Fitting a quadratic using regression when the y-intercept needs to be 0 [closed]

Question

I'm trying to fit a quadratic $a_0 + a_1x + a_2x^2$ by Polynomial Regression:

$$ \begin{pmatrix} n & \Sigma x_i & \Sigma x_i\\ \Sigma x_i & \Sigma x_i^2 & \Sigma x_i^3\\ \Sigma x_i^2 & \Sigma x_i^3 & \Sigma x_i^4\\ \end{pmatrix} \begin{pmatrix} a_0\\ a_1\\ a_2\\ \end{pmatrix}= \begin{pmatrix} \Sigma y_i\\ \Sigma x_iy_i\\ \Sigma x_i^2y_i\\ \end{pmatrix} $$ by Gaussian Elimination on the augmented matrix: $$ \left[ \begin{array}{ccc|c} n & \Sigma x_i & \Sigma x_i & \Sigma y_i\\ \Sigma x_i & \Sigma x_i^2 & \Sigma x_i^3 & \Sigma x_iy_i\\ \Sigma x_i^2 & \Sigma x_i^3 & \Sigma x_i^4 & \Sigma x_i^2y_i\\ \end{array} \right] $$

I'm not a matrix whizz, so how do I solve this using the same method when I know I want the y-intercept of the curve, $a0$, to be 0?

Answer 1

If you want $a_0$ to be $0$, then the curve you are trying to fit has equation $y = a_1 x + a_2 x^2$. If you derive the normal equations starting from this equation, you get

$a_1 \sum x_i^2 + a_2 \sum x_i^3 = \sum x_i y_i\\ a_1 \sum x_i^3 + a_2 \sum x_i^3 = \sum x_i^2 y_i$

Or

$\left[\begin{array}{cc|c} \sum x_i^2 & \sum x_i^3 & \sum x_i y_i\\ \sum x_i^3 & \sum x_i^4 & \sum x_i^2 y_i \end{array}\right]$

Derivation of Normal Equations
$y = a_1 x + a_2 x^2$

The error $\epsilon_i$ in the $i$^th term is the difference between the predicted value $y(x_i)$ and the observed value $y_i$:
$\epsilon_i = y(x_i) - y_i = a_1 x_i + a_2 x_i^2 - y_i$

The total error is the sum of squares of these errors:
$E = \displaystyle\sum_i \epsilon_i^2 = \displaystyle\sum_i(a_1x_i + a_2x_i^2 - y_i)^2$

To find the values of the parameters $a_1$ and $a_2$ that minimize the total error, let $\dfrac{\partial E}{\partial a_1} = \dfrac{\partial E}{\partial a_2} = 0$:

$ \dfrac{\partial E}{\partial a_1} = \sum 2x_i(a_1 x_i + a_2 x_i^2 - y_i) = 0 \Rightarrow a_1 \sum x_i^2 + a_2 \sum x_i^3 = \sum x_i y_i\\ \dfrac{\partial E}{\partial a_2} = \sum 2x_i^2(a_1 x_i + a_2 x_i^2 - y_i) = 0 \Rightarrow a_1 \sum x_i^3 + a_2 \sum x_i^4 = \sum x_i^2 y_i\\ $