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Can someone explain the intuition behind the Haussman test for exogenity and the equation for H. what does it mean, that "it is normalised by the covariance of the difference". enter image description here

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I think it means: $\bf d \in R^p$ is asymptotically multivariate normal with mean vector $\bf 0$ and some covariance matrix $Var[\bf d]$. That is, $\bf d \rightarrow N_p(\bf 0,Var[\bf d])$ (by CLT).

Normalizing means dividing by the square root of the variance (or, in matrix form, multiplying $\bf d$ by the variance matrix raised to $-1/2$) so that the asymptotic distribution because multivariate standard normal:

$$ Var[\bf d]^{-1/2}\bf d \rightarrow N_p(\bf 0, I_p)$$ Above, $\bf I_p$ is the $p\times p$ identity matrix and arrow indicates convergence in distribution. This is similar to "normalizing" any variable $x \sim N(\mu, \phi)$...you normalize by subtracting by the mean and dividing by the standard deviation. The resulting distribution is $\phi^{-1/2}(x-\mu) \sim N(0,1)$. Hope that's clear.

By the way, squaring the left hand side just gets you from a Normal to a $\chi^2$ since the square of standard normal random variable is $\chi^2$.

Specifically, $$(Var[\bf d]^{-1/2}\bf d)^TVar[\bf d]^{-1/2}\bf d = d^TVar[\bf d]^{-1/2}Var[\bf d]^{-1/2} \bf d = d^TVar[\bf d]^{-1} \bf d$$

This is the $\chi^2$ test statistics - equivalent to the standard normal test statistic.

If the variance is unknown, you can substitute in any consistent estimator of the variance matrix and maintain the convergence in distribution result due to Slutsky's theorem.

Note the degrees of freedom of the $\chi^2$ is determined by how many null hypotheses you are testing.

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  • $\begingroup$ If only all answers were like yours: non-elitist, fully-understood and explained from A to Z. +1. $\endgroup$ – keepAlive Oct 8 '18 at 20:07

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