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I understand that the homoscedasticity assumption is one of the Gauss Markov assumptions to get a BLUE estimator.

Why is homoscedasticity crucial for justifying the usual t and F statistics?

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2 Answers 2

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Why is homoscedasticity crucial for justifying the usual t and F statistics?

Homoscedasticity is not crucial for the use of the t and F statistics, because they can be used, at least asymptotically, even in the presence of heteroscedasticity.

Homoskedasticity is required for the usual OLS parameters estimator. However, even with heteroscedasticity, it can be retained for pointwise estimates, but using the robust standard errors. Note that the t and F statistics still apply.

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  • $\begingroup$ That is what the Wooldridge says. I guess the author means for usual t and F statistics those different from the heteroskedasticity-robust ones. My question was (maybe silly) why do we need heteroscedasticity-robust statistics in presence of heteroscedasticity? $\endgroup$
    – Dimitru
    Commented Apr 21, 2023 at 20:17
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    $\begingroup$ Under heteroskedasticity what is not more valid is the OLS estimator of standard errors $\sigma^2(X'X)^{-1}$. You should to put his elements into t or F stat, but those value are incorrect. You need to replace them with the correct ones (robust). However t and F remain. Speak about of usual/modified t and F sound me misleading. $\endgroup$
    – markowitz
    Commented Apr 21, 2023 at 20:36
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    $\begingroup$ The consistency of tests of hypothesis under heteroscedasticity is not true without bounds on the design of the test. For instance, let the design of size $n$ be given by $X_i = i$ for $i = 1, 2, \ldots, n$ and $Y_i = X_i + e_i$ where $e_i \sim N(0, i^2)$. Then $SE(\hat{\beta}) = \sum_{i \le n} r_i^2 / \sum_{i \le n} i^2$ it's quick the corresponding Wald test statistic will not diverge. $\endgroup$
    – AdamO
    Commented Apr 21, 2023 at 21:12
  • $\begingroup$ @AdamO, You rightly underscore that some special cases can appear. Moreover several version of robust SE exist. Even for these reasons I remained quite vague in my reply. I gave the simplest possible key message to the asker. $\endgroup$
    – markowitz
    Commented Apr 22, 2023 at 8:05
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Homoscedasticity is NOT an assumption of the Gauss Markov theorem.

The Gauss Markov theorem establishes OLS as the BLUE estimator for $\beta$ when the error is homoscedastic.

The GM theorem also established WLS as the BLUE estimator when the error is heteroscedastic. WLS is weighted least squares. The weight is specifically chosen to be the inverse of the variance-covariance matrix of the error vector. We don't even require that the data are independent.

It's of course nearly impossible to know the actual variance structure in practice, but the theorem should be understood in the precise terms in which it is presented.

The justification for the $t$ and $F$ tests come from the asymptotic distribution of the Wald statistic $W = \frac{\hat{\beta} - \beta}{SE(\hat{\beta})}$ which has a limiting $N(0, I_p)$ distribution when the null hypothesis is true. The same test statistics and limiting distributions are obtained when, under heteroscedasticity with $V$ known, the estimator is replaced $\hat{\beta} = (X^T V^{-1}X)^{-1} X^T V^{-1} Y$.

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    $\begingroup$ You mean The Gauss Markov theorem establishes OLS as the BLUE estimator for β when the error is NOT heteroscedastic, right? What about the homoscedasticity for justifying the t and F stats? $\endgroup$
    – Dimitru
    Commented Apr 21, 2023 at 19:34
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    $\begingroup$ @Adamo, you are wrong, Gauss Markov Theorem require homoscedastic errors. See here: en.wikipedia.org/wiki/Gauss%E2%80%93Markov_theorem $\endgroup$
    – markowitz
    Commented Apr 21, 2023 at 19:49
  • $\begingroup$ @markowitz Please reference a more credible source. I will paste a snip from Longitudinal Data Analysis by Diggle Heagerty Liang Zeger. Consider Gauss and Markov did not intersect in their lives, yet Gauss already established the unweighted OLS as optimal in his personal work. Neyman & David published on "Markoff[sic]'s" contribution to least squares in 1936, which I think became the basis for the famously named theorem. $\endgroup$
    – AdamO
    Commented Apr 21, 2023 at 21:00
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    $\begingroup$ @Dimitru no, the unweighted estimator will not be BLUE when the error is not IID. $\endgroup$
    – AdamO
    Commented Apr 21, 2023 at 21:03
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    $\begingroup$ Also I just realized they mispelled "Gauss"! Too funny, I will mail the editors. $\endgroup$
    – AdamO
    Commented Apr 21, 2023 at 21:32

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