I am estimating the following logistic regression (binomial family) by maximum likelihood:

$$ \ln\left(\frac{Y}{1-Y}\right) = \beta_{0} + \beta_{1}D + \beta_{2}X + \epsilon$$

where D is a dummy. I am interested in the marginal effect at the mean of $D$, i.e. $\beta_{1}$ is of interest. What does the marginal effect at the mean look like in equation form? Would it be:

$$ \frac{1}{\bar{Y}} \left(\frac{e^{\beta_{0}+\beta_{1}+\beta_{2}\bar{X}}}{1+e^{\beta_{0}+\beta_{1}+\beta_{2}\bar{X}}} - \frac{e^{\beta_{0}+\beta_{2}\bar{X}}}{1+e^{\beta_{0}+\beta_{2}\bar{X}}}\right) $$


$$\left(\frac{e^{\beta_{0}+\beta_{1}+\beta_{2}\bar{X}}}{1+e^{\beta_{0}+\beta_{1}+\beta_{2}\bar{X}}} - \frac{e^{\beta_{0}+\beta_{2}\bar{X}}}{1+e^{\beta_{0}+\beta_{2}\bar{X}}}\right) $$

My intuition is that Y corresponds to the probability of the event occurring, and the e/(1+e) terms also correspond to the probability of the event occurring, but at specific values of Y. What would the rationale for dividing by $\bar{Y}$ be? Thank you.

  • $\begingroup$ despite its interest to economists, this question is likely to attract more answers at the Stats StackExchange (stats.stackexchange.com) $\endgroup$
    – emeryville
    Dec 29, 2019 at 10:41
  • $\begingroup$ The marginal effect of $D$ on which indicator? What is $Y$? Is it really "the probability of the event occurring"? You should indicate the "LHS" function whose variations you want to measure after a change of $D$ by one unit. $\endgroup$
    – Bertrand
    Jan 30, 2020 at 7:34

1 Answer 1



OP's second expression corresponds to $$\Delta p = P(Y=1|D=1,X=\bar{X}) - P(Y=1|D=0,X=\bar{X}),$$ which is $$ \Delta p = \Lambda(\beta_0 + \beta_1 + \beta_2 \bar{X}) - \Lambda(\beta_0 + \beta_2 \bar{X}), $$ where $\Lambda(z) = e^z/(1+e^z)$.

Rationale for dividing by p

OP's first expression corresponds to $(\Delta p)/p$, where $p$ is the proportion of 1's in the sample $P(Y=1)$. This expresses the MEA $\Delta p$ as a ratio to $p$. This is informative as it lets us have better ideas on how large $\Delta p$ is. (For example, if we just have $\Delta p=0.01$, we don't know how large it is. But if $p=0.02$ is also given, then we know $\Delta p=0.01$ is a quite large effect as it is 50% of $p$.) One might call it "MEA as a ratio to Y-bar". Anyway, MEA usually refers to $\Delta p$, but if you mean $(\Delta p)/p$ by it, it should be OK as long as you explain it.


The MEA associated with $D$ can also mean $\dot{p}=\partial P(Y=1|D=\bar{D},X=\bar{X})/\partial \bar{D}$. As $$P(Y=1|D=\bar{D},X=\bar{X}) = \Lambda(\beta_0+\beta_1 \bar{D} + \beta_2 \bar{X})$$ for your logit model, we have $$\dot{p} = \frac{\partial \Lambda(\beta_0+\beta_1 \bar{D} + \beta_2 \bar{X})}{\partial \bar{D}} = \beta_1 \lambda(\beta_0 + \beta_1 \bar{D} + \beta_2 \bar{X}),$$ where $\lambda(z) = \Lambda'(z) = e^z / (1+e^z)^2$. Note that $D$ is binary but $\bar{D}$ is continuous so differentiation with respect to $\bar{D}$ makes sense. $\Delta p$ and $\dot{p}$ have different interpretations. I personally think $\Delta p$ in Part A is easier to understand as the notion of $P(Y=1|D=\bar{D}, X=\bar{X})$ is a bit awkward.


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