# Tag Info

9

Both models penalise the inclusion of a non-zero coefficient, using a penalty function. LASSO regression penalises in a way that is proportionate to the absolute magnitude of the coefficient, and ridge regression penalises in a way that is proportionate to the square of the coefficient. Neither model penalises inputs in the likelihood function where the ...

4

This is our objective function, composed of a loss function and a regularizer. $$\mathcal O(w,x,y) = \mathcal L(w,x,y)+\mathcal R(w)$$ So $\mathcal R(w)=\|w\|_2^2=\sum{w_i^2}$ in the case of $\ell_2$ regularization. Let's perform gradient-based minimization, i.e. we will update params based on the negative partial derivatives of the objective function. ...

3

Let's examine the coefficient estimates of ridge as a function of the OLS estimates. The matrix expression which minimize the loss is $$\hat{\beta}^{\text {ridge }}=\left(\mathbf{X}^{T} \mathbf{X}+\lambda \mathbf{I}\right)^{-1} \mathbf{X}^{T} \mathbf{y}$$ According to the authors of Elements of Statistical Learning, each element of that vector can be ...

3

One of the factor to consider is computational simplicity. By not introducing the square root, the gradient has a more elegance form. Also, minimizing $MSE$ subject to $\|\theta\|_2\le c$ is equivalent to minimizing $MSE$ subject to $\|\theta\|_2^2\le c^2$ is equivalent to

2

L2 regularization adds $w_i^2$ term to the loss function. In iterative approaches using gradients, we subtract the gradient of the loss function not the magnitude of the weight itself. And in the loss function, the regularization part's derivative with respect to $w_i$ is going to be ${d\over dw_i}(w_i^2)=2w_i$. Typically, this part is multiplied with a ...

2

Why should we expect there to be a square root involved? This would simply be a type of scaling which means $\alpha$ values are much larger than without the square root. The total sum $\sum_{k=1}^K \theta_k^2$ is obviously the same in both forms. Since the alpha parameter is there to scale the importance of the regularisation part relative to the loss ...

1

l2 norm of $\bf {\vec x} = {\lvert\lvert \bf {x \rvert\rvert}}_2^2$ = $x_1^2+x_2^2+\ldots+x_n^2$, which is present in the objective function of ridge regression. The $\frac{\alpha}{2}$ that precedes the l2-norm is a tuning parameter in the objective function of ridge regression which is always $\ge0$. If I'm not mistaken, you are referring this from Andrew ...

1

Multiple sources suggest you can, and this has been done. You can combine the predictions of multiple neural networks as an ensemble estimator, even though the RF approach tends to be the classic estimator because of the thematic congruity between random forests and multiple forests. You can also boost neural networks as well.

1

Ridge was originally designed for correlated variables, and that's where it's best. Consider examinations to determine a degree. ( Which supposedly is measuring ability) Which do you think is more reliable: taking the average of all the exams or picking a single exam most correlated with ability (if there is one)? Averaging over the different exams removes ...

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