Logistic Regression

I recently gave a talk on Logistic Regression and its usage using scikit-learn as part of the Austin SIGKDD Advanced Machine Learning Meetup. The slides can be found here.

Quasi Newton Methods - Part II (Symmetric Rank 1)

In the previous post we saw that the "smallest" update to a matrix so that it satisfies a given linear constraint is a rank-1 update. However, the updated matrix may lose it symmetric and positive definite characteristics. In this post we will find a matrix update that maintains the symmetric nature of the matrix and we will stick to a rank-1 update.

A symmetric rank-1 update can be expressed as \(\alpha xx^T\) where \(x\) is some vector and \(\alpha\) is \(\pm 1\). So, compared to our previous post, instead of the matrix \(X\) we have this symmetric rank-1 update and we still have the same constraint, namely,
\[ \alpha xx^T a = b\]
which can be rewritten as
\[ \alpha \left( x^Ta \right) x = b\]
and straightaway tells us that \(x\) is linearly dependent on \(b\). Plugging, \(x=\beta b\) into the original constraint we get:
\[ \alpha \beta^2 (b^Ta) b = b\]
which implies that
\[ \alpha \beta^2 = \frac{1}{b^Ta} \]
giving us our symmetric rank-1 update as \(\frac{bb^T}{b^Ta}\). Mapping, the symbols back to the original problem we get:
\[J'_n = J'_{n-1} + \frac{ (\Delta f - J_{n-1}x_{n-1})(\Delta f - J_{n-1}x_{n-1})^T}{(\Delta f - J_{n-1}x_{n-1})^T(\Delta x)}\]
This update is known by the TLA SR1 (Symmetric Rank-1 update) and is the update for a symmetric rank-1 update that satisfies the secant condition. Depending on the sign of the denominator this update may or may not be positive definite. Since SR1 was a result of just requiring the update to be symmetric and rank-1 and no other constraint we cannot hope to obtain a symmetric positive definite rank-1 update that satisfies the secant condition. We will again use the approach of Part I and look for a "smallest" update that gives us both symmetry and positive definiteness and we will end up with a rank-2 update and the famous DFP and BFGS formulas.

Quasi Newton Methods - Part I (Rank 1)

This time we will be looking at Quasi Newton Methods. But lets start off with an introduction to Newton's Method, firstly, looking at its application at root finding. Let \(f : \mathbb{R}^n \rightarrow \mathbb{R}\) be a function and we want to find an \(x\) such that \(f(x)=0\). If we start from an initial guess \(x_0\) we can use Newton's method to iteratively update the initial guess. In its most basic form we look at the Jacobian matrix of \(f\) and create a simple first order approximation of the function and use the zero of the approximate function as the initial guess for the next iteration of the method. So,

\[ f_{approx}(x) = f(x_0) + J_0(x-x_0)\]

where \(f_{approx}\) represents the first order approximation to \(f\) and \(J_0\) is the Jacobian matrix of \(f\) at \(x_0\). So, our next guess will be the zero of \(f_{approx}\) which can be expressed as (assuming \(J_0\) is invertible)

\[ x_1 = x_0 - J_{0}^{-1}f(x_0)\]

and is known as the "full" Newton step. In general, going from the \(n-1^{th}\) iterate to \(n^{th}\) iterate we have

\[ x_{n} = x_{n-1} - J_{n-1}^{-1}f(x_{n-1})\]

Computing, the Jacobian matrix at each iterate is a computationally expensive step and "Quasi" Newton methods attempt to approximate the Jacobian matrix based on the Jacobian used to obtain the iterate. Lets impose the condition that the approximation we construct should match the function at the previous iterate, in addition to the current iterate (as in the very first equation in this post). This gives,

\[f_{n-1} = f_n + J'_n (x_{n-1}-x_{n})\]

where \(J'\) is our approximation to the Jacobian. The constraint above can be succinctly represented as \(J'_n\Delta x = \Delta f\) and is known as the Secant Condition. The constraint is not enough to approximate the Jacobian as we have \(n^2\) variables and only \(n\) constraints. Since, we want to use the previous Jacobian to approximate the new Jacobian lets add an objective function that minimizes the Frobenius norm of the change. This given us our overall optimization problem as:

\[\mathrm{min} : \frac{1}{2}\|J'_n-J'_{n-1}\|_F\]

\[\mathrm {s.t.} :  J'_n\Delta x = \Delta f\]

