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I was reading the book introductory econometrics by "Wooldridge", and in Chapter 8 (Heteroscedasticity), it is stated that (see pink part)

enter image description here

I could not understand, if $u$ and $x$ are uncorrelated, then how weighted $u/\sqrt{h(x)}$ and $x/\sqrt{h(x)}$ can be uncorrelated. Since both terms are divided by common denominator ie $h(x)$, then it should cause some correlation between the both. I just wanted to know which statistical properties is applied in this case ?

I also simulated the same, and found that when divided by common denominator, then both variables are highly correlated:

#R Code:
rm(list = ls())

n = 5000
x1 <- rnorm(n)
x2 <- rnorm(n)
u <- runif(n)

#correlation between u and x
cor(x1, u)
[1] -0.01413626

cor(x2, u)
[1] -0.02338137

#taking weighted x and u; h(x) is x1 + x2
x1w <- x1/(x1 + x2)
x2w <- x2/(x1 + x2)
uw <- u/(x1 + x2)

#correlation
cor(uw, x1w)
[1] 0.9085537

cor(uw, x2w)
[1] -0.9085537
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  • $\begingroup$ because $$\Bbb E[ux_j/h(x)]=\Bbb E\{\Bbb E[ux_j/h(x)|x]\}=\Bbb E\{\Bbb E[u|x]x_j/h(x)\}=0$$ $\endgroup$ – Xi'an May 13 at 8:55
  • $\begingroup$ Regarding your second paragraph, $\mathbb{E}(u|x)$ is stronger than $\text{Corr}(u,x)=0$. $\endgroup$ – Richard Hardy May 13 at 9:08
  • $\begingroup$ I got your point. Then, why the simulating results shows very high correlation ? $\endgroup$ – Neeraj May 13 at 9:08
  • $\begingroup$ @RichardHardy Rightly said. Because $E(u|x) = 0$ assumes $u$ and $x$ are independent. $\endgroup$ – Neeraj May 13 at 9:09
  • $\begingroup$ Mean-independent, to be precise. We do not have information about the dependence or lack thereof between other distributional characteristics of $u$ and $x$. $\endgroup$ – Richard Hardy May 13 at 9:43

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