# The Azimuth Project Blog - Exploring regression on the El Ni&ntilde;o data (Rev #4)

## Regression using an $l_{1/2}$ prior

We can write the regression cost function—adding an $l_{1/2}$ prior—using explicit summations as

$cost_2 = \sum_{e=1}^{E} \left( \sum_{i=1}^P M^{(e)}_{i} x_{i} - y^{(e)}\right)^2 + \lambda \sum_{i=1}^P |x_i|^2$
$cost_1 = \sum_{e=1}^{E} \left( \sum_{i=1}^P M^{(e)}_{i} x_{i} - y^{(e)}\right)^2 + \lambda \sum_{i=1}^P |x_i|$
$cost_{1/2} = \sum_{e=1}^{E} \left( \sum_{i=1}^P M^{(e)}_{i} x_{i} - y^{(e)}\right)^2 + \lambda \sum_{i=1}^P \sqrt{|x_i|}$

If we restrict to one variable from the numerous vectors and denote this variable by $x$, we get

$\frac{\partial cost}{\partial x} = A x + B + \frac{\lambda sgn(x)}{2\sqrt{|x|}} = 0$

where $A$ and $B$ don’t depend on $x$. If we denote $\lambda sgn(x)/2$ by $C$, we can multiply through by $y =\sqrt{|x|}$ to find the minimum (along this co-ordinate) is

$\pm A y^3 + B y + C = 0$

where the $\pm$ depends whether $x$ is positive/negative and all subject to needing to ensure the solutions in $y$ are also consistent with the original equation. Since this is a cubic equation we have a simple closed form for the solutions to this equation and hence can efficiently solve the original equation.