# Formula of quantile regression?

We already know that the estimator of quantile regression defined by LAD (Least Absolut Deviation), minimizes sigma |e_i|.

But I also found a formula that minimize with an integral, not a sigma

Actually I can't understand why there are 2 different formulas. In what case that use a sigma or integral ? Are the distributions effect these formulas ?

• Where do the formulas come from, and what context are they presented in? What have they defined the symbols as in each case? Jan 27, 2017 at 6:29
• Conceptually, integration is a summation. Summing over a region and integrating over some region are conceptually similar. The top formula looks like it's estimating a linear function to give the quantile $q$. The bottom formula like it's estimating the value $q$ for quantile $\theta$ of a random variable with CDF $F_Y$. In the top, you're summing over your observations (basically, using the probability measure implicitly defined by your sample) and in the bottom, you're integrating over $dF$. Jan 27, 2017 at 7:07
• See stats.stackexchange.com/questions/251600 for an explanation of the integral expression and stats.stackexchange.com/questions/73623 for the connection between the sums and the integrals.
– whuber
Apr 17, 2018 at 13:02

The integral is used to define the (population) quantile of the random variable $Y$, while the summation is for the linear quantile regression, which aims at estimating the regression quantile coefficients $\beta_q$ using data $(y_i,x_i),i=1,\ldots,N$. Both employ the quantile check loss $\rho_{\theta}(u)=u(\theta-I(u<0))$. Indeed, the integral can be viewed as $E[\rho_{\theta}(Y-q)]$ and the summation is $\sum_{i=1}^N\rho_{\theta}(y_i-x_i'\beta_q)$. An analogy is the definition of the expectation vs. that of the least squares estimation. Specifically, $E(Y)$ can be viewed as the minimizer of the integral $\int_{-\infty}^{\infty}(y-m)^2\mathrm{d}F_Y(y)$ and the least squares estimator (of the coefficient vector) is obtained by minimizing $\sum_{i=1}^N(y_i-x_i'\beta_m)^2$.