Proof about the diagonal element of the hat matrix I'd really appreciate it if you could help me find the proof for the following formula:
$$h_{ii}=1/n + \frac{(x_{i}-\bar{x})^2}{\sum(x_{j}-\bar{x})^2},$$
where $j=1,\ldots,n$.
I don't really know where to start or what to do, so any help would be very much appreciated. Thanks!
 A: First note that this formula applies just to simple linear regression where you're modeling $y_i = \beta_0 + \beta_1 x_i + \varepsilon_i$. 
$\newcommand{\1}{\mathbf 1}$We can represent our regression as $y = X\beta + \varepsilon$ with $X = (\1 \mid x)$ where $x \in \mathbb R^n$ is the non-intercept univariate predictor; by assumption $X$ is full rank and this is equivalent to $x$ not being constant. This means
$$
H = X(X^TX)^{-1}X^T = (\1 \mid x)\left(\begin{array}{cc}n & x^T\1 \\ x^T\1 & x^Tx\end{array}\right)^{-1}{\1^T\choose x^T}.
$$
We can use the formula for the explicit inverse of a $2\times 2$ matrix to find
$$
(X^TX)^{-1} = \frac{1}{nx^Tx - (x^T\1)^2}\left(\begin{array}{cc}x^Tx & -x^T\1 \\ -x^T\1 & n\end{array}\right)
$$
so all together we can do the multiplication to get
$$
H = \frac{1}{n x^Tx - (\1^T x)^2}\left(x^Tx\cdot \1\1^T - x^T\1 \cdot (\1 x^T + x \1^T) + n xx^T\right).
$$
This means
$$
h_i = \frac{x^Tx - 2x^T\1\cdot x_i + nx_i^2}{n x^Tx - (\1^T x)^2}.
$$
For the numerator, I can use the fact that $\1^Tx = n \bar x$ to rewrite it as
$$
x^Tx - 2nx_i\bar x + n x_i^2 = x^Tx + n(x_i^2 - 2 x_i\bar x + \bar x^2 - \bar x^2) \\
= x^Tx - n\bar x^2 + n(x_i - \bar x)^2.
$$
Can you finish from here?

(later update) For the sake of completeness I'll finish the proof now.
$(\1^T x)^2 = n^2(\1^T x / n)^2 = n^2{\bar x}^2$ so
$$
h_i = \frac{x^Tx - n\bar x^2 + n(x_i - \bar x)^2}{n x^Tx - (\1^T x)^2} \\
= \frac{x^Tx - n\bar x^2 + n(x_i - \bar x)^2}{n x^Tx - n^2{\bar x}^2} \\
= \frac 1n + \frac{(x_i - \bar x)^2}{x^Tx - n{\bar x}^2}
$$
and then it's well known that 
$$
x^Tx - n{\bar x}^2 = \sum_{i}(x_i - \bar x)^2
$$
so we're done.
