## Lecture 13: Kernels

Video II

Linear classifiers are great, but what if there exists no linear decision boundary? As it turns out, there is an elegant way to incorporate non-linearities into most linear classifiers.

### Handcrafted Feature Expansion

We can make linear classifiers non-linear by applying basis function (feature transformations) on the input feature vectors. Formally, for a data vector $\mathbf{x}\in\mathbb{R}^d$, we apply the transformation $\mathbf{x} \rightarrow \phi(\mathbf{x})$ where $\phi(\mathbf{x})\in\mathbb{R}^D$. Usually $D \gg d$ because we add dimensions that capture non-linear interactions among the original features.

Advantage: It is simple, and your problem stays convex and well behaved. (i.e. you can still use your original gradient descent code, just with the higher dimensional representation)

Disadvantage: $\phi(\mathbf{x})$ might be very high dimensional.

Consider the following example: $\mathbf{x}=\begin{pmatrix}x_1\\ x_2\\ \vdots \\ x_d \end{pmatrix}$, and define $\phi(\mathbf{x})=\begin{pmatrix}1\\ x_1\\ \vdots \\x_d \\ x_1x_2 \\ \vdots \\ x_{d-1}x_d\\ \vdots \\x_1x_2\cdots x_d \end{pmatrix}$.

Quiz: What is the dimensionality of $\phi(\mathbf{x})$?

This new representation, $\phi(\mathbf{x})$, is very expressive and allows for complicated non-linear decision boundaries - but the dimensionality is extremely high. This makes our algorithm unbearable (and quickly prohibitively) slow.

### The Kernel Trick

#### Gradient Descent with Squared Loss

The kernel trick is a way to get around this dilemma by learning a function in the much higher dimensional space, without ever computing a single vector $\phi(\mathbf{x})$ or ever computing the full vector $\mathbf{w}$. It is a little magical.

It is based on the following observation: If we use gradient descent with any one of our standard loss functions, the gradient is a linear combination of the input samples. For example, let us take a look at the squared loss:

$$\ell(\mathbf{w}) = \sum_{i=1}^n (\mathbf{w}^\top \mathbf{x}_i-y_i)^2\label{eq:c15:sql}$$ The gradient descent rule, with step-size/learning-rate $s>0$ (we denoted this as $\alpha>0$ in our previous lectures), updates $\mathbf{w}$ over time, $$w_{t+1} \leftarrow w_t - s(\frac{\partial \ell}{\partial \mathbf{w}})\ \textrm{ where: } \frac{\partial \ell}{\partial \mathbf{w}}=\sum_{i=1}^n \underbrace{2(\mathbf{w}^\top \mathbf{x}_i-y_i)}_{\gamma_i\ :\ \textrm{function of \mathbf{x}_i, y_i}} \mathbf{x}_i = \sum_{i=1}^n\gamma_i \mathbf{x}_i$$ We will now show that we can express $\mathbf{w}$ as a linear combination of all input vectors, $$\mathbf{w}=\sum_{i=1}^n \alpha_i {\mathbf{x}}_i.\label{eq:c15:alphas}$$ Since the loss is convex, the final solution is independent of the initialization, and we can initialize $\mathbf{w}^0$ to be whatever we want. For convenience, let us pick $\mathbf{w}_0=\begin{pmatrix}0 \\ \vdots \\ 0\end{pmatrix}$. For this initial choice of $\mathbf{w}_0$, the linear combination in $\mathbf{w}=\sum_{i=1}^n \alpha_i {\mathbf{x}}_i$ is trivially $\alpha_1=\dots=\alpha_n=0$. We now show that throughout the entire gradient descent optimization such coefficients $\alpha_1,\dots,\alpha_n$ must always exist, as we can re-write the gradient updates entirely in terms of updating the $\alpha_i$ coefficients: \begin{align} \mathbf{w}_1=&\mathbf{w}_0-s\sum_{i=1}^n2(\mathbf{w}_0^\top \mathbf{x}_i-y_i)\mathbf{x}_i=\sum_{i=1}^n \alpha_i^0 {\mathbf{x}}_i-s\sum_{i=1}^n\gamma_i^0\mathbf{x}_i=\sum_{i=1}^n\alpha_i^1\mathbf{x}_i&(\textrm{with $\alpha_i^1=\alpha_i^0-s\gamma_i^0$})\nonumber\\ \mathbf{w}_2=&\mathbf{w}_1-s\sum_{i=1}^n2(\mathbf{w}_1^\top \mathbf{x}_i-y_i)\mathbf{x}_i=\sum_{i=1}^n \alpha_i^1\mathbf{x}_i-s\sum_{i=1}^n\gamma_i^1\mathbf{x}_i=\sum_{i=1}^n\alpha_i^2\mathbf{x}_i&(\textrm{with $\alpha_i^2=\alpha_i^1\mathbf{x}_i-s\gamma_i^1$})\nonumber\\ \mathbf{w}_3=&\mathbf{w}_2-s\sum_{i=1}^n2(\mathbf{w}_2^\top \mathbf{x}_i-y_i)\mathbf{x}_i=\sum_{i=1}^n \alpha_i^2\mathbf{x}_i-s\sum_{i=1}^n\gamma_i^2\mathbf{x}_i=\sum_{i=1}^n\alpha_i^3\mathbf{x}_i&(\textrm{with $\alpha_i^3=\alpha_i^2-s\gamma_i^2$})\nonumber\\ \cdots & \qquad\qquad\qquad\cdots &\cdots\nonumber\\ \mathbf{w}_t=&\mathbf{w}_{t-1}-s\sum_{i=1}^n2(\mathbf{w}_{t-1}^\top \mathbf{x}_i-y_i)\mathbf{x}_i=\sum_{i=1}^n \alpha_i^{t-1}\mathbf{x}_i-s\sum_{i=1}^n\gamma_i^{t-1}\mathbf{x}_i=\sum_{i=1}^n\alpha_i^t\mathbf{x}_i&(\textrm{with $\alpha_i^t=\alpha_i^{t-1}-s\gamma_i^{t-1}$})\nonumber \end{align}

