I'm working on a problem in which we consider a simple regression with no regressors and equi-correlated disturbances. So we have

$y_i = \alpha + \varepsilon_i$

where $E[\varepsilon_i, \varepsilon_j] = 0$

and $Cov[\varepsilon_i, \varepsilon_j]=\left\{\begin{matrix} \rho \sigma^2 & i\neq j\\ \sigma^2 & i=j \end{matrix}\right.$

The above model can be rewritten as $y = \alpha\iota_n + u$

Since $\hat\beta = (X'X)^{-1}X'y$ we can say $\hat\alpha_{OLS}=(\iota_n'\iota_n)^{-1}\iota_n'y$. In Baltagi (2010) I found that $\hat\alpha_{OLS}=(\iota_n'\iota_n)^{-1}\iota_n'y = \sum_{i=1}^{n}\frac{y_i}{n}=\bar{y}$. How is this last transformation derived? I fail to understand how the sum comes into play here as well as the n.

Further, it is stated that

$\Psi = E[\varepsilon\varepsilon']=\begin{pmatrix} \sigma^2 & \rho\sigma^2 & {...} &\rho\sigma^2 \\ \rho\sigma^2 &\sigma^2 & {...} & \rho\sigma^2\\ {...} & {...} & {...} &{...}\\ \rho\sigma^2 &{...} &{...} &\sigma^2 \end{pmatrix}= \sigma^2(1-\rho)I+\rho\sigma^2\iota\iota'$.

The matrix makes sense to me but here I also fail to understand the last step how one derives the expression

$\sigma^2(1-\rho)I+\rho\sigma^2\iota\iota'$ from the matrix. Maybe someone can explain, help would be much appreciated!


You can understand both if you consider that $\iota_n$ is a vector of $1$s of dimension $(n \times 1)$.

In the formula for $\alpha_{OLS}$ the $(\iota_n' \iota_n)$ is the inner product of two vectors of $1$s of dimension $(1 \times n)$ and $(n \times 1)$, since the first one is transposed, which is the sum of the products of their $n$ elements: $\underbrace{1*1+1*1+...1*1}_{n \text{ times}} = n$. Since it is taking the inverse you get the $n$ in the denominator.

The second part is again the inner product between the vector of $1$s of dimension $(1 \times n)$ and the vector $y$ of dimension $(n \times 1)$, hence $\iota_n y = \sum_{i=1}^n 1*y_i = \sum_{i=1}^n y_i$.

Regarding the matrix $\Psi$, $\sigma^2(1-\rho)I$ is a diagonal $(n \times n)$ matrix with $\sigma^2(1-\rho)$ in the diagonal and $0$ elsewhere. The second part is a $(n \times n)$ matrix where every element is $\rho \sigma^2$, since $\iota \iota'$ is a matrix of $1$s.

  • $\begingroup$ Now I understand, thanks a lot! $\endgroup$ – Phil Nov 12 '20 at 9:37

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