Lets rewrite the above by using \(X = J'_n-J'_{n-1}\) and \(a = \Delta x\) and \(b = \Delta f - J'_{n-1}\Delta x\) as

\[\mathrm{min} : \frac{1}{2}\|X\|_F\]

\[\mathrm {s.t.} :  Xa = b\]

Lets, first square the objective function (which does not change the optimization problem as we have a positive objective function) and solve the resulting problem by introducing the Lagrangian and using Matrix derivatives:

\[L(X, \lambda) = \frac{1}{2}\mathbb{Tr}(X^TX) + \lambda^T(Xa-b) \]

\[\mathrm{d}L = \mathbb{Tr}\left((X^T\mathrm + a\lambda^T)\mathrm{d}X\right) \]

Which implies that at the optimal point \((X^*, \lambda^*)\), \(X^* + \lambda^* a^T=0\). Plugging, this into the original constraint of the optimization problem we get \(\lambda^* = \frac{-b}{a^Ta}\). Plugging, this back into the expression for \(X^*\), finally, we get

\[ X^* = \frac{ba^T}{a^Ta}\]

So we see that the update is Rank-1 and using this we can now write the expression for the updated Jacobian as

\[J'_n = J'_{n-1} + \frac{ (\Delta f - J_{n-1}x_{n-1})(\Delta x)^T}{(\Delta x)^T(\Delta x)}\]

which is known as Broyden's good method. The Broyden's bad method is obtained if we try to minimize the Frobenius norm of the update to the inverse of the Jacobian and can be easily written down by using the result for our simplified optimization problem (by flipping the roles of \(a\) and \(b\)) as

\[J'^{-1}_n = J'^{-1}_{n-1} + \frac{ (\Delta x - J^{-1}_{n-1}f_{n-1})(\Delta f)^T}{(\Delta f)^T(\Delta f)}\]

Using the Sherman–Morrison formula we can write the update to the inverse of the Jacobian in the good Broyden case and the update to the Jacobian in the bad Broyden case as rank-1 updates as well.

There are two shortcomings of this result:

1) The update is not symmetric. So even if we start with a symmetric Jacobian we will lose symmetry. This is important if we look at Newton's method as an optimization method where the function \(f\) will be a gradient of an objective function and the Jacobian will be the Hessian of the objective function and, therefore, symmetric. The next post will be about a symmetric update to the Jacobian though still rank-1.

2) The update is not positive definite. Again a problem when we are looking to optimize say a convex function. The third post in this series will be on the DFP and BFGS updates that will get us the desired SPD (symmetric positive definite) update, but we will need to go one rank up to a rank-2 update.

Bishop PRML Chapter 1 (1.15-1.16)

Finding \( n(D,M) \) (problem 1.15) is finding the number of ways we can sum \(D\) non-negative numbers to M. This is a basic combinatorics problem with a number of interesting solution methods. I will outline three approaches that feel somewhat different and find application in solving other problems.

\[ \mathrm{Solve:} \quad x_1+x_2+...+x_D=M \]
\[ x_i \gt =0 \quad 1 \le i \le D\]

Method 1: Using generating functions

Coefficient of \(x^M\) in \( \left(1+x+x^2+....\right)^D \)

=Coefficient of \(x^M\) in \( \left(1-x\right)^{-D} \)

Using -ve binomial  series the coefficient is \( \left( \begin{matrix} D+M-1\\M  \end{matrix} \right) \)

Method 2: Bars and Balls

Lets say we have a set of \(M\) 1’s, and a set of \(D-1\) bars. The \(D-1\) bars split the 1’s into the number of 1's representing each of the \(D x_i\)’s . So in total we have \(D+M-1\) positions out of which we select \(M\) position for 1’s and the rest are for the bars.

