# In the presence of qualitative effect modification, additive effect modification if and only if multiplicative effect modification?

I am reading Robin's What If Causal Inference book Chapter 4 section 1. In the book, it says the following assertion where $$V$$ denotes subset of the groups. I could not figure out the reason why "In the presence of qualitative effect modification, additive effect modification implies multiplicative effect modification, and vice versa."

"That is, there is qualitative effect modification because the average causal effects in the subsets $$V = 1$$ and $$V = 0$$ are in the opposite direction. In the presence of qualitative effect modification, additive effect modification implies multiplicative effect modification, and vice versa. In the absence of qualitative effect modification, however, one can find effect modification on one scale (e.g., multiplicative) but not on the other (e.g., additive)."

The statement sounds like stating a proposition without proof. Consider the following. Let $$Y^1,Y^0$$ denote (treatment and control) outcomes, $$A$$ treatment(=1 for treatment and =0 for control) and $$V$$ subgroups(=0 or =1).

Assume there is qualitative effect modification with $$E[Y^1-Y^0|V=1]=\alpha$$ and $$E[Y^1-Y^0|V=0]=-\alpha$$ with $$\alpha\neq 0$$. So there is additive effect modification. Denote $$E[Y^1|V=1]=x_1,E[Y^1|V=0]=x_2$$. Then $$\frac{E[Y^1|V=1]}{E[Y^0|V=1]}=\frac{x_1}{x_1-\alpha}$$ and $$\frac{E[Y^1|V=0]}{E[Y^0|V=0]}=\frac{x_2}{x_2+\alpha}$$. We can ask whether multiplicative effect the same here. Choose $$x_2=\alpha$$ and $$x_1=-\alpha$$. However, the statement says there will be multiplicative effect modification. I think I cooked up a counter example.

$$Q1:$$ What is the formal statement on effect modification here and what is the proof of the statement?

$$Q2:$$ This is not related to $$Q1$$. The book says in causal inference, we do not consider odds ratio in general. Why odds ratio is not big concerned here? This might have something to do with collapsibility. However, one might be still interested in odds ratios.

Without loss of generality, let's say $$E[Y^1|V=1]-E[Y^0|V=1]<0$$ and $$E[Y^1|V=0]-E[Y^0|V=0]>0$$. We have additive effect modification because across strata of $$V$$, the risk differences differ; indeed, the effect modification is qualitative because in one stratum the risk difference is negative and in the other it is positive. From $$E[Y^1|V=1]-E[Y^0|V=1]<0$$, we have \begin{align} E[Y^1|V=1]-E[Y^0|V=1]&<0 \\ E[Y^1|V=1] &< E[Y^0|V=1] \\ \frac{E[Y^1|V=1]}{E[Y^0|V=1]} &< 1 \end{align} That is, the risk ratio is less than 1 in the $$V=1$$ stratum (assuming $$E[Y^0|V=1]>0$$, which is necessary to identify the risk ratio).

From $$E[Y^1|V=0]-E[Y^0|V=0]>0$$, we have \begin{align} E[Y^1|V=0]-E[Y^0|V=0]&>0 \\ E[Y^1|V=0] &> E[Y^0|V=0] \\ \frac{E[Y^1|V=0]}{E[Y^0|V=0]} &> 1 \end{align} That is, the risk ratio is greater than 1 in the $$V=0$$ stratum. So, we have qualitative effect modification on the multiplicative scale as well. These derivations use simple identities so they are reversible, giving the "implies [...] and vice versa".

It's kind of trivial if you think about it. If one risk difference is negative, its risk ratio will be less than 1. If one risk difference is positive, its risk ratio will be greater than 1. Clearly this implies effect modification on both scales.

Your example fails because you set one of the risks to be negative, which is impossible. The statement is true because the expected potential outcomes are nonnegative.

Related to Q2, I would guess it's because odds ratios are so complicated and fraught with misunderstanding and confusion that including them would only get in the way of a book that is focused on explaining the basics of causal inference. Odds ratios are so counterintuitive that including them would only serve to confuse the reader. This is not to say that understanding odds ratios isn't important; just that it is really an advanced concept outside of the scope of the book and serves no useful illustrative purpose for the concepts being explained.

• A dumb follow up. Are we assuming the outcome is binary here or the equivalence holds only for binary case? It seems that you want to rule out $E[Y^i|V=v]<0$ case and $Y^i$ could be transformed response. I was thinking $Y^i$ as continuous r.v. here though. Thanks for the clarification. Oct 18, 2022 at 0:24
• It doesn't have to be binary for the above to hold, but it does need to be nonnegative (e.g., count). In H&R, they are discussing binary outcomes. For negative outcomes, a risk ratio of 2 could correspond to going from -1 to -2 (i.e., getting worse) or 1 to 2 (i.e., getting better), so it really only makes sense to talk about treatment effects on a multiplicative scale when outcomes are nonnegative.
– Noah
Oct 18, 2022 at 1:30
• Thanks a lot for the clarification. Oct 18, 2022 at 1:32

In response to Q2, the book "What If" indeed says that the odds ratio is not a parameter of interest in causal inference and does not really discuss effect modification on this scale. However see this paper that suggests that the OR scale is the only one on which effect modification should be assessed: Redefining effect modification