# Quantile Regression loss function/ check function proof

Can anyone help me to show that this statement is true. I have looked in Koenker's Quantile Regression (2000) and a load of other sources but I cannot find a solution. There seems to be a trick required that I am unaware of.

$$\rho _ { u } ( z ) = [ u 1 ( z \geq 0 ) + ( 1 - u ) 1 ( z < 0 ) ] \times | z | = [u-1(z<0)]z$$

note $$u$$ is the quantile and is a number between 0 and 1, $$1 ( z \geq 0 )$$ is a dummy variable which is 1 if $$z \geq 0$$ and 0 otherwise, and $$z$$ is a real number.

Thank you for any help.

So far this is my best effort

$$[ u 1 ( z \geq 0 ) + ( 1 - u ) 1 ( z < 0 ) ]$$

because $$u\in[0,1]$$ $$( 1 - u )=|( u - 1 )|$$ so $$[ u 1 ( z \geq 0 ) + |( u - 1 )| 1 ( z < 0 ) ]$$ then $$[ u (1-1 ( z < 0 )) + |( u - 1 )| 1 ( z < 0 ) ]$$

if i am then able to able to remove the modulus from $$|(u-1)|$$ so it becomes $$(u-1)$$, I will get

$$[ u - u1 ( z < 0 )) + u1 ( z < 0 ) - 1 ( z < 0 ) ]$$ which is equal to

$$[ u - 1 ( z < 0 ) ]$$

Is there a way that I can justify removing the modulus? Considering the modulus has been removed from $$z$$ in the original expression I think this can be done but I am not sure why?

• Have you perused this question? If that does not help, perhaps you can define what the terms mean to improve the odds of getting a better answer. – Dimitriy V. Masterov Apr 16 '19 at 2:56