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Normally we have $ y= X \beta + u $ and we find $ \hat{\beta}_{MLE} $

and $\hat{\sigma^2}_{MLE}$ from the function of:

$$ l(\theta,y) = -\dfrac {n}{2}\cdot \ln (2 \pi) -\dfrac {n}{2} \cdot \ln (\sigma^2) -\dfrac {1}{2 \sigma^2} \cdot (y - X \beta)^T (y - X \beta) $$

and we take partial derivative w.r.t $\beta$ and $\sigma^2$ such that $ \dfrac {\partial l(\theta,y)}{\partial \beta} $ and $\dfrac {\partial l(\theta,y)}{\partial \sigma^2}$

so we obtain $2\times 1$ matrix we take $E(S(\theta,y))$ of this matrix (as you know $S(\theta,y) = \dfrac {\partial l(\theta,y)}{\partial \theta}$ ) and we go on Variance too.

so for that regression $y= X_1 \beta_1 + X_2 \beta_2 + u $ should we write $ y -X_1 \beta_1 -X_2 \beta_2$ instead of $y - X \beta$ and take partial derivative w.r.t $\beta_1$ and $\beta_2$ and get a $3\times 1$ dimension matrix ? or did I miss a point here? then in what form information matrix should be in this case?

Edit:

if I should process with this, then does it make sense to construct Information Matrix in this way:

$$ -E\left[\dfrac{\partial^2 l\left( \theta | y\right) }{\partial \theta \partial \theta ^{T}}\right] = \begin{pmatrix} \dfrac{\partial^2 l\left( \theta | y\right) }{\partial \beta _{1}\partial \beta _{1}^{T}} & \dfrac{\partial^2 l\left( \theta | y\right) }{\partial \beta _{1}\partial \beta _{2}} & \dfrac{\partial^2 l\left( \theta | y\right) }{\partial \beta _{1}\partial{\sigma^2}} \\ \dfrac{\partial^2 l\left( \theta | y\right) }{\partial \beta _{2}\partial \beta _{1}} & \dfrac{\partial^2 l\left( \theta | y\right) }{\partial \beta _{2}\partial \beta _{2}^T} & \dfrac{\partial^2 l\left( \theta | y\right) }{\partial \beta _{2}\partial \sigma^2} \\ \dfrac{\partial^2 l\left( \theta | y\right) }{\partial \sigma^2\partial \beta_1} & \dfrac{\partial^2 l\left( \theta | y\right) }{\partial \sigma^2\partial \beta_2} & \dfrac{\partial^2 l\left( \theta | y\right) }{\partial \sigma^2\partial \sigma^2} \end{pmatrix} $$

if so, after taking partial derivatives, I will get $3\times 3$ matrix, so what is the variance of $\beta_1, \beta_2$?

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  • $\begingroup$ What is $In$? Is it supposed to be $\ln$? $\endgroup$ Commented Jan 6, 2023 at 6:58
  • $\begingroup$ Oh yes, It is I thought It won't be a problem. I fixed. $\endgroup$
    – Tatanik501
    Commented Jan 6, 2023 at 7:01
  • $\begingroup$ Your current derivative w.r.t. $\beta$ does the same, ie, collects the partial derivatives with respect to all elements of $\beta$, such as $\beta_1$ and $\beta_2$ $\endgroup$ Commented Jan 6, 2023 at 11:41
  • $\begingroup$ I edited the question and tried an information matrix construction, is it correct? $\endgroup$
    – Tatanik501
    Commented Jan 6, 2023 at 12:57

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