1
$\begingroup$

For that regression:

$$ y =X_1 \beta_1 +X_2 \beta_2 + u $$ with $u \sim N(0,\sigma^2 I)$

and I want to derive Wald for the hypothesis $\beta_2 = 0$

as you know, for the regression of $ y = y =X \beta + u$ and hypothesis $ H_0:R\beta =r$

Wald is equal to $W$ = $ \left( R\widehat{\beta }-r\right) ^{T} \left( R \cdot I^{-1} \left( \widehat{\beta }\right)\cdot R^{T}\right) \left( R\widehat{\beta }-r\right) $

where "inverse of information matrix" $I^{-1} ( \widehat{\beta })$ $ = \sigma^2 (X^ T X) ^{-1} $

so Wald becomes:

$W$ = $ \dfrac{1}{\widehat {\sigma} ^2} \cdot \left( R\widehat{\beta }-r\right) ^{T} \left( R\left( X^{T}X\right) ^{-1}R ^T\right) ^{-1} \left( R\widehat{\beta }-r\right) $ has an asymptotically Chi-Square distribution $\chi_q^2$

but I want to derive this Wald in the form of $n \cdot \dfrac{SSR_R - SSR_U}{SSR_U} $ (with the help of LM test)

$SSR_R$ and $SSR_U$ stand for the restricted and the unrestricted.

because I want to build that relationship: $$ W = \dfrac{n \cdot k_2}{n-k} \cdot F $$ where F is the F-test in the form of $\dfrac{(SSR_R - SSR_U)/k_2}{SSR_U/(n-k)} $

How can I do this from the perspective of Maximum Likelihood Estimation? I really couldn't understand how I should use the restriction.

$\endgroup$
4
  • $\begingroup$ I think Chapter 5 of Johnston and DiNardo's Econometric Methods has this worked out. $\endgroup$
    – dimitriy
    Commented Jan 2, 2023 at 20:23
  • $\begingroup$ It covers the case $ y = X \beta + u$ on the page of 148 as I see but but I couldn't understand how can I relate this to partitioned form $\endgroup$
    – Tatanik501
    Commented Jan 3, 2023 at 7:10
  • $\begingroup$ I work out the result for the F-statistic here: stats.stackexchange.com/questions/258461/… $\endgroup$ Commented Jan 3, 2023 at 10:45
  • $\begingroup$ Thanks I will check $\endgroup$
    – Tatanik501
    Commented Jan 3, 2023 at 12:54

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.