Regression Equivalence

Setup

Purpose: Show the univariate OLS estimate for the regression coefficient is equivalent to the AP method assuming 1) there are an even number of observations \(n\), and 2) the observations are evenly spaced by a length \(L\) such that \(x_{i+1} - x_i=L\) for all \(i\) in \(\left\{i=1, \ldots, n-1 \right\}\).

Begin by expressing the regression coefficient as a sum of the pairwise combinations of observations \[\begin{align*} \hat{\beta} = \frac{\sum_{i,j} (y_i - y_j)(x_i - x_j) }{\sum_{i,j} (x_i - x_j)^2} \end{align*}\]

The proof proceeds by re-expressing the numerator and denominator separately, and finally combining these results to show equivalence to the AP method.

Result (1)

First note that \[\begin{align*} \sum_{i=1}^n \sum_{j=i}^n y_j (x_i - x_j) = \sum_{i=1}^n \sum_{k=1}^i y_i(x_k - x_i) \end{align*}\]

Result (2), numerator

Re-express the numerator as \[\begin{align*} \sum_{i,j} (y_i - y_j)(x_i - x_j) & = \sum_{i=1}^n \left[ \sum_{j=i}^n y_i(x_i-x_j) - \sum_{j=i}^n y_j(x_i-x_j) \right] \\ & = \sum_{i=1}^n \left[ \sum_{j=i}^n y_i(x_i-x_j) - \sum_{k=1}^i y_i(x_k - x_i) \right] \qquad \mbox{(Using result (1))} \\ & = \sum_{i=1}^n \left[ y_i \sum_{j=i}^n (x_i-x_j) - y_i \sum_{k=1}^i (x_k - x_i) \right] \\ & = \sum_{i=1}^n \left[ y_i \left( \frac{ (n-i+1)(-(n-i)) }{2} \right)L - y_i \left( \frac{i(-(i-1))}{2} \right)L \right] \qquad \mbox{($\sum_{j=i}^n (x_i-x_j)$ and $\sum_{k=1}^i (x_k - x_i)$ are arithmetic sequences)} \\ & = \sum_{i=1}^n -y_i \left[ \frac{n(n-2i+1)}{2} \right] L \\ & = \frac{n}{2} \sum_{i=1}^n -y_i \left( n-2i+1\right) L \\ & = \frac{n}{2} \left[ \sum_{i=1}^{n/2} -y_i \left( n-2i+1\right) L - \sum_{i=n/2+1}^n y_i \left( n-2i+1\right) L\right] \\ & = \frac{n}{2} \left[ \sum_{i=1}^{n/2} (y_{n-i+1} -y_i) (n-2i+1) L \right] \end{align*}\]

Result (3), denominator

\[\begin{align*} \sum_{i,j} (x_i - x_j)^2 & = \sum_{i=1}^{n-1} \left[ (n-i) L\right]^2 i \\ & = L^2 \sum_{i=1}^{n-1} (n-i)^2 i \\ & = \frac{L^2}{12}n^2 (n^2-1) \\ & = \frac{n}{2} \frac{L^2}{6} n (n^2-1) \\ & = \frac{n}{2} L^2 \sum_{i=1}^{n/2} (n-2i+1)^2 \end{align*}\]

AP method

The AP method uses \[\begin{align*} \hat{\beta_{AP}} & = \frac{\sum_{k=1}^{n/2} \frac{y_{n-k+1} - y_k}{x_{n-k+1} - x_k} (x_{n-k+1} - x_k)^2}{\sum_{k=1}^{n/2} (x_{n-k+1} - x_k)^2 } \\ & = \frac{\sum_{k=1}^{n/2} (y_{n-k+1} - y_k) (x_{n-k+1} - x_k)}{\sum_{k=1}^{n/2} (x_{n-k+1} - x_k)^2 } \\ & = \frac{\sum_{k=1}^{n/2} (y_{n-k+1} - y_k) (n-2k+1) L}{\sum_{k=1}^{n/2} \left( (n-2k+1)L \right)^2 } \\ & = \frac{\sum_{k=1}^{n/2} (y_{n-k+1} - y_k) (n-2k+1) L}{L^2 \sum_{k=1}^{n/2} (n-2k+1)^2 } \\ \end{align*}\]

Dividing the final expression in result (2) by the final expression in result (3) shows that \(\hat{\beta}=\hat{\beta_{AP}}\).