I'm trying to understand how the odds ratio can approximate the relative risk under the rare disease assumption.

I'm given the relative risk for disease $B$ and risk factor $A$ is 2 and that 20% of the population has been exposed to the risk factor. Assuming the probability of having the disease among those without the risk factor is $d$, how do I express the odds ratio in terms of this $d$?


Using law of total probability, you can get $$ P(B=1)=d(1-0.2)+2d \times 0.2 = 1.2d. $$

$$ OR=\frac{P(B=1|A=1)/P(B=0|A=1)}{P(B=1|A=0)/P(B=0|A=0)} = RR \times \frac{P(B=0|A=0)}{P(B=0|A=1)} = 2 \frac{1-d}{1-2d}. $$

In general, $$P(B=1) = d(1-P(A=1))+dRRP(A=1)=d(1+(RR-1)P(A=1));\\ OR=RR\frac{1-d}{1-dRR}$$ If $P(B=1)$ close to 0, $d$ is close to 0. Then if $dRR$ is not large,$\frac{1-d}{1-dRR}$ is close to 1, RR and OR are approximately equal. However there are some extreme cases. For example, $RR=100,d=0.009,P(A=1)$ very small, then $OR=9.91RR$.


An intuitive way to understand this is to consider the basic definitions of odds and risk. Odds are simply the ratio of the number of events ($E$) to the number of non-events ($NE$) whereas risk is the the ratio of events to total events ($E+NE$). As an event becomes more rare, the number of non-events approaches the number of total events. Take a simple example:

The odds of this event are $E/NE$ while the risk is $E/(E+NE)$

It is easy to see that as an event gets rarer ($E$ approaches zero), odds and risk both approach the same value (zero), whereas, as the event gets more common ($NE$ approaches zero), they diverge (odds approaches infinity and risk approaches unity).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.