I find myself puzzled by a passage about marginal effects of discrete variables in quantile regression. On p. 217 of Cameron and Trivedi's MUS book, the authors write:

For the $j$th (continuous) regressor, the ME is $$\frac{\partial Q_\tau(y\vert x)}{\partial x_j}=\beta_{\tau j}$$ As for linear least-squares regression, the ME is given by the slope coefficient and is invariant across individuals, simplifying analysis. The interpretation is somewhat delicate for discrete changes that are more than infinitessimal, however, because the partial derivative measures the impact of a change in $x_j$ under the assumption that the in­dividual remains in the same quantile of the distribution after the change. For larger changes in a regressor, the individual may shift into a different quantile.

Above, $y$ is the outcome, $\tau$ is the quantile, $x$ is the vector of covariates.

I am not sure what the bold part means. Is that just a different way of saying that while we can estimate the average treatment effect for each individual, $\mathbb E(Y_1-Y_0)$, with $\mathbb E(Y_1)-\mathbb E(Y_0)$ with quality experimental data, the median treatment effect $$median(Y_1-Y_0)\ne median(Y_1)- median(Y_0)?$$

The only other passage I've found on this topic is on p. 48 of Roger Koenker's Quantile Regression:

The interpretation of the partial derivative itself, $\partial Q_\tau(y\vert x)/\partial x_j$, often requires considerable care. We emphasized earlier in the context of the two-sample problem that the Lehmann–Doksum quantile treatment effect is simply the response necessary to keep a respondent at the same quantile under both control and treatment regimes. Of course, this is not to say that a particular subject who happens to fall at the τth quantile initially, and then receives an increment $\Delta x_j$ , say, another year of education, will necessarily fall on the $\tau$th conditional quantile function following the increment. Indeed, as much of the recent literature on treatment effects has stressed (see, e.g., Angrist, Imbens, and Rubin, 1996), we are typically unable to identify features of the joint distribution of control and treatment responses because we do not observe responses under both regimes for the same subjects.

  • $\begingroup$ Quantile regression is stated in terms of increments in quantiles (your right hand side) and not quantiles of differences (your left hand side). So you can estimate increments in quantiles. More commonly I plot predicted quantiles varying one $X$ hold other $X$s to their median/mode. $\endgroup$ Nov 12, 2014 at 20:06

1 Answer 1


The partial derivative does not necessarily identify the causal impact of a binary treatment. Some assumptions about the joint distribution of the treated and untreated outcomes are needed for this. If a treatment causes rank reversals in the distribution, then knowing the difference of $\tau$th quantile for two distributions is not enough to calculate the $\tau$th quantile treatment effect. You can say that if any of the QTEs are negative, then the treatment effect must also be negative for some non-degenerate interval of the counterfactual outcome distribution. When changes in $x$ are small, rank reversals are unlikely, so we don't worry about them. If you make the assumption of rank preservation, the the partial with respect to a binary $x$ can be interpreted as the quantile treatment effect. This assumption means that the treated outcomes are a monotone transformation of the untreated outcomes and that transformation need not be the same at all quantiles. Bitler, Gelbach and Hoynes (2005) develop a test of rank preservation. If rank preservation holds, the distributions of $x$s not affected by the treatment should be the same in the treatment and control group at each quantile.

Another assumption is the common effect model, where the impact of treatment is the same at all quantiles. The whole outcome distribution shifts over by the impact of the treatment. This one is much more restrictive, but will give you the same interpretation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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