A following similar approach yields the wrong solution: So we have \(M\) 1’s so we can select each bar in \(M+1\) ways. The mistake here is that selecting the first and second bars in positions 1 and 2 is the same as selecting them in positions 2 and 1. We need each new bar to come to the right of the ones we have already positioned. Okay, so can't we just select the position of \(D-1\) bars from \(M+1\) positions. This gives us the solution to the problem where each \(x_i \gt 0\) because we cannot choose the same position for two bars. However, when \(x_i \ge 0\) we are allowed to choose bars from the same position between balls so we start off with a set of positions for bars and balls and then select the positions of the bars.

Method 3: Indices

Let each \(x_i\) represent the number of balls in box \(i\). The number of boxes is \(D\) and the number of balls is \(M\). Represent each solution as a non-decreasing list of box numbers. E.g. 11335 represents 2 balls in box 1, two balls in box 3 and one ball in box 5. Now create another sequence which is the indices of the balls 01234. Add the two sequences 12569. We can see that all increasing sequences can be generated with a smallest value of 1 and a largest value of \(D+M-1\). So the number of ways is again \( \left( \begin{matrix} D+M-1\\M  \end{matrix} \right) \).

The extension to problem 1.16 is then fairly straightforward. We just add another variable giving us a total of \(D+1\) dimensions and solve the same problem as above. The newly added variable will take over the leftover whenever the other \(D\) variables sum to less than \(M\). So, in this case we have the number of ways as \( \left( \begin{matrix} D+M\\M  \end{matrix} \right) \)

Puzzles: Languages and locks

I recently heard a fun puzzle. Puzzle: How many languages are needed so that we can have that for any two people in a group of 70 each one knows a language that the other does not.






Solution: Basically we need to assign to each person a unique subset of languages. If there are K languages number of such such subsets will be maximum if each person knows K/2 languages. So, we will need the number of languages K to be such that

\[ \left( \begin{matrix} K\\K/2  \end{matrix} \right) \ge 70 \]

Another, variant of this puzzle is regarding bank locks. Puzzle: Suppose, you have \(n\) people working at a bank and you want that at least \(m\) should be present to be able to open a lock. How many locks do you need, and how many keys are needed for each lock.




Solution: For each subset of \(m-1\) people there should be a lock that they cannot open so  we need

\[ \left( \begin{matrix} n\\m-1  \end{matrix} \right) \]

locks. Also, for any lock we cannot find m people that together cannot open it. So each lock should have \(n-(m-1)\) keys.

Trace and transpose trick: Computing derivatives

I am not sure why but none of the college courses on linear algebra or calculus ever introduced me to working with derivatives of functions of matrices and vectors. This comes up on a daily basis in optimization and is a lot cleaner than dealing with summations. Here is a trick that comes in very handy when taking derivatives of a functions that maps vectors or matrices to scalars. Basically, one can replace a scalar results of matrix vector products by its transpose and in some cases by their trace (and then use the cyclic property of  trace to rewrite expressions) in which the desired terms appear at the end and are correctly transposed.

Simplest case: Linear function takes a vector to a scalar.
\[f : \mathbb{R}^n \rightarrow \mathbb{R}\]
\[ \mathrm{d} \left(c^Tx \right) = c^T \mathrm{d}x\]

Next level: Function takes a vector to a scalar using a quadratic form. We will use the transpose trick here:
\[f : \mathbb{R}^n \rightarrow \mathbb{R}\]
\[ \mathrm{d} \left(x^TAx \right) = (\mathrm{d}x)^TAx + x^TA(\mathrm{d}x) \]

The first term on the RHS is a scalar so we can take its transpose and get

\[ \mathrm{d} \left(x^TAx \right) = x^TA^T(\mathrm{d}x) + x^TA(\mathrm{d}x)  = (x^TA^T + x^TA) \mathrm{d}x\]