Formally, the argument is by induction. $\mathbf{w}$ is trivially a linear combination of our training vectors for $\mathbf{w}_0$ (base case). If we apply the inductive hypothesis for $\mathbf{w}_t$ it follows for $\mathbf{w}_{t+1}$.

The update-rule for $\alpha_i^t$ is thus $$\alpha_i^t=\alpha_i^{t-1}-s\gamma_i^{t-1}, \textrm{ and we have } \alpha_i^t=-s\sum_{r=0}^{t-1}\gamma_i^{r}.$$ In other words, we can perform the entire gradient descent update rule without ever expressing $\mathbf{w}$ explicitly. We just keep track of the $n$ coefficients $\alpha_1,\dots,\alpha_n$. Now that $\mathbf{w}$ can be written as a linear combination of the training set, we can also express the inner-product of $\mathbf{w}$ with any input ${\mathbf{x}}_i$ purely in terms of inner-products between training inputs: $$\mathbf{w}^\top {\mathbf{x}}_j=\sum_{i=1}^n \alpha_i {\mathbf{x}}_i^\top{\mathbf{x}}_j.\nonumber$$ Consequently, we can also re-write the squared-loss from $\ell(\mathbf{w}) = \sum_{i=1}^n (\mathbf{w}^\top \mathbf{x}_i-y_i)^2$ entirely in terms of inner-product between training inputs: $$\ell(\mathbf{\alpha}) = \sum_{i=1}^n \left(\sum_{j=1}^n\alpha_j\mathbf{x}_j^\top \mathbf{x}_i-y_i\right)^2\label{eq:c15:sql:ip}$$ During test-time we also only need these coefficients to make a prediction on a test-input $x_t$, and can write the entire classifier in terms of inner-products between the test point and training points: $$h({\mathbf{x}}_t)=\mathbf{w}^\top {\mathbf{x}}_t=\sum_{j=1}^n\alpha_j{\mathbf{x}}_j^\top {\mathbf{x}}_t.$$ Do you notice a theme? The only information we ever need in order to learn a hyper-plane classifier with the squared-loss is inner-products between all pairs of data vectors.

### Inner-Product Computation

Let's go back to the previous example, $\phi(\mathbf{x})=\begin{pmatrix}1\\ x_1\\ \vdots \\x_d \\ x_1x_2 \\ \vdots \\ x_{d-1}x_d\\ \vdots \\x_1x_2\cdots x_d \end{pmatrix}$.

