Finding the mean ($\bar{y}$ vector) given $y$ vector

Question

Let $y$ be an $n \times 1$ vector of observations. Consider the $n \times 1$ vector $\bar{y}$, which contains the mean of $y$ vector. How does one show that the $\bar{y}$ vector is given by $$\bar{y}=\vec{1} \left( \vec{1}^T \vec{1} \right)^{-1} \vec{1}^Ty$$ where $\vec{1}$ is the vector of ones (also $n \times 1$).

Answer 1

$\displaystyle\color{blue}{\vec{1}^Ty} = \sum_{i=1}^n 1\cdot y_i=\sum_{i=1}^n y_i$

$\displaystyle\color{red}{\vec{1}^T\vec{1}}=\sum_{i=1}^n 1\cdot 1 =n$

$\displaystyle\color{green}{({\vec{1}^T\vec{1}})^{-1}\,{\vec{1}^Ty}}=\frac{1}{\color{red}{\vec{1}^T\vec{1}}}\,\color{blue}{\vec{1}^Ty} =\frac{1}{n}\sum_{i=1}^n y_i=\bar y$

$\displaystyle\vec{1}\,\color{green}{({\vec{1}^T\vec{1}})^{-1}\,{\vec{1}^Ty}}=\vec{1}\bar y$

Answer 2

Perhaps it's cleaner when written with inner products:

$$\overline{y} = \frac1n \begin{bmatrix} y_1+ \cdots + y_n \\ \vdots \\ y_1+ \cdots + y_n\end{bmatrix} = \left\langle y, \frac{\vec{1}}{\|\vec{1}\|}\right\rangle\frac{\vec{1}}{\|\vec{1}\|} = \frac1{\|\vec{1}\|^2}\langle y, \vec{1}\rangle \vec{1} = \frac1{\vec{1}^T\vec{1}}(\vec{1}^Ty)\vec{1} = \vec{1}(\vec{1}^T\vec{1})^{-1}(\vec{1}^Ty)$$