Index: The Book of Statistical ProofsStatistical Models ▷ Univariate normal data ▷ Simple linear regression ▷ Projection of data point to regression line

Theorem: Consider simple linear regression and an estimated regression line specified by

$\label{eq:slr-regline} y = \hat{\beta}_0 + \hat{\beta}_1 x \quad \text{where} \quad x,y \in \mathbb{R} \; .$

For any given data point $O(x_o \vert y_o)$, the point on the regression line $P(x_p \vert y_p)$ that is closest to this data point is given by:

$\label{eq:slr-proj} P\left(w \mid \hat{\beta}_0 + \hat{\beta}_1 w\right) \quad \text{with} \quad w = \frac{x_0 + (y_o - \hat{\beta}_0) \hat{\beta}_1}{1 + \hat{\beta}_1^2} \; .$

Proof: The intersection point of the regression line with the y-axis is

$\label{eq:S} S(0 \vert \hat{\beta}_0) \; .$

Let $a$ be a vector describing the direction of the regression line, let $b$ be the vector pointing from $S$ to $O$ and let $p$ be the vector pointing from $S$ to $P$.

Because $\hat{\beta}_1$ is the slope of the regression line, we have

$\label{eq:a} a = \left( \begin{matrix} 1 \\ \hat{\beta}_1 \end{matrix} \right) \; .$

Moreover, with the points $O$ and $S$, we have

$\label{eq:b} b = \left( \begin{matrix} x_o \\ y_o \end{matrix} \right) - \left( \begin{matrix} 0 \\ \hat{\beta}_0 \end{matrix} \right) = \left( \begin{matrix} x_o \\ y_o - \hat{\beta}_0 \end{matrix} \right) \; .$

Because $P$ is located on the regression line, $p$ is collinear with $a$ and thus a scalar multiple of this vector:

$\label{eq:p} p = w \cdot a \; .$

Moreover, as $P$ is the point on the regression line which is closest to $O$, this means that the vector $b-p$ is orthogonal to $a$, such that the inner product of these two vectors is equal to zero:

$\label{eq:a-b-p-orth} a^\mathrm{T} (b-p) = 0 \; .$

Rearranging this equation gives

$\label{eq:w} \begin{split} a^\mathrm{T} (b-p) &= 0 \\ a^\mathrm{T} (b - w \cdot a) &= 0 \\ a^\mathrm{T} b - w \cdot a^\mathrm{T} a &= 0 \\ w \cdot a^\mathrm{T} a &= a^\mathrm{T} b \\ w &= \frac{a^\mathrm{T} b}{a^\mathrm{T} a} \; . \end{split}$

With \eqref{eq:a} and \eqref{eq:b}, $w$ can be calculated as

$\label{eq:w-qed} \begin{split} w &= \frac{a^\mathrm{T} b}{a^\mathrm{T} a} \\ w &= \frac{\left( \begin{matrix} 1 \\ \hat{\beta}_1 \end{matrix} \right)^\mathrm{T} \left( \begin{matrix} x_o \\ y_o - \hat{\beta}_0 \end{matrix} \right)}{\left( \begin{matrix} 1 \\ \hat{\beta}_1 \end{matrix} \right)^\mathrm{T} \left( \begin{matrix} 1 \\ \hat{\beta}_1 \end{matrix} \right)} \\ w &= \frac{x_0 + (y_o - \hat{\beta}_0) \hat{\beta}_1}{1 + \hat{\beta}_1^2} \end{split}$

Finally, with the point $S$ \eqref{eq:S} and the vector $p$ \eqref{eq:p}, the coordinates of $P$ are obtained as

$\label{eq:P-qed} \left( \begin{matrix} x_p \\ y_p \end{matrix} \right) = \left( \begin{matrix} 0 \\ \hat{\beta}_0 \end{matrix} \right) + w \cdot \left( \begin{matrix} 1 \\ \hat{\beta}_1 \end{matrix} \right) = \left( \begin{matrix} w \\ \hat{\beta}_0 + \hat{\beta}_1 w \end{matrix} \right) \; .$

Together, \eqref{eq:P-qed} and \eqref{eq:w-qed} constitute the proof of equation \eqref{eq:slr-proj}.

Sources:

Metadata: ID: P283 | shortcut: slr-proj | author: JoramSoch | date: 2021-11-09, 10:16.