I have been looking at quantile regression (since it is a much better method when trying to quantify welfare effects), and I am struggling with the following, standard model:


$y=x'\beta(u)$ where $u|x\text{~}Uniform\,[0,1]$ and for any $x,\, x'\beta(\tau)$ is a strictly increasing function in $\tau$.

For simplicity, assume $x=(1,x_1)$ meaning that we have one regressor and an intercept


What is $Var(y|x)$?

  • $\begingroup$ The way you've written it, your coefficient vector is a random variable, as in a Bayesian setting. Is this indeed the case? $\endgroup$ Mar 23, 2016 at 14:33
  • $\begingroup$ @AlecosPapadopoulos yes, the coefficient vector is a function of u which is a random variable with a uniform conditional distribution $\endgroup$
    – DornerA
    Mar 23, 2016 at 14:39

1 Answer 1


I cannot say how helpful this is for you but if your model is

$$y_i = a(u_i) + b(u_i)x_i, \;\; u_i|x_i \sim U[0,1] \;\;\forall i$$


$$\text{Var}(y_i\mid x_i) = \text{Var}[a(u_i)\mid x_i] + x_i^2\text{Var}[b(u_i)\mid x_i] +2x_i\text{Cov}[a(u_i),b(u_i)\mid x_i]$$

which tells you that $y_i$ is conditionally heteroskedastic. This can acquire a complicated integral expression if you use the "Law of unconscious statistician" and the fact the the denisty function of a $U[0,1]$ random variable is just unity:

$$\text{Var}(y_i\mid x_i) = \int_0^1[a(u)]^2du - \left(\int_0^1a(u)du \right)^2 \\ + x_i^2\cdot \left[\int_0^1[b(u)]^2du - \left(\int_0^1b(u)du\right)^2\right] \\ +2x_i\cdot \left[\int_0^1a(u)b(u)du - \int_0^1a(u)du\int_0^1b(u)du \right]$$

  • $\begingroup$ Thanks! This is definitely a start, so I'll see if I can work something out and get a less messy result $\endgroup$
    – DornerA
    Mar 23, 2016 at 22:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.