​Watch this. Also explains about ISO surfaces, lp norm, sparseness.
(what is?) Regularization (in linear regression) - to find the best model we define a loss or cost function that describes how well the model fits the data, and try minimize it. For a complex model that fits even the noise, i.e., over fitted, we penalize it by adding a complexity term that would add BIGGER LOSS for more complex models.
    Bigger lambda -> high complexity models (deg 3) are ruled out, more punishment.
    Smaller lambda -> models with high training error are rules out. I.e., linear model on non linear data?, i.e., deg 1.
    Optimal is in between (deg 2)
​Rehearsal on vector normalization - for l1,l2,l3,l4 etc, what is the norm? (absolute value in certain cases)
(Difference between? And features of) L1 vs L2 as loss function and regularization.
    L1 - moves the regressor faster, feature selection by sparsing coefficients (zeroing them), with sparse algorithms it is computationally efficient, with others no, so use L2.
    L2 - moves slower, doesn't sparse, computationally efficient.
Why does L1 lead to sparity?
    ​Intuition + some mathematical info​
    L1 & L2 regularization add constraints to the optimization problem. The curve H0 is the hypothesis. The solution is a set of points where the H0 meets the constraints.
    In L2 the the hypothesis is tangential to the ||w||_2. The point of intersection has both x1 and x2 components. On the other hand, in L1, due to the nature of ||w||_1, the viable solutions are limited to the corners of the axis, i.e., x1. So that the value of x2 = 0. This means that the solution has eliminated the role of x2 leading to sparsity.
    This can be extended to a higher dimensions and you can see why L1 regularization leads to solutions to the optimization problem where many of the variables have value 0.
    In other words, L1 regularization leads to sparsity.
    Also considered feature selection - although with LibSVM the recommendation is to feature select prior to using the SVM and use L2 instead.
    For simplicity, let's just consider the 1-dimensional case.
    L2-regularized loss function F(x)=f(x)+Ξ»βˆ₯xβˆ₯^2 is smooth.
    This means that the optimum is the stationary point (0-derivative point).
    The stationary point of F can get very small when you increase Ξ», but it will still won't be 0 unless fβ€²(0)=0.
      regularized loss function F(x)=f(x)+Ξ»βˆ₯xβˆ₯ is non-smooth, i.e., a min knee of 0.
      It's not differentiable at 0.
      Optimization theory says that the optimum of a function is either the point with 0-derivative or one of the irregularities (corners, kinks, etc.). So, it's possible that the optimal point of F is 0 even if 0 isn't the stationary point of f.
      In fact, it would be 0 if Ξ» is large enough (stronger regularization effect). Below is a graphical illustration.
In multi-dimensional settings: if a feature is not important, the loss contributed by it is small and hence the (non-differentiable) regularization effect would turn it off.
(did not watch) but here is andrew ng talks about cost functions.
L2 regularization equivalent to Gaussian prior​
L1 regularization equivalent to a Laplacean Prior(same link as above) - β€œSimilarly the relationship between L1 norm and the Laplace prior can be undestood in the same fashion. Take instead of a Gaussian prior, a Laplace prior combine it with your likelihood and take the logarithm.β€œ How does regularization look like in SVM - controlling β€˜C’
Last modified 2mo ago
Copy link