I'm working on a difference-in-differences (DiD) analysis for a healthcare study measuring # hospitalizations (X) per # eligible months (Y).

Simplifying the math a bit - my actuary colleagues like to measure the DiD in the aggregate as sum(X)/sum(Y) = E(X)/E(Y) vs. the member-level rate I prefer (the average treatment effect), average of (X/Y) = E(X/Y).

I know via Jensen's Inequality that E(X)/E(Y) <= E(X/Y) given 1/Y is convex and X and Y are independent (actually, a fair assumption in this case).

Question: Which method is preferred for estimating the DiD treatment effect? Is one more "correct" than the other, or are they simply addressing different questions?


1 Answer 1


Recall that the expectation is linear, so

$$ E[a + b X] = a + c E[X] $$

Let's say that $E[X] = 0$, in such a case if we have $Y = cX$, where $c$ is some constant, $E[Y] = E[cX] = c E[X] = c \times 0$, so $E[X]\,/\,E[Y]$ would be undefined regardless of what $c$ is. But we can shift $X$ and $Y$ by a constant $a$, so that $E[X + a] = 0 + a$, and $E[Y + a] = 0 + a$, then $E[X+a]\,/\,E[Y+a] = a \,/\, a = 1$, regardless of what $c$ is. This example shows that comparing expectations to learn $c$ easily breaks apart.

Another problem with comparing expectations is ecological fallacy and the related Simpson's paradox that show how the relations in aggregated vs raw data (or differently aggregated) may be completely different.

I am not saying that comparing expected values does not make sense. I am trying to say that it doesn't necessarily tell you how $X$ and $Y$ are related to each other, just how their averages are related.

  • $\begingroup$ Thanks - so in your opinion you can't compare E(X)/E(Y) to E(X/Y), it's not as simple as E(X)/E(Y) = c * E(X/Y)? I'm also curious if E[E(X)/E(Y)] and Var(E(X)/E(Y)) can be defined. $\endgroup$
    – RobertF
    Commented Aug 25, 2022 at 15:28
  • $\begingroup$ @RobertF I don't understand your comment. As for E[E(X)/E(Y)], E(X) and E(Y) are just numbers, so there is no expected value or variance for their ratio. $\endgroup$
    – Tim
    Commented Aug 26, 2022 at 8:55
  • $\begingroup$ Link to another similar post that may answer my question I think: stats.stackexchange.com/questions/172196/… $\endgroup$
    – RobertF
    Commented Aug 29, 2022 at 12:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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