# Nonlinear quantile regression SSReg analogue

I have recently remembered that $$SSTot = SSRes + SSReg$$ fails to hold in the case of nonlinear regression.

$$y_i-\bar{y} = (y_i - \hat{y_i} + \hat{y_i} - \bar{y}) = (y_i - \hat{y_i}) + (\hat{y_i} - \bar{y})$$

$$( y_i-\bar{y})^2 = \Big[ (y_i - \hat{y_i}) + (\hat{y_i} - \bar{y}) \Big]^2 = (y_i - \hat{y_i})^2 + (\hat{y_i} - \bar{y})^2 + 2(y_i - \hat{y_i})(\hat{y_i} - \bar{y})$$

$$\sum_i ( y_i-\bar{y})^2 = \sum_i(y_i - \hat{y_i})^2 + \sum_i(\hat{y_i} - \bar{y})^2 + 2\sum_i\Big[ (y_i - \hat{y_i})(\hat{y_i} - \bar{y}) \Big]$$

Therefore, for a regression in general that predicts conditional expected value, $$SSTot =$$ $$SSRes + SSReg + Other$$. In the case of linear regression, the $$Other$$ term drops out due to orthogonality.

Consequently, I think that we can write the $$R^2$$ for nonlinear regression as:

$$R^2 =\dfrac{SSReg}{SSTot} = 1 - \dfrac{SSRes + Other}{SSTot}.$$

This becomes the regular $$R^2$$ when the regression is linear and $$Other=0$$. When the regression is nonlinear, we still have a calculation that R or Python can handle just fine.

However, I want something akin to $$R^2$$ for quantile regression. I have reasoned through a way to get something for the case of a linear quantile regression, which is what Koenker and Machado use in their 1999 JASA paper, according to another post on CrossValidated. (I can't access the paper, but this equation makes sense to me.) Let $$q$$ be the quantile being estimated by quantile regression, and let $$\bar y$$ be the marginal/pooled quantile (coming from an intercept-only quantile regression model, following Koenker and Machado (1999)).

$$R^2_q = 1 - \frac{\sum_{y_i \ge \hat{y_i}} q \vert y_i - \hat{y_i} \vert + \sum_{y_i < \hat{y_i}} (1-q) \vert y_i - \hat{y_i} \vert}{\sum_{y_i \ge \bar{y}} q \vert y_i - \bar{y} \vert + \sum_{y_i < \bar{y}} (1-q) \vert y_i - \bar{y} \vert}$$

This uses the loss function for the model divided by the loss function for the naive model that takes the pooled $$q^{th}$$ quantile as the estimate, analogous to how $$R^2$$ divides the loss function of the model by the loss function of the naive model that takes the estimate as the pooled mean.

This assumes that the analogous $$Other$$ term is zero for the linear quantile regression, bringing me to my question.

What is that $$Other$$ component for a nonlinear quantile regression?

Proceeding analogous to my first set of equations for the regular regression case, I start with $$\vert y_i - \bar{y} \vert = \vert y_i - \hat{y_i} + \hat{y_i} - \bar{y} \vert$$. Breaking this into something related to $$\vert y_i - \hat{y_i} \vert$$ and $$\vert \hat{y_i} - \bar{y} \vert$$ is eluding me.

REFERENCE

Koenker, Roger, and Jose AF Machado. "Goodness of fit and related inference processes for quantile regression." Journal of the American Statistical Association 94.448 (1999): 1296-1310.

• While I posted a self-answer, I remain interested in seeing a decomposition of the sum of absolute deviations or a proof that no such nice decomposition exists.
– Dave
Commented Jun 2, 2023 at 0:19

## 1 Answer

This assumes that the analogous $$Other$$ term is zero for the linear quantile regression, bringing me to my question.

I was mistaken about this. It is totally acceptable to compare the performance on this "pinball loss" of a model of interest to some kind of baseline model, such as prediction the marginal/pooled quantile every time. Thus, that $$R^2_q$$ equation I gave four years ago would describe the reduction of pinball loss compared to predicting the marginal/pooled quantile every time. This is quite analogous to the usual $$R^2$$ for square loss and seems to satisfy the original desire for something $$R^2$$-like for quantile regression.

• Thank you for answering this! (+1) Commented Jun 1, 2023 at 21:09