1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Dimitriy V. Masterov Apr 16 '19 at 2:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.