Next level: Function takes a matrix to a scalar. We will use the trace trick here: 

\[f : \mathbb{R}^{nxn} \rightarrow \mathbb{R}\]
\[ \mathrm{d} \left(a^TXb \right) = a^T (\mathrm{d}X)b \]
Now since the RHS is a scalar we can think of it as a trace and rotate the terms to get:
\[ \mathrm{d} \left(a^TXb \right) = \mathrm{Tr}(ba^T \mathrm{d}X) \].
Now, we will use the symbol (:) that rewrites the matrix as a concatenation of its vectors and rewrite the above as:
\[ \mathrm{d} \left(a^TXb \right) = (ab^T): (\mathrm{d}X): \].
Giving us \( \frac{\mathrm{d} \left(a^TXb \right)}{\mathrm{d}X} = ab^T \).

Lets use this machinery to attempt something more complicated and find \( \frac{\mathrm{d}(a^TX^TCXb)}{\mathrm{d}X} \)

\[ \mathrm{d} (a^TX^TCXb)  = a^T(\mathrm{d}X^T)CXb + a^TX^TC(\mathrm{d}X)b \]

\[ = b^TX^TC^T(dX)a + a^TX^TC(dX)b \]

\[ = \mathrm{Tr}( b^TX^TC^T(\mathrm{d}X)a + a^TX^TC(\mathrm{d}X)b ) \]

\[ = \mathrm{Tr}( ab^TX^TC^T(\mathrm{d}X) + ba^TX^TC(\mathrm{d}X)) \]

\[ = (CXba^T+C^TXab^T): (\mathrm{d}X): \]

\[ \Rightarrow \mathrm{d}(a^TX^TCXb)/\mathrm{d}X = (CXba^T+C^TXab^T) \]

Bell states to test nature's inherent uncertainity

This post is to succinctly describe the key ideas that are presented by Prof. Vazirani in his QMQC course on EdX (I saw this when it was originally offered on coursera) related to Bell's idea to show that quantum randomness is (or is not) inherent in nature.

So, as usual, we have two players Alice and Bob who can initially agree on a strategy and are then separated by a huge distance over which we will assume that they cannot communicate. Now, the game is that both Alice and Bob receive a binary input and produce a binary input. They win when they generate different inputs only when both their inputs are 1, otherwise they should produce the same output. Classically, we can see that the best that Alice and Bob can do is win with a probability of 0.75.

However, if they can share an entangled Bell state then they can achieve a better probability of success. An entangled state of two particles is one which cannot be expressed as a product of the states of the two particles. So, as an example,

\[ \lvert \Psi \rangle = \frac{1}{\sqrt{2}} \left(   \lvert 00 \rangle +  \lvert 11 \rangle \right) \]

is an entangled state. It is instructive to see that this state cannot be written as a tensor product of the individual states of the two particles. Now, the next fun thing to note is that the representation of this state when written in any other orthonormal basis \(u, u^{\perp} \) does not change (again, expand out the tensor products to see this). So, our state above can be expressed as


\[ \lvert \Psi \rangle = \frac{1}{\sqrt{2}} \left(   \lvert uu \rangle +  \lvert u^{\perp}u^{\perp} \rangle \right) \]

Now, let Alice and Bob share the entangled state with one particle each before separation and are then taken to two corners of the universe. They can choose their basis (x for Alice and y for Bob) as shown in the figure below and announce their result depending on whether they see the particle oriented along their basis vector or orthogonal to it. In, all cases except when \(x=y=1\) their basis vectors are oriented at an angle of \( \frac{\pi}{8} \), and only when \(x=y=1\) there basis vectors are orientated at an angle of \( \frac{3\pi}{8} \). In both cases the probability of winning is \( \cos^2\frac{\pi}{8} \approx 0.85 \), much better than can be obtained classically.

A nice article discussing the questions still surrounding the experimental results with Bell's states appeared here recently.

