# Transformation in regression with no regressors and equi-correlated disturbances

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!

## 1 Answer

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.

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