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$\newcommand{\var}{\mathrm{Var}}$In my textbook I am studying about heteroskedasticity in a linear regression model, and for doing the hypothesis test, it says our hypothesis is: $H_0: \var(u|x_1,x_2,...,x_k)=\sigma ^2$, however than it says:

Because we are assuming that $u$ has a zero conditional expectation, $\var(u|\mathbf{x})=E(u^2|\mathbf{x})$ (where $\mathbf{x}$ is all the regressors $x_1,x_2,x_3,...,x_n$), and so the null hypothessis of homoskedasticity is equivalent to: $$H_0: E(u^2|x_1,x_2,...,x_k)=E(u^2)=\sigma^2$$

Why is this true? I dont understand why we use $u^2$ and then regress the independent variables on it.

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  • $\begingroup$ Recall that $Var(u|x)=E(u^2|x)-E(u|x)^2$. If the second term is zero, as is assumed here, then we must have $Var(u|x)=E(u^2|x)$, for any $x$. Under $H_0$ we assume that $Var(u|x)=\sigma^2$. Putting these together, under $H_0$ we must have that $E(u^2|x)=\sigma^2$, for any $x$. $\endgroup$ – guest Nov 22 '12 at 17:28
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Becasue $Var(u|x)=E(u^2|x)-E^2(u|x)=E(u^2|x)-0^2=E(u^2|x)=E(u^2)=\sigma^2$, since $u$ has a zero conditional expectation. Here I used the fact that $u$ is independent of $x$. Therefore, (roughly speaking) each function of $u$ is independent of $x$ as well. So in particular $u^2$ is independent of $x$. This means that $E(u^2|x)=E(u^2)$.

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