$ and $z$, rather than measuring them from the experiment. Often we choose 1:1 groups, e.g. 1000 cases and 1000 controls. This would make $\frac{y}{y+z}=0.5$ even when the actual disease prevalence could be different. Similarly, we cannot estimate $P({Exposure}^+)\not\approx \frac{w+x}{y+z}$, because that requires having the correct mix of cases and control.

If you know the disease and exposure prevalences from previous studies, then you can calculate risk and RR using them; but that information does not come from your case-control study (and isn't in the tables above).]

It may help to think of both cohort and case-control studies being limited, just in opposite ways. A case-control study cannot give you risk of disease, but a cohort study cannot give you the probability of exposure (it's not really a "risk" of exposure, because studies usually require the exposure to occur before the outcome). Only a cross-sectional study can give you both the probability of exposure and the probability of outcome (though, as Alexis mentioned below in a comment, these are prevalences not risks in a cross sectional study. The math is the same.).

This is because risk is calculated from probabilities, and those probabilities in your study need to match what's in the actual population in order for your study to give good estimates. The moment you fix these probabilities to something other than reality, that limits what can be calculated from your study. For example, we force the probability of disease to 50% when we do a 1:1 case-control study. We force the probability of exposure to 50% when we do a 1:1 cohort study. These rarely match the true proportion in the population.

Here's an example. The true probabilities per group are shown in the table below from a cross-sectional study.

In a cohort study, you are fixing the number in each exposure group, say 1200 in each group. Thus, your study can't estimate the probability of being exposed anymore: you forced it to be 50%. Notice how the risk of disease is unchanged (far column), but the probabilities of exposure are all different now.

In a case-control study, you are fixing the number of cases and controls, say 1200 in each group. Thus, your study can't estimate the probability of disease (risk). Notice how the probability of exposure is the correct, but the risk of disease is not.

The simplest "risk" to calculate here is the probability of exposure given disease status: $P({Exposure}^+|{Disease}^+)\approx \frac{w}{y}$. Compare to the cohort study, where the simplest risk was the risk of disease given exposure: $P({Disease}^+|{Exposure}^+)\approx \frac{a}{b}$. The previous answers have explained why $\frac{w}{y}\neq \frac{a}{b}$.

[Back into the statistics, unlike $b$, $w+x$ is not constant. Therefore, the distribution in the top row is not $Bin(w+x,Risk_{{Exposure}^+})$. Since we cannot measure directly the risk, we'd have to use Bayes Theorem:$$Risk_{{Exposure}^+}=P({Disease}^+|{Exposure}^+)=\dfrac{P({Exposure}^+|{Disease}^+)P({Disease}^+)}{P({Exposure}^+)}.$$ We can estimate $P({Exposure}^+|{Disease}^+)\approx \frac{w}{y}$, because the binomial distribution holds. However, we cannot estimate $P({Disease}^+)\not\approx \frac{y}{y+z}$, because we chose $y$ and $z$, rather than measuring them from the experiment. Often we choose 1:1 groups, e.g. 1000 cases and 1000 controls. This would make $\frac{y}{y+z}=0.5$ even when the actual disease probability could be different. Similarly, we cannot estimate $P({Exposure}^+)\not\approx \frac{w+x}{y+z}$, because that requires having the correct mix of cases and control.

If you know the disease and exposure probabilities from previous studies, then you can calculate risk and RR using them; but that information does not come from your case-control study (and isn't in the tables above).]

[Back into the statistics, unlike $b$, $w+x$ is not constant. Therefore, the distribution in the top row is not $Bin(w+x,Risk_{{Exposure}^+})$. Since we cannot measure directly the risk, we'd have to use Bayes Theorem:$$Risk_{{Exposure}^+}=P({Disease}^+|{Exposure}^+)=\dfrac{P({Exposure}^+|{Disease}^+)P({Disease}^+)}{P({Exposure}^+)}.$$ We can estimate $P({Exposure}^+|{Disease}^+)\approx \frac{w}{y}$ and $P({Exposure}^+)\approx \frac{w+x}{y+z}$, because in both cases the binomial distribution holds. However, we cannot estimate $P({Disease}^+)\not\approx \frac{y}{y+z}$, because we chose $y$ and $z$, rather than measuring them from the experiment. Often we choose 1:1 groups, e.g. 1000 cases and 1000 controls. This would make $\frac{y}{y+z}=0.5$ even when the actual disease prevalence could be rare. If you know the disease prevalence from previous studies, then you can calculate risk and RR using it; but that information does not come from your case-control study (and isn't in the tables above).]

[Back into the statistics, unlike $b$, $w+x$ is not constant. Therefore, the distribution in the top row is not $Bin(w+x,Risk_{{Exposure}^+})$. Since we cannot measure directly the risk, we'd have to use Bayes Theorem:$$Risk_{{Exposure}^+}=P({Disease}^+|{Exposure}^+)=\dfrac{P({Exposure}^+|{Disease}^+)P({Disease}^+)}{P({Exposure}^+)}.$$ We can estimate $P({Exposure}^+|{Disease}^+)\approx \frac{w}{y}$, because the binomial distribution holds. However, we cannot estimate $P({Disease}^+)\not\approx \frac{y}{y+z}$, because we chose $y$ and $z$, rather than measuring them from the experiment. Often we choose 1:1 groups, e.g. 1000 cases and 1000 controls. This would make $\frac{y}{y+z}=0.5$ even when the actual disease prevalence could be different. Similarly, we cannot estimate $P({Exposure}^+)\not\approx \frac{w+x}{y+z}$, because that requires having the correct mix of cases and control.

E.g. these two tables have the same $P({Exposure}^+|{Disease}^+)=1$ and $P({Exposure}^+|{Disease}^-)=0$, but different $P({Exposure}^+)$, 50% and 20% respectively.

If you know the disease and exposure prevalences from previous studies, then you can calculate risk and RR using them; but that information does not come from your case-control study (and isn't in the tables above).]