The inner product $\phi(\mathbf{x})^\top \phi(\mathbf{z})$ can be formulated as: $$\phi(\mathbf{x})^\top \phi(\mathbf{z})=1\cdot 1+x_1z_1+x_2z_2+\cdots +x_1x_2z_1z_2+ \cdots +x_1\cdots x_dz_1\cdots z_d=\prod_{k=1}^d(1+x_kz_k)\text{.}\label{eq:c15:poly}$$ The sum of $2^d$ terms becomes the product of $d$ terms. We can compute the inner-product from the above formula in time $O(d)$ instead of $O(2^d)$! We define the function $$\underbrace{\mathsf{k}(\mathbf{x}_i,\mathbf{x}_j)}_{\text{this is called the} \textbf{ kernel function}}=\phi(\mathbf{x}_i)^\top \phi(\mathbf{x}_j).$$ With a finite training set of $n$ samples, inner products are often pre-computed and stored in a Kernel Matrix: $$\mathsf{K}_{ij}=\phi(\mathbf{x}_i)^\top \phi(\mathbf{x}_j).$$ If we store the matrix $\mathsf{K}$, we only need to do simple inner-product look-ups and low-dimensional computations throughout the gradient descent algorithm. The final classifier becomes: $$h(\mathbf{x}_t)=\sum_{j=1}^n\alpha_j\mathsf{k}(\mathbf{x}_j,\mathbf{x}_t).$$

During training in the new high dimensional space of $\phi(\mathbf{x})$ we want to compute $\gamma_i$ through kernels, without ever computing any $\phi(\mathbf{x}_i)$ or even $\mathbf{w}$. We previously established that $\mathbf{w}=\sum_{j=1}^n\alpha_j \phi(\mathbf{x}_j)$, and $\gamma_i=2(\mathbf{w}^\top \phi(\mathbf{x}_i)-y_i)$. It follows that $\gamma_i=2(\sum_{j=1}^n \alpha_jK_{ij})-y_i)$. The gradient update in iteration $t+1$ becomes $$\alpha_i^{t+1}\leftarrow \alpha_i^t-2s(\sum_{j=1}^n \alpha_j^tK_{ij})-y_i).$$ As we have $n$ such updates to do, the amount of work per gradient update in the transformed space is $O(n^2)$ --- far better than $O(2^d)$.

### General Kernels

Below are some popular kernel functions:

Linear: $\mathsf{K}(\mathbf{x},\mathbf{z})=\mathbf{x}^\top \mathbf{z}$.

(The linear kernel is equivalent to just using a good old linear classifier - but it can be faster to use a kernel matrix if the dimensionality $d$ of the data is high.)

Polynomial: $\mathsf{K}(\mathbf{x},\mathbf{z})=(1+\mathbf{x}^\top \mathbf{z})^d$.

Radial Basis Function (RBF) (aka Gaussian Kernel): $\mathsf{K}(\mathbf{x},\mathbf{z})= e^\frac{-\|\mathbf{x}-\mathbf{z}\|^2}{\sigma^2}$.

The RBF kernel is the most popular Kernel! It is a Universal approximator!! Its corresponding feature vector is infinite dimensional and cannot be computed. However, very effective low dimensional approximations exist (see this paper).

Exponential Kernel: $\mathsf{K}(\mathbf{x},\mathbf{z})= e^\frac{-\| \mathbf{x}-\mathbf{z}\|}{2\sigma^2}$

Laplacian Kernel: $\mathsf{K}(\mathbf{x},\mathbf{z})= e^\frac{-| \mathbf{x}-\mathbf{z}|}{\sigma}$

Sigmoid Kernel: $\mathsf{K}(\mathbf{x},\mathbf{z})=\tanh(\mathbf{a}\mathbf{x}^\top + c)$

#### Kernel functions

Can any function $\mathsf{K}(\cdot,\cdot)\rightarrow{\mathcal{R}}$ be used as a kernel?

No, the matrix $\mathsf{K}(\mathbf{x}_i,\mathbf{x}_j)$ has to correspond to real inner-products after some transformation ${\mathbf{x}}\rightarrow \phi({\mathbf{x}})$. This is the case if and only if $\mathsf{K}$ is positive semi-definite.

Definition: A matrix $A\in \mathbb{R}^{n\times n}$ is positive semi-definite iff $\forall \mathbf{q}\in\mathbb{R}^n$, $\mathbf{q}^\top A\mathbf{q}\geq 0$.

Remember $\mathsf{K}_{ij}=\phi(\mathbf{x}_i)^\top \phi(\mathbf{x}_j)$. So $\mathsf{K}=\Phi^\top\Phi$, where $\Phi=[\phi(\mathbf{x}_1),\dots,\phi(\mathbf{x}_n)]$. It follows that $\mathsf{K}$ is p.s.d., because $\mathbf{q}^\top\mathsf{K}\mathbf{q}=(\Phi^\top \mathbf{q})^2\geq 0$. Inversely, if any matrix $\mathbf{A}$ is p.s.d., it can be decomposed as $A=\Phi^\top\Phi$ for some realization of $\Phi$.

You can even define kernels over sets, strings, graphs and molecules.

Figure 1: The demo shows how kernel function solves the problem linear classifiers can not solve. RBF works well with the decision boundary in this case.