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With a linear model, estimating an OLS regression of y on a constant only will give us the mean of y. I was wondering whether the same would not hold with a quantile regression?

In other words, is running a quantile regression of y on a constant equivalent to estimating the unconditional quantile? My intuition was telling me that yes, but a quick example in R shows me this is not necessarily the case? Or are they supposed to be equivalent, yet there is some small machine imprecision between algorithms?

Maybe this is linked to the post on What is the difference between conditional and unconditional quantile regression? My motivation for wondering about the quantile regression comes from the quantile pseudo $R^2$, which contains precisely a quantile regression on a constant, see R-squared in quantile regression

> set.seed(1)
> y <- rnorm(100)
> 
> ## Numerically equivalent for linear regression
> all.equal(coef(lm(y~1)), mean(y), check.attributes = FALSE)
> #> [1] TRUE
> sapply(1:9, function(i) quantile(y, probs=0.3, type=i))
       30%        30%        30%        30%        30%        30%        30%        30%        30% 
-0.3942900 -0.3807557 -0.3942900 -0.3942900 -0.3807557 -0.3861694 -0.3753420 -0.3825603 -0.3821091 
> coef(quantreg::rq(y~1, method="fn", tau=0.3))
(Intercept) 
 -0.3672215 
> coef(quantreg::rq(y~1, method="br", tau=0.3))
(Intercept) 
 -0.3672215 
Warning: In rq.fit.br(x, y, tau = tau, ...) : Solution may be nonunique

Created on 2021-06-08 by the reprex package (v2.0.0)

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  • $\begingroup$ I took the liberty of inserting set.seed(1) and updating the output, so we all talk about the same thing here. $\endgroup$ – Stephan Kolassa Jun 9 at 6:30
  • $\begingroup$ If you use a sample size where all quantiles are uniquely defined, then you should get the same. $\endgroup$ – Michael M Jun 9 at 6:34
  • $\begingroup$ Try an odd number for your sample size, not $100$. I think the quantile regression package and the default settings in the quantile function disagree about what to do with an even sample size. $\endgroup$ – Dave 2 days ago
  • $\begingroup$ dave you got it, and now @MichaelM I see what you meant! Want to add that as an answer? $\endgroup$ – Matifou yesterday
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You can generally reproduce the desired behaviour

  1. if you use the inverse ECDF method to calculate the empirical quantiles and
  2. if $np$ is not an integer ($n$ is the sample size, $p$ the probability).

Here the code:

library(quantreg)

#====================================
# n*probs is an integer
#====================================

n <- 100
set.seed(1)
y <- rnorm(n)
p <- 0.3

if (n*p == trunc(n*p)) {
  print("quantile not uniquely defined")
}

# Comparison: FALSE
all.equal(
  coef(rq(y ~ 1, tau = p)),
  quantile(y, probs = p, type = 1),
  check.attributes = FALSE
)

#====================================
# n*probs not an integer
#====================================

n <- 99
set.seed(1)
y <- rnorm(n)
p <- 0.3

if (n*p == trunc(n*p)) {
  print("quantile not uniquely defined")
}

# Comparison: TRUE
all.equal(
  coef(rq(y ~ 1, tau = p)),
  quantile(y, probs = p, type = 1),
  check.attributes = FALSE
)

#====================================
# Factor as covariable
#====================================

set.seed(1)
x <- rep(0:1, each = 99)
y <- rnorm(99 * 2) + x

# Comparison: TRUE
all.equal(
  coef(rq(y ~ 0 + factor(x), tau = 0.3)),
  c(by(y, x, quantile, type = 1, probs = 0.3)),
  check.attributes = FALSE
)

The last snippet shows how to calculate quantiles within subgroups with the same trick.

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