Linear Regression

I recently gave a talk on Linear Regression and its extensions (Ridge regression, LASSO, LARS etc.) as part of the Austin SIGKDD Advanced Machine Learning Meetup. The slides can be found here.

One of the interesting things that I learnt about while preparing for this talk was the fact that coordinate descent works for function where the non-differentiable part of the objective function is separable (slides 29-30). This is why we can use gradient optimization for minimizing the LASSO objective function. The proof works out and the only background that is needed is subgradients. Also, it is instructive to make sure that the proof "really" uses the property of separability and why the proof breaks when we do not have separability.

Hint: For a separable function the subgradient set is a direct product of the subgradient sets of the individual components. However, if the non-differentiable function is not separable then the negative gradient of the differentiable part may not lie in the subgradient set of the non-differentiable function.

Another, interesting thing was the connection between LASSO and LARS. LASSO minimizes the quadratic residual norm with a \(\mathrm{L1}\) penalty and requires that the sign of the coefficient related to a vector be the same as the sign of the inner product of the residual with the vector (easy to show using the Lagrangian). LARS does not enforce this and can therefore produce different solutions. LARS can modified to drop a vector from the active set when this sign change happens and can thus be tailored to generate a LASSO solution. However, then the number of steps taken by LARS is not bounded by \(n\), the dimensionality of the problem.

PCA vs Least Squares

We all know that least squares minimizes the sum of the squared errors. The error is defined as the error between the measured and the predicted value of the dependent variable, where the predicted value is a linear function of the independent variables. If both the dependent and the independent variable are measured quantities and may be noisy, then we might be more interested in minimizing the error of the orthogonal projection of the data points on a linear hyperplane, and this turns out to be the first principal component of the data. The page here has a couple of good images that get this point across. But, lets see that is indeed the case, again, with matrix algebra :)

Let's start with the basics. The projection of a vector \(x\) on a vector \(w\) can be expressed as \(w^Tx \frac{w}{\| w \|^2}\), which in the case of a unit-norm vector \(w\) reduces to \((w^Tx) w\). So, if we have a data matrix \(X\), where the columns of the data matrix form the data points, then the projection of the entire data matrix can be written as \((w^TX) w\). Basically, the projections are all pointing along \(w\) with lengths given by \(w^TX\). In PCA, we seek to find a direction that has the maximum spread (or variance) and we can continue to consider the direction as a vector through the origin if we assume that the data matrix \(X\) has \(0\) mean. Equivalently, we want to find a direction \(w\) that maximizes the following:

\[\mathrm{Cov}\left( w^TX\right) = w^T\mathrm{Cov}\left(X\right)w = w^TXX^Tw\]

Now, lets see what we get for the sum of the squared orthogonal error of the data points on to the line given by the vector \(w\). Each data points has an error given by the 2-norm of the vector that represents the orthogonal error. The sum of the squared errors for each data point can then be represented by a Frobenius norm of a matrix that is composed of these error vectors.

\[\|X - w (w^TX)\|_F = \mathrm{Tr}( (X-ww^TX)^T (X - ww^TX) )\]

\[= \mathrm{Tr}\left(X^TX - X^Tww^TX-X^Tww^TX+X^Tww^Tww^TX\right)\]

\[= \mathrm{Tr}\left(X^TX - X^Tww^TX-X^Tww^TX+X^Tww^TX\right)\] (as \(w\) is unit norm)

\[=\mathrm{Tr}\left( X^TX \right) - \mathrm{Tr}\left(X^Tww^TX\right)\]

\[=\mathrm{Tr}\left( X^TX \right) - \mathrm{Tr}\left(w^TXX^Tw\right)\] (Since trace of cyclic rotations of matrix products is the same)

\[=\mathrm{Tr}\left( X^TX \right) - \left(w^TXX^Tw\right)\] (The trace of 1x1 matrix is the matrix itself)

which is the same as the case above for PCA except for a constant term (\(\mathrm{Tr} (X^TX)\) that is independent of \(w\)) and that we seek to maximize the PCA variance and minimize the orthogonal error. Now, the matrix \(A=XX^T\) represents the covariance matrix of the data points and one seeks to find a unit-norm direction that minimizes the quadratic form \(w^TAw\) for a positive semi-definite symmetric matrix \(A\). One can either construct the Lagrangian or write \(w\) in terms of the eigenbasis to see that we need to choose \(w\) as the eigenvector with the largest eigenvalue.

