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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:

$\textbf{Model:}$

$y=x'\beta(u)$ 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

$\textbf{Question:}$

What is $Var(y|x)$?

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  • $\begingroup$ The way you've written it, your coefficient vector is a random variable, as in a Bayesian setting. Is this indeed the case? $\endgroup$ – Alecos Papadopoulos Mar 23 '16 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 '16 at 14:39
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I cannot say how helpful this is for you but if your model is

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

Then

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

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:

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

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  • $\begingroup$ 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 '16 at 22:57

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