Bounding residual variance with distance from mean For a linear regression $Y = X\beta + \varepsilon$ with $\varepsilon \sim \mathcal N(0,\sigma^2 I)$, we have $\hat Y = H Y$ for $H = X(X^TX)^{-1}X^T$. This means that $Var(Y - \hat Y) = \sigma^2(I-H)$ so in particular $Var(Y_i - \hat Y_i) = \sigma^2(1-h_i)$.
Suppose my predictor matrix $X$ has rows $x_1,\dots,x_n\in\mathbb R^p$. I want to bound this residual variance, and therefore $1-h_i$, in terms of $\|x_i - \frac 1n\mathbf 1^TX\|^2$, the distance of $x_i$ to the mean of the $x_i$. Is there something nice I can do here?
If this is a simple linear regression I know 
$$
1 - h_i = 1 - \frac 1n - \frac{(x_i - \bar x)^2}{\sum_j (x_j - \bar x)^2}
$$
so I am handed such a bound via the appearance of the $(x_i - \bar x)^2$ term. But what about in a multiple regression?
 A: I was able to come up with a bound although it's not very tight.
Let $X = UDV^T$ be the SVD of $X$, so that $H = UU^T$. Let $u_1,\dots,u_n \in \mathbb R^p$ be the rows of $U$ (as column vectors) which means that $h_i = \|u_i\|^2$.
Let $s_i^2 = \|x_i - \frac 1n X^T\mathbf 1\|^2$. If $e_i$ is the $i$th standard basis vector then $x_i = X^Te_i$ so I can write 
$$
s_i^2 = \|X^Te_i - \frac 1n X^T\mathbf 1\|^2 = \|X^T(e_i - \frac 1n\mathbf 1)\|^2 \\
= (e_i - \frac 1n\mathbf 1)^TXX^T(e_i - \frac 1n\mathbf 1) \\
= (e_i - \frac 1n\mathbf 1)^TUD^2U^T(e_i - \frac 1n\mathbf 1) .
$$
I don't like the $D^2$ since it seems like it'll be helpful to get a quadratic form with $H=UU^T$ instead of $XX^T$ so I'll use the fact that $d_1^2$ is the largest squared singular value so
$$
s_i^2 \leq d_1^2(e_i - \frac 1n\mathbf 1)^TUU^T(e_i - \frac 1n\mathbf 1) \\
= d_1^2 \left(e_i^TUU^Te_i - \frac 2n \mathbf 1^TUU^Te_i + \frac 1{n^2}\mathbf 1^TUU^T\mathbf 1\right) \\
= d_1^2 \left(h_i - \frac 2n \mathbf 1^TUU^Te_i + \frac 1{n^2}\mathbf 1^TUU^T\mathbf 1\right).
$$
Now I'll assume that $\mathbf 1$ is in the column space of $X$ (i.e. it has an intercept, which shouldn't be too controversial). This means that $H\mathbf 1 = UU^T\mathbf 1 = \mathbf 1$ so
$$
s_i^2 \leq d_1^2 \left(h_i - \frac 2n \mathbf 1^Te_i + \frac 1{n^2}\mathbf 1^T\mathbf 1\right) \\
= d_1^2 \left(h_i - \frac 1n \right)
$$
which implies that
$$
h_i \geq \frac{s_i^2}{d_1^2} + \frac 1n \\
\iff 1 - h_i \leq 1 - \frac 1n - \frac{s_i^2}{d_1^2}.
$$
Thus as $s_i^2$ increases this upper bound on the variance of a residual does go up, which is nice to see, but I'm not particularly happy with this as $d_1^2$ is often massive in the experiments I've done so this is close to just saying $1 - h_i \leq 1 - \frac 1n$.
Is there a better bound available?
