It is known that odds ratios enjoy a certain symmetry. For example, the odds ratio of outcome $Y$ is the inverse of the odds ratio of outcome $\neg Y$. Risk ratios, on the other hand, do not enjoy this symmetry. However, risk ratios have the property of collapsibility. So adjusting for a covariate that is not a confounder does not change the magnitude of the risk ratio. Consider the more formal definition of collapsibility:

Definition. Let $g[P(x,y)]$ be any functional that measures the association between $Y$ and $X$ in the joint distribution $P(x,y)$. Then $g$ is collapsibile on a variable $Z$ if $$E_{z}g[P(x,y|z)] = g[P(x,y)]$$

So in the case of the risk ratio, $g[P(x,y)]$ would be the risk ratio? What if we don't know $P(x,y)$? A risk ratio is the ratio of two incidence densities which doesn't seem to depend on any probability distribution.

  • 1
    $\begingroup$ Where did this definition come from? I'm not sure if it's right. $\endgroup$ – AdamO Jan 22 '18 at 15:13

Yes and no. The risk ratio is collapsible, so adjusting for any variable that is not associated with either the exposure or outcome should not change the magnitude of the risk ratio. In addition, the summary risk ratio across strata should be between the values of the stratum-specific risk ratios (something that does not hold with the odds ratio).

But there are many ways in which a risk ratio can be biased after adjusting for variables that are not technically confounders. For example, adjusting for an intermediate will cause bias, despite collapsibility, since some of the association will be removed. Also, adjusting for a common effect of the exposure and outcome will cause bias in the risk ratio - this is sometimes referred to as collider bias, or selection bias. Finally, adjusting for variables associated only with the outcome can under some conditions change the magnitude of the risk ratio - in this case the risk ratio is not technically biased, it is simply estimating a different quantity (the effect of exposure on the outcome among people with a certain distribution of these third variables).

In terms of the formal definition you give for collapsibility, I am not entirely familiar with this, but I believe you are correct that g(P(x,y)) would be the risk ratio. Also, it is my understanding that the risk ratio (and more importantly the variance and confidence intervals for the risk ratio) is calculated under the assumption that the risks each follow independent Binomial distributions.

  • 3
    $\begingroup$ It may be worth pointing out that the cost of using risk ratios is that they are not constant over the conditioning variables, a problem worse than non-collapsibility in my view. For example a risk ratio of 2 cannot apply to a high risk individual whose background risk is 0.51. $\endgroup$ – Frank Harrell Jul 5 '13 at 23:20

It is well known, that the risk ratio (RR) is collapsible. Consequently, if RR(|z)<1 for all z, then E(RR(|)<1 for any distribution of z. Let us consider the initial Simpson example

       x  0 1        x    0  1            x   0  1
z=0: Y=0: 5 8   z=1:Y=0: 15 12   Total: Y=0: 20 20 
     Y=1  3 4       Y=1  3  2           Y=1   6  6. 

Then RR(Y=1|z=0) = (4/12)/(3/8) = 8/9, RR(Y=1|z=1) = (2/14)/(3/18) = 6/7 Marginal RR(Y=1) = (6/26)/(6/26) = 1 How it can be if RR is collapsible?

Also, RR is not strictly collapsible i.e. it may be that RR(|z) is constant over z, but differs from marginal RR. Example

       x  0 1        x    0  1            x   0  1
z=0: Y=0: 3 2   z=1:Y=0:  2  5   Total: Y=0:  5  7 
     Y=1  7 1       Y=1   3  2          Y=1: 10  3. 

Then RR(Y=1|z=0) = (1/3)/(7/10) = 10/21, RR(Y=1|z=1) = (2/7)/(3/5) = 10/21 Marginal RR(Y=1) = (3/10)/(10/15) = 9/20

Ilya Novikov


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.