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I am attempting to write a program that will (among other things) use the F-test in multivariate regression under standard robust errors. I am having trouble finding a specific formula for the F-statistic using robust errors.

I understand the purpose of robust errors. I simply need the modified formula for the F-statistic.

Can anyone help?

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  • $\begingroup$ Please note, not to be confused with testing the test for heteroskedasticity using the F-distribution. I'm talking about restricting a number of variables in the model under robust standard errors. $\endgroup$
    – Alex
    Apr 14, 2014 at 23:20

1 Answer 1

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In the linear regression model $$\mathbf y = \mathbf X\beta + \mathbf u$$ with $K$ regressors and a sample of size $N$, if we have $q$ linear restrictions on the parameters that we want to test,

$$\mathbf R\beta = \mathbf r$$

where $\mathbf R$ is $q \times K$, then we have the Wald statistic

$$W= (\mathbf R\hat \beta - \mathbf r)'(\mathbf R \mathbf {\hat V} \mathbf R')^{-1}(\mathbf R\hat \beta - \mathbf r) \sim_{asymp.} \mathcal \chi^2_q$$

and where $\mathbf {\hat V}$ is the consistently estimated heteroskedasticity-robust asymptotic variance-covariance matrix of the estimator,

$$\mathbf {\hat V}=(\mathbf X'\mathbf X)^{-1}\left(\sum_{i=1}^n \hat u_i^2\mathbf x_i'\mathbf x_i\right)(\mathbf X'\mathbf X)^{-1}$$ or

$$\mathbf {\hat V}=\frac {N}{N-K}(\mathbf X'\mathbf X)^{-1}\left(\sum_{i=1}^n \hat u_i^2\mathbf x_i'\mathbf x_i\right)(\mathbf X'\mathbf X)^{-1}$$

(there is some evidence that this degrees-of-freedom correction improves finite-sample performance).

If we divide the statistic by $q$ we obtain an approximate $F$-statistic

$$W/q \sim_{approx} F_{q, N-K}$$

but why add one more layer of approximation?

ADDENDUM 2-8-2014

The reason why we obtain an approximate $F$-statistic if we divide a chi-square by its degrees of freedom is because

$$\lim_{N-K \rightarrow \infty} qF_{q, N-K} = \chi^2_q$$

see this post.

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  • $\begingroup$ How do you prove that $W/q \sim^{approx} F_{q,(N-k)}$ ? I trust this is correct, but I have never actually seen a proof. $\endgroup$ Feb 10, 2022 at 8:52
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    $\begingroup$ @MichaelGmeiner Please follow the link in the Addendum. $\endgroup$ Feb 15, 2022 at 1:59

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