Logistic Regression: Gradients computed the matrix way

I really enjoy performing calculus with matrix/vector operations rather than messing around with tons of summations. The other nice thing about calculations with matrices/vectors is that the results come out cleanly in a form that can be used directly in Matlab or a similar high-level language and where the matrix/vector multiplication based implementations are typically much faster than using your own for-loops to compute the result. One of the nice examples of this approach is computing the gradients for logistic regression without using any summations at all!

Before we get to logistic regression let's build up one small piece of the machinery (vectorized functions) that will help us when we get there. Lets begin with vectors \(p, z \in \mathbb{R}^n\) and let \(f_v\) be a vectorized version of function \(f\) that applies the function \(f\) component-wise.

\[p = f_v(z)\]
\[\mathrm{d}p = \mathrm{Diag}((f')_v(z))\mathrm{d}z\].

Now, our logistic regression objective function is

\[J(\theta) = \frac{1}{m} \left(   -y^T(log(p)) -(1-y)^T(log(1-p))   \right)\]

where

\[p=g_v(z)\]
\[g(x)=\frac{1}{1+\exp^{-x}}\]
\[z=X^T\theta\]

Lets start with the simplest derivatives and then use those to compute the gradient of the objective function

\[\mathrm{d}g=\frac{\exp^{-x}}{(1+\exp^{-x})^2}\mathrm{d}x=g(x)(1-g(x))\mathrm{d}x\]

Now, lets get to \(p\) and make use of machinery we developed at the very beginning:

\[\mathrm{d}p = \mathrm{Diag}((g')_v(z)) \mathrm{d}z\]
\[\mathrm{d}p = \mathrm{Diag}(g_v(z)(1-g_v(z))\mathrm{d}z=\mathrm{Diag}(g_v(z)(1-g_v(z))X^T\mathrm{d}\theta\]

Now, we come to our original goal of computing the gradient of the objective function \(J\)

\[\mathrm{d}J = \frac{1}{m} \left(   -y^T(\mathrm{Diag}(1/p)) + (1-y)^T(\mathrm{Diag}(1/(1-p)))   \right)\mathrm{d}p\]

\[\mathrm{d}J = \frac{1}{m} \left(   -y^T(\mathrm{Diag}(1/g_v(z))) + (1-y)^T(\mathrm{Diag}(1/(1-g_v(z))))   \right)\mathrm{Diag}(g_v(z)(1-g_v(z))X^T\mathrm{d}\theta\]

\[\mathrm{d}J = \frac{1}{m} \left(   -y^T(\mathrm{Diag}(1-g_v(z))) + (1-y)^T(\mathrm{Diag}(g_v(z)))   \right)X^T\mathrm{d}\theta\]

\[\mathrm{d}J = \frac{1}{m} \left(   g_v(z)^T -y ^T    \right)X^T\mathrm{d}\theta\]

So, finally we have

\[\nabla J = X(g_v(z)-y)\]

A good reference for matrix calculus can be found here.

Kernelizing the SVM

Recently, I learnt a few examples of the kernel trick. Earlier, I used to believe that the kernel trick was as simple as adding a few nonlinear features and running your standard machine learning algorithm that works with linear features and getting out a nonlinear model. Its a little more than that :)
The kernel trick is really the above idea with the additional goal that the computations should not be performed in the higher dimensional feature space that includes both the linear and the nonlinear features, and we get away with performing computations in a space that has the same dimensionality as the original set of linear features.

