Tag Info

New answers tagged

2

I agree with vanguard2k's comment: A few more details on the notation would be helpful. But, as far as I can tell, the second equality is a simple expansion. First, $\mathbf{1}'\mathbf{1} = T$ (assuming the vectors are elements of $\mathbb{R}^T$). The expression $\mathbf{1} (\mathbf{1}'\mathbf{1})^{-1} \mathbf{1}'$ is therefore nothing else than a $T\times ...



Top 50 recent answers are included