Does the "divide by 4 rule" give the upper bound marginal effect? In the logisitic regression chapter of "Data Analysis Using Regression and
Multilevel/Hierarchical Models" by Gelman and Hill, The "Divide by 4" rule is presented to approximate average marginal effects.

Essentially, dividing the estimated log-odds ratio gives the maximum slope (or the maximum change in probabilities) of the logistic function.

Since the text above states that the "divide by 4 rule" gives the maximum change in $P(y=1)$ with a unit change in x, why is the estimated 8% less than the 13% calculated from actually taking the derivative of the logistic function in the example given?
Does the "divide by 4 rule" actually give the upper bound marginal effect?
Other "divide by 4" resources:


*

*Using the "Divide by 4 Rule" to Interpret Logistic Regression Coefficients

*Divide by 4 Rule for Marginal Effects - Econometric Sense

*http://vulstats.ucsd.edu/pdf/Gelman.ch-05.logistic-regression.pdf
 A: For a continuous variable $x$, the marginal effect of $x$ in a logit model is
$$\Lambda(\alpha + \beta x)\cdot \left[1-\Lambda(\alpha + \beta x)\right]\cdot\beta = p \cdot (1 - p) \cdot \beta,$$ where the inverse logit function $\Lambda$ is
$$\Lambda(z)=\frac{\exp{z}}{1+\exp{z}}.$$
Here $p$ is a probability, so the factor $p\cdot (1-p)$ is maximized when $p=0.5$ at $0.25$, which is where the $\frac{1}{4}$ comes from. Multiplying by the coefficient gives you the upper bound on the marginal effect. Here it is 
$$0.25\cdot0.33 =0.0825.$$
Calculating the marginal effect at the mean income yields,
$$\mathbf{invlogit}(-1.40 + 0.33 \cdot 3.1)\cdot \left(1-\mathbf{invlogit}(-1.40 + 0.33 \cdot3.1)\right)\cdot 0.33 = 0.07963666$$
These are pretty close, with the approximate maximum marginal effect bounding the marginal effect at the mean.
A: I think it's a typo.
The derivative of the logistic curve with respect to $x$ is:
$$
\frac{\beta\mathrm{e}^{\alpha + \beta x}}{\left(1 + \mathrm{e}^{\alpha + \beta x}\right)^{2}}
$$
So for their example where $\alpha = -1.40, \beta = 0.33$ it is:
$$
\frac{0.33\mathrm{e}^{-1.40 + 0.33 x}}{\left(1 + \mathrm{e}^{-1.40 + 0.33 x}\right)^{2}}
$$
Evaluated at the mean $\bar{x}=3.1$ gives:
$$
\frac{0.33\mathrm{e}^{-1.40 + 0.33 \cdot 3.1}}{\left(1 + \mathrm{e}^{-1.40 + 0.33\cdot 3.1}\right)^{2}} = 0.0796367
$$
This result is very close to the maximum slope of $0.33/4 = 0.0825$ which is attained at $x=-\frac{\alpha}{\beta}=4.24$, supporting their claim.
On page 82, they write

But $0.33\mathrm{e}^{-0.39}/\left(1+\mathrm{e}^{-0.39}\right)^{2}\neq 0.13$. Instead, it's around $0.08$, as shown above.