In this post I will go over the kernel trick as applied to the SVM and in a later post to linear least squares (my favourite thing of all).

SVM 


The basic idea of SVM is to find a linear hyperplane that separates two classes of data points. So, ideally, we want to find a normal vector \(w\) and an offset \(b\) such that for points \(x\) with the class \(+1\) we have \(w^Tx + b \ge 1\) and for points \(x\) with the class \(-1\) we have \(w^Tx + b \le -1\).


To make the problem realistic we allow for situations where the data might not be linearly separable by adding positive slacks \(s_i\) by which the above constraints can be relaxed. We need to minimize \(\sum s_i\) under the linear classification constraints above and the additional constraint that the norm of the normal vector of the classifier be 1. The quadratic constraint can be introduced into the objective with a user-provided weighting \(c\).

\[\mathrm{min}_{w,b,s}:   \frac{1}{2}w^Tw + c (\mathbf{1}^Ts)\]
\[\mathrm{s.t.}: \mathrm{Diag}(y) \left(X^Tw + b\mathbf{1} \right) \ge \mathbf{1} - s\]
\[\mathrm{s.t.}: s \ge 0\]

where \(y\) represents the vector of data point classes and \(X\) represents the matrix of data points forming its columns.
The Lagrangian can then be expressed as:

\[\frac{1}{2}w^Tw + c(\mathbf{1}^Ts) - \lambda^T(\mathrm{Diag}(y)\left(X^Tw+b\mathbf{1} \right)-1+s) - \mu^Ts\]
where \(\lambda\) and \(\mu\) are positive Lagrange multipliers.

Minimizing over the primal variables gives us the following constraints:
\[b: \lambda^T\mathrm{Diag}(y)1=\lambda^Ty=0\]
\[s: \lambda + \mu=c\mathbf{1}\]
\[w: w = X\mathrm{Diag}(y)\lambda\]

So, \(w\) turns out to be a weighted sum of the data points in the \(X\) matrix. Also, by complementary slackness, the \(\lambda\)'s can be strictly positive only if the corresponding constraint in the primal is satisfied with equality and these data points will form our support vectors.

We will solve the dual problem which can be obtained by simplifying the Lagrangian using the constraints obtained above to get

\[\mathrm{min}_{\lambda}: \frac{1}{2}\lambda^T\mathrm{Diag}(y)\left(X^TX\right)\mathrm{Diag}(y)\lambda - \lambda^T\mathbf{1}\]
\[\mathrm{s.t.}: \lambda^Ty=0\]
\[\mathrm{s.t.}: 0 \le \lambda \le c\mathbf{1}\]

Now, the kernel trick is to observe that in the dual problem the data matrix \(X\) only appears in a multiplication with its transpose. The composite \(X^TX\) can be thought of as a matrix where its \((i,j)\) entry is the dot-product \((x_i^Tx_j)\) of the \(i\) and \(j\) data points. Using the kernel trick we replace each entry by a non-linear function \(K(x_i, x_j)\) which can capture the inner product in the higher dimensional feature space. More on kernel functions later.

Another, important thing to note is that is the kernel trick should carry over to the prediction step as well. In the SVM case the prediction is \(\mathrm{sgn}(w^Tx_{test}+b)\). Now, \(w^Tx_{test}=\lambda^T\mathrm{Diag}(y)X^Tx_{test}\) and we see that the data points again only appear as a dot-product which we will replace with the kernel function. The value of \(b\) can be obtained with the data points involved only in dot-products by observing that the the primal classification constraints are satisfied with equality and \(s_i=0\) when we have \(0 \lt \lambda_i \lt c\). When \(0 \lt \lambda\) the constraint \(y_i(w^Tx_i+b)\ge1-s_i\) is satisfied with equality and when \(\lambda \lt c\) we have non-zero value for \(\mu\) which forces \(s_i\) to be zero. In that case we can evaluate \(b=y_i-w^Tx_i\). We already saw that the \(w^Tx_i\) term  involves only dot-products when we plug-in the expression for \(w\).