The "check function" in quantile regression is defined as

$\rho_\tau(u) = u(\tau-1_{\{u<0\}})$

I do understand the basic princible of quantile regression. Now I tried to dig a bit deeper to understand the basic algebra behind it. Now my, probably very trivial question regarding the above mention function:

I read the original paper by Koenker and the $u$ is not defined formally. The same goes for a few other papers I have read. The only comment I found was: "We should think of $u$ as an individual error $u=y-r$ and $ρ_τ(u)$ as the loss associated with $u$." (Arellano). What does this mean? What is $u$? Is it simply the residual?

I am aware that my question might be rather trivial, but I tried my best in finding a proper explanation and I just dont get it.


1 Answer 1


The check function stems from applying an optimization view of expressing the $\tau$-th sample quantile of a sample $\{Y_1, \ldots, Y_n\}$.

Conventionally, given an observed sample $Y_1, \ldots, Y_n$, the $\tau$-th sample quantile $\hat{Q}_Y(\tau)$ is defined by ranking, i.e., $\hat{Q}_Y(\tau)$ is the $\lfloor n\tau \rfloor$-th order statistic of $(Y_1, \ldots, Y_n)$. In a completely different point of view, it can be shown that it is also the solution of the following optimization problem: $$\hat{Q}_Y(\tau) = \text{argmin}_{\xi} \sum_{i = 1}^n \rho_\tau(Y_i - \xi). \tag{1}$$

An intuitive proof of this fact can be found in Qunantile Regression (2005), Section 1.3, by Roger Koenker. In view of $(1)$, the $\tau$-th sample quantile receives a new interpretation as the minimizer of some loss function which is determined by the check function $\rho_\tau(\cdot)$. This is in agreement with more standardized results for the least-squares estimate and the least-absolute-deviation estimate, as the following chart shows:

enter image description here

The extension from the one-sample problem above to the regression setting is straightforward, which simply replaces the $\xi$ in $(1)$ by the regression function $x'b$ (yes, the aim here is to minimize the total "loss" of residuals, where "loss" is clearly defined by the $\rho_\tau(\cdot)$:

$$\hat{\beta}(\tau) = \text{argmin}_{b \in \mathbb{R}^p} \sum_{i = 1}^n \rho_\tau(Y_i - x_i'b).$$

$\hat{\beta}(\tau)$ is referred to as $\tau$-th regression quantile, which, by virtue of the property of $\rho_\tau(\cdot)$, also bears some interesting ordering interpretation relative to the fitted regression quantile surface $y = x'\hat{\beta}(\tau)$, for details, see remark on page 40 of Regression Quantiles (1978) by Koenker and Bassett.

In summary, the check function is a loss function that retrieves the $\tau$-th sample quantile, and more importantly, that makes the generalization from the one-sample problem (where ordering is possible) to the regression problem (where ordering is rather awkward) practical.

  • $\begingroup$ Alright, thank you. But the u-variable which is part of the check function, which I described in my question - what does it stand for? Does it stand for the residuals in the case of the regression setting? $\endgroup$
    – shenflow
    Jan 6, 2018 at 22:37
  • 1
    $\begingroup$ $u$ is just the argument of the function $\rho_\tau$ (it can be whatever letter you can use). Itself doesn't have any particular meaning (I don't think the comment you listed in your question meaningful). When defining the regression quantile, the argument of $\rho_\tau$ is the residual, defined as the discrepancy between the observation $Y_i$ and the fitted value $x_i'b$. What really matters is not $u$, but the "check pattern" of $\rho_\tau$, whose property is described in my answer. $\endgroup$
    – Zhanxiong
    Jan 6, 2018 at 23:04
  • $\begingroup$ Hello @Zhanxiong I'm wondering, is it possible to derive a gradient for this function analytically and solve using gradient descent? $\endgroup$
    – Bar
    Nov 2, 2018 at 15:41
  • $\begingroup$ @Bar: As it is linear, it is usually solved via linear programing. By drawing it, you will see it misses a derivative at one point. For code see stats.stackexchange.com/questions/260504/… $\endgroup$ Nov 10, 2021 at 11:08

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