The linear model is

$$y_{i}= x_{i}'\beta+u_{i}$$

When written in vector notation such that $y_{i}$ is a $1$ x $1$ matrix of outcomes, $x_{i}'$ is a $1$ x $k$ matrix of control variables, $\beta$ is a $k$ x $1$ matrix of population parameters for the coefficients of the control variables, and $u_{i}$ is a $1$ x $1$ matrix for the residual error term.

$F_{R}$ is the robust F statistic for the linear model.

$$F_{R}= (R\hat\beta_{ols}- c)'\Big[R[\sum{x_{i}x_{i}'}]^{-1}[\sum{\hat u^2_{i}x_{i}x_{i}'}][\sum{x_{i}x_{i}'}]^{-1}R'\Big]^{-1}(R\hat\beta_{ols}- c)/J$$

where $R$ is a $j$ x $k$ matrix of restrictions, $c$ is a $j$ x $1$ matrix of null hypothesis, $\hat\beta_{ols}$ is a $k$ x $1$ matrix of estimates of coefficients of $x_{i}$, $\hat u_{i}$ is a $1$ x $1$ matrix of the estimated residual error, $J$ is the degrees freedom, I think it is a scalar but I am not 100% certain.

How do you prove that $F_{R}$ is asymptotically approximately chi squared distributed with $J$ degrees of freedom.

$$ F_{R} \sim \chi^{2}(J) $$

  • $\begingroup$ Please tell us what asymptotics you have in mind and what procedure the F statistic is used for, for otherwise this result is not generally true. $\endgroup$
    – whuber
    Commented Dec 28, 2018 at 15:13
  • $\begingroup$ If the number of degrees of freedom in the denominator remains fixed while that in the numerator stays constant, then the distribution approaches a chi-square distribution. The question should get rephrased to make it clear that that is what is meant. $\endgroup$ Commented Dec 28, 2018 at 19:44
  • $\begingroup$ Following the math writing, apart from $R$ and $c$, other symbols also need to be defined, otherwise, people need to guess what you say, which is not acceptable in scientific writing. $\endgroup$
    – user158565
    Commented Dec 30, 2018 at 5:38

1 Answer 1


I won't give you the full proof, just some key hints:

  • Recall that, if $X \sim N(A,\Sigma_{l*l}) $, then $ (X-A)'\Sigma^{-1}_{l*l}(X-A) \sim \chi_l^2 $

  • Under the appropiate assumptions we have asymptotic normality i.e. $ \sqrt n (b-\beta)\sim N(0,Var(b) ) $ where b is our OLS estimate

  • $H0:R\beta=c$, then under $H0$ we have $ \sqrt n (Rb-R\beta) = R \sqrt n (b-\beta) \sim N(0,Rvar(b)R') $

  • Under the appropiate assumptions, in large samples the sample variance converges in probability to the true population variance so it's not random anymore


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