I am using quantile regression to find predictors of 90th percentile of my data. I am doing this in R using the quantreg package. How can I determine $r^2$ for quantile regression which will indicate how much of variability is being explained by predictor variables?

What I really want to know: "Any method I can use to find how much of variability is being explained?". Significance levels by P values is available in output of command: summary(rq(formula,tau,data)). How can I get goodness of fit?

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    $\begingroup$ $R^2$ is not relevant to quantile regression. $\endgroup$
    – whuber
    Commented Dec 15, 2014 at 17:24
  • $\begingroup$ @whuber : Any alternative method I can use to find how much of variability is being explained? $\endgroup$
    – rnso
    Commented Dec 15, 2014 at 17:27
  • 2
    $\begingroup$ That would be a good thing to ask in the body of your question, rather than buried in a comment! "Variability explained" (as measured in terms of variances, anyway) is essentially a least-squares concept; perhaps what you want is an appropriate measure of statistical significance or possibly goodness of fit. $\endgroup$
    – whuber
    Commented Dec 15, 2014 at 17:32
  • $\begingroup$ For any figure of merit you need to consider what would be good performance, what would be poor performance and what would be irrelevant. For example, it is no criticism of the 90th percentile if that is a lousy predictor of the 10th percentile. Your benchmark might be whatever you might use if you were not using quantile regression. If your predictors are continuous, that could be hard to define. $\endgroup$
    – Nick Cox
    Commented Dec 15, 2014 at 17:40
  • 1
    $\begingroup$ @whuber : I have added that in body of the question. Significance level by P value is available in summary(rq(formula,tau,data)) output. How can I get goodness of fit? $\endgroup$
    – rnso
    Commented Dec 15, 2014 at 17:51

2 Answers 2


Koenker and Machado$^{[1]}$ describe $R^1$, a local measure of goodness of fit at the particular ($\tau$) quantile.

Let $V(\tau) = \min_{b}\sum \rho_\tau(y_i-x_i'b)$

Let $\hat{\beta}(\tau)$ and $\tilde{\beta}(\tau)$ be the coefficient estimates for the full model, and a restricted model, and let $\hat{V}$ and $\tilde{V}$ be the corresponding $V$ terms.

They define the goodness of fit criterion $R^1(\tau) = 1-\frac{\hat{V}}{\tilde{V} }$.

Koenker gives code for $V$ here,

rho <- function(u,tau=.5)u*(tau - (u < 0))
V <- sum(rho(f$resid, f$tau))

So if we compute $V$ for a model with an intercept-only ($\tilde{V}$ - or V0 in the code snippet below) and then an unrestricted model ($\hat{V}$), we can calculate an R1 <- 1-Vhat/V0 that's - at least notionally - somewhat like the usual $R^2$.

Edit: In your case, of course, the second argument, which would be put in where f$tau is in the call in the second line of code, will be whichever value of tau you used. The value in the first line merely sets the default.

'Explaining variance about the mean' is really not what you're doing with quantile regression, so you shouldn't expect to have a really equivalent measure.

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. You need to be clear about what properties you need and what you don't -- in some cases it may be possible to have a measure that does what you want.


$[1]$ Koenker, R and Machado, J (1999),
Goodness of Fit and Related Inference Processes for Quantile Regression,
Journal of the American Statistical Association, 94:448, 1296-1310

  • $\begingroup$ Should tau=0.9 be rather than 0.5? $\endgroup$
    – dimitriy
    Commented Dec 16, 2014 at 0:27
  • $\begingroup$ Yes, it should, but if you supply the correct second argument (as is done in the second line I quoted above), that's how it works. The value of 0.5 in the first line is simply a default argument if you don't specify tau when you call the function. I will clarify in the post. $\endgroup$
    – Glen_b
    Commented Dec 16, 2014 at 0:36
  • $\begingroup$ @Glen_b Thanks for the explanation. Unless I am doing something stupid, V appears to be the sum of weighted deviations about estimated quantile, rather than a pseudo-$R^2$. $\endgroup$
    – dimitriy
    Commented Dec 16, 2014 at 6:18
  • $\begingroup$ @Dimitriy Uh, you're right, I left something out. I will fix this up shortly. $\endgroup$
    – Glen_b
    Commented Dec 16, 2014 at 7:39
  • $\begingroup$ @Dimitriy I think I have fixed it now. $\endgroup$
    – Glen_b
    Commented Dec 16, 2014 at 12:08

The pseudo-$R^2$ measure suggested by Koenker and Machado (1999) in JASA measures goodness of fit by comparing the sum of weighted deviations for the model of interest with the same sum from a model in which only the intercept appears. It is calculated as

$$R_1(\tau) = 1 - \frac{\sum_{y_i \ge \hat y_i} \tau \cdot \vert y_i-\hat y_i \vert +\sum_{y_i<\hat y_i} (1-\tau) \cdot \vert y_i-\hat y_i \vert}{\sum_{y_i \ge \bar y} \tau \cdot \vert y_i-\bar y \vert +\sum_{y_i<\bar y_i} (1-\tau) \cdot \vert y_i-\bar y \vert},$$

where $\hat y_i =\alpha_{\tau}+\beta_{\tau}x$ is the fitted $\tau$th quantile for observation $i$, and $\bar y=\beta_{\tau}$ is the fitted value from the intercept-only model.

$R_1(\tau)$ should lie in $[0,1]$, where 1 would correspond to a perfect fit since the numerator which consists of the weighted sum of deviations would be zero. It a local measure of fit for QRM since it depends on $\tau$, unlike the global $R^2$ from OLS. That is arguably the source of the warnings about using it: if you model fits in the tail, there's not guarantee that it fits well anywhere else. This approach could also be used to compare nested models.

Here's an example in R:


fit0 <- rq(foodexp~1,tau=0.9,data=engel)
fit1 <- rq(foodexp~income,tau=0.9,data=engel)

rho <- function(u,tau=.5)u*(tau - (u < 0))
R1 <- 1 - fit1$rho/fit0$rho

This could probably be accomplished more elegantly.

  • $\begingroup$ Your formula does not display well. After the minus sign in: R_1(\tau) = 1 - 􀀀 the last character is some kind of mess. Could you check that? Maybe you pasted some non-standard character instead of using Tex. $\endgroup$
    – Tim
    Commented Dec 16, 2014 at 8:02
  • $\begingroup$ @Tim I don't see anything strange, either in the tex or on the screen. $\endgroup$
    – dimitriy
    Commented Dec 16, 2014 at 17:01
  • $\begingroup$ It looks like this on both linux and windows: snag.gy/ZAp5T.jpg $\endgroup$
    – Tim
    Commented Dec 16, 2014 at 17:04
  • $\begingroup$ @Tim That box doesn't correspond to anything, so it can be ignored. I will try to edit it out later from another machine. $\endgroup$
    – dimitriy
    Commented Dec 16, 2014 at 17:06
  • 2
    $\begingroup$ Your function rho is not used after, and is actually more confusing?! $\endgroup$
    – Matifou
    Commented Jun 6, 2021 at 19:42

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