英文:
Find Pseudo R-squared for quantile regression models
问题
I have tried to implement quantile regression for the Boston dataset.
library(MASS)data(Boston)attach(Boston)qr_res_0.9 <- rq(medv ~ lstat + rm + crim + dis,tau = 0.9,data = Boston)
Now, results of model summary is shown below
summary(qr_res_0.9,se="boot")#### Call: rq(formula = medv ~ lstat + rm + crim + dis, tau = 0.9, data = Boston)#### tau: [1] 0.9#### Coefficients:## Value Std. Error t value Pr(>|t|)## (Intercept) -20.75975 13.81979 -1.50218 0.13368## lstat -0.25857 0.22411 -1.15378 0.24914## rm 9.01335 1.58000 5.70464 0.00000## crim -0.04028 0.11367 -0.35440 0.72319## dis -0.94489 0.29403 -3.21355 0.00140
This does not include Psuedo R-squared/McFadden R-squared value? How can I estimate this?
What I have tried?
Referring to the discussion in https://stackoverflow.com/questions/19861194/extract-r2-from-quantile-regression-summary
I have implemented the following
rho <- function(u, tau = 0.5) u * (tau - (u < 0))V <- sum(rho(qr_res_0.9$resid, qr_res_0.9$tau))V## [1] 558.4133
The R-squared should be between 0 to 1?
英文:
I have tried to implement quantile regression for the Boston dataset.
library(MASS)data(Boston)attach(Boston)qr_res_0.9 <- rq(medv ~ lstat + rm + crim + dis,tau = 0.9,data = Boston)
Now, results of model summary is shown below
summary(qr_res_0.9,se="boot")#### Call: rq(formula = medv ~ lstat + rm + crim + dis, tau = 0.9, data = Boston)#### tau: [1] 0.9#### Coefficients:## Value Std. Error t value Pr(>|t|)## (Intercept) -20.75975 13.81979 -1.50218 0.13368## lstat -0.25857 0.22411 -1.15378 0.24914## rm 9.01335 1.58000 5.70464 0.00000## crim -0.04028 0.11367 -0.35440 0.72319## dis -0.94489 0.29403 -3.21355 0.00140
This does not include Psuedo R-squared/McFadden R-squared value? How can I estimate this?
What I have tried?
Referring to the discussion in https://stackoverflow.com/questions/19861194/extract-r2-from-quantile-regression-summary
I have implemented the following
rho <- function(u,tau=.5)u*(tau - (u < 0))V <- sum(rho(qr_res_0.9$resid, qr_res_0.9$tau))V## [1] 558.4133
The R-squared should be between 0 to 1?
答案1
得分: 1
以下是翻译的内容：
显然，您正在尝试计算Koenker和Machado R1：
链接：https://stats.stackexchange.com/a/129246/11849
library(MASS)library(quantreg)data(Boston)# 不要使用 `attach`qr_res_0.9 <- rq(medv ~ lstat + rm + crim + dis,tau = 0.9,data = Boston)qr_res_0.9_0 <- rq(medv ~ 1,tau = 0.9,data = Boston)rho <- function(u, tau = 0.5) u * (tau - (u < 0))Vhat <- sum(rho(qr_res_0.9$resid, qr_res_0.9$tau))V0 <- sum(rho(qr_res_0.9_0$resid, qr_res_0.9_0$tau))R1 <- 1 - Vhat / V0#[1] 0.4659297
从链接的回答中：
我认为R^2的概念不太适用于分位数回归。您可以定义各种更或少类似的量，如此处所示，但无论您选择什么，都不会具有OLS回归中真实R^2的大多数属性。
英文:
Apparently, you are trying to calculate the Koenker and Machado R1:
https://stats.stackexchange.com/a/129246/11849
library(MASS)library(quantreg)data(Boston)#don't use `attach`qr_res_0.9 <- rq(medv ~ lstat + rm + crim + dis,tau = 0.9,data = Boston)qr_res_0.9_0 <- rq(medv ~ 1,tau = 0.9,data = Boston)rho <- function(u,tau=.5)u*(tau - (u < 0))Vhat <- sum(rho(qr_res_0.9$resid, qr_res_0.9$tau))V0 <- sum(rho(qr_res_0.9_0$resid, qr_res_0.9_0$tau))R1 <- 1-Vhat/V0#[1] 0.4659297
From the linked answer:
> I don't think the concept of R^2 translates well to quantile
> regression. You can define various more-or-less analogous quantities,
> as here, but no matter what you choose, you won't have most of the
> properties real R^2 has in OLS regression.
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