Why not always present logistic regression estimates in the response scale (probablity)?

Question

There's a lot of discussion in epidemiology around the relative merits of odds ratios vs risk ratios. Proponents of the former cite the mathematical qualities of odds (not constrained to be between 0 and 1) and their suitability for examining common outcomes. Proponents of the latter believe that RR's are often more interpretable, and match how clinicians and the public tend to think.

Some have suggested methods to model risk ratios directly (in multivariable models). For instance, the log binomial model and poisson regression with robust standard errors.

What I don't understand is - why don't we just fit the model using standard logistic regression and perform the inverse logit transformation on the fitted estimates? For instance, lets say we are trying to predict coronary heart disease using BMI (exposure) and age (confounder). We could fit the model and estimate the odds of CHD across the range of values of BMI (adjusted for age), according to the model. We then take the inverse-logit of those fitted odds to transform them onto the probability scale.

Seemingly this way we now have an idea of the risk of the outcome across all values of the exposure, which we can plot or otherwise describe. But i've never seen this method recommended. Which makes me think there must be a conceptual issue with it??

p.s. lets assume we are performing a cross-sectional or cohort study (not a case-control study).

Answer 1

Let's look at an example model, where we use $\hat{\eta}$ to denote the (estimated) log-odds and $\hat{p}$ to denote the (estimated) probability.

$$g(\hat{p}) := \hat{\eta} = \hat{\beta}_0 + \hat{\beta}_1x_{1} + \hat{\beta}_2x_{2} $$

If we want to know the change in log-odds for each shift of $1$ in $x_1$, we take the derivative with respect to $x_1$.

$$\dfrac{\partial\hat{\eta}}{\partial x_1} = \hat{\beta}_1 $$

So regardless of the value of $x_1$, an increase of $1$ in $x_1$ results in a change in log-odds of $\hat{\beta}_1$.

If we want to look at the change in probability, however, then we have to use the inverse link function to isolate $\hat{p}$.

$$\hat{p} = \dfrac{\exp(\hat{\beta}_0 + \hat{\beta}_1x_{1} + \hat{\beta}_2x_{2})} {\exp(\hat{\beta}_0 + \hat{\beta}_1x_{1} + \hat{\beta}_2x_{2}) + 1} $$

Now take the partial derivative with respect to $x_1$.

$$ \dfrac{\partial \hat{p}} {\partial x_1} = \dfrac{\hat{\beta}_1\exp(\hat{\beta}_0 + \hat{\beta}_1x_{1} + \hat{\beta}_2x_{2})} {(\exp(\hat{\beta}_0 + \hat{\beta}_1x_{1} + \hat{\beta}_2x_{2}) + 1)^2} $$

This derivative depends on $x_1$, so the effect of $x_1$ on $\hat{p}$ is not constant.