# Viewing a kernel as a probability density: Why are we only integrating with respect to $x$ and not $x^\prime$?

I am currently studying the textbook Learning with kernels: support vector machines, regularization, optimization and beyond by Schölkopf and Smola. Chapter 1.2 A Simple Pattern Recognition Algorithm says the following:

We are now in the position to describe a pattern recognition learning algorithm that is arguably one of the simplest possible. We make use of the structure introduced in the previous section; that is, we assume that our data are embedded into a dot product space $$\mathcal{H}$$. Using the dot product, we can measure distances in this space. The basic idea of the algorithm is to assign a previously unseen pattern to the class with closer mean.
We thus begin by computing the means of the two classes in feature space; $$\mathbf{c}_+ = \dfrac{1}{m_+} \sum_{\{i \vert y_i = +1\}} \mathbf{x}_i, \tag{1.7}$$ $$\mathbf{c}_- = \dfrac{1}{m_-} \sum_{\{i \vert y_i = -1\}} \mathbf{x}_i, \tag{1.8}$$ where $$m_+$$ and $$m_-$$ are the number of examples with positive and negative labels, respectively. We assume that both classes are non-empty, thus $$m_+, m_- > 0$$. We assign a new point $$\mathbf{x}$$ to the class whose mean is closest (Figure 1.1). This geometric construction can be formulated in terms of the dot product $$\langle \cdot, \cdot \rangle$$. Half way between $$\mathbf{c}_+$$ and $$\mathbf{c}_-$$ lies the point $$\mathbf{c} := (\mathbf{c}_+ + \mathbf{c}_-)/2$$. We compute the class of $$\mathbf{x}$$ by checking whether the vector $$\mathbf{x} - \mathbf{c}$$ connecting $$\mathbf{c}$$ to $$\mathbf{x}$$ encloses an angle smaller than $$\pi / 2$$ with the vector $$\mathbf{w} := \mathbf{c}_+ - \mathbf{c}_-$$ connecting the class means. This leads to \begin{align} y &= \text{sgn} \langle (\mathbf{x} - \mathbf{c}), \mathbf{w} \rangle \\ &= \text{sgn} \langle ( \mathbf{x} - ( \mathbf{c}_+ + \mathbf{c}_- ) / 2), ( \mathbf{c}_+ - \mathbf{c}_- ) \rangle \\ &= \text{sgn}( \langle \mathbf{x}, \mathbf{c}_+ \rangle - \langle \mathbf{x}, \mathbf{c}_- \rangle + b). \tag{1.9} \end{align} Here, we have defined the offset $$b := \dfrac{1}{2} ( \left| \right| \mathbf{c}_- \left| \right|^2 - \left| \right| \mathbf{c}_+ \left| \right|^2), \tag{1.10}$$ with the norm $$\left| \right| \mathbf{x} \left| \right| := \sqrt{\langle \mathbf{x}, \mathbf{x} \rangle}$$. If the class means have the same distance to the origin, then $$b$$ will vanish.
Note that (1.9) induces a decision boundary which has the form of a hyperplane (Figure 1.1); that is, a set of points that satisfy a constraint expressible as a linear equation.
It is instructive to rewrite (1.9) in terms of the input patterns $$x_i$$, using the kernel $$k$$ to compute the dot products. Note, however, that (1.6) only tells us how to compute the dot products between vectorial representations $$\mathbf{x}_i$$ of inputs $$x_i$$. We therefore need to express the vectors $$\mathbf{c}_i$$ and $$\mathbf{w}$$ in terms of $$\mathbf{x}_1, \dots, \mathbf{x}_m$$.
To this end, substitute (1.7) and (1.8) into (1.9) to get the decision function \begin{align} y &= \text{sgn} \left( \dfrac{1}{m_+} \sum_{\{i \vert y_i = +1 \}} \langle \mathbf{x}, \mathbf{x}_i \rangle - \dfrac{1}{m_-} \sum_{\{i \vert y_i = -1 \}} \langle \mathbf{x}, \mathbf{x}_i \rangle + b \right) \\ &= \text{sgn} \left( \dfrac{1}{m_+} \sum_{\{i \vert y_i = +1 \}} k( x, x_i ) - \dfrac{1}{m_-} \sum_{\{i \vert y_i = -1 \}} k( x, x_i ) + b \right). \tag{1.11} \end{align} Similarly, the offset becomes $$b := \dfrac{1}{2} \left( \dfrac{1}{m^2_-} \sum_{\{(i, j) \vert y_i = y_j = -1 \}} k( x_i, x_j ) - \dfrac{1}{m^2_+} \sum_{\{(i, j) \vert y_i = y_j = +1 \}} k( x_i, x_j ) \right). \tag{1.12}$$ Surprisingly, it turns out that this rather simple-minded approach contains a well-known statistical classification method as a special case. Assume that the class means have the same distance to the origin (hence $$b = 0$$, cf. (1.10)), and that $$k$$ can be viewed as a probability density when one of its arguments is fixed. By this we mean that it is positive and has unit integral, $$\int_\mathcal{X} k(x, x^\prime) \ dx = 1 \ \text{for all} \ x^\prime \in \mathcal{X}. \tag{1.13}$$

In chapter 1.1 Data Representation and Similarity, $$k$$ was defined as follows:

Let us consider a similarity measure of the form \begin{align} k: &\mathcal{X} \times \mathcal{X} \to \mathbb{R} \\ &(x, x^\prime) \mapsto k(x, x^\prime), \tag{1.2} \end{align} that is, a function that, given two patterns $$x$$ and $$x^\prime$$, returns a real number characterizing their similarity. Unless stated otherwise, we will assume that $$k$$ is symmetric, that is, $$k(x, x^\prime) = k(x^\prime, x)$$ for all $$x, x^\prime \in \mathcal{X}$$. For reasons that will become clear later (cf. Remark 2.16), the function $$k$$ is called a kernel.

What is the reasoning behind

$$\int_\mathcal{X} k(x, x^\prime) \ dx = 1 \ \text{for all} \ x^\prime \in \mathcal{X} \tag{1.13}$$

only being integrated with respect to $$x$$? What about the $$x^\prime$$?

• The statement below (1.2) states $k$ is symmetric, which implies integrating over $x$ will give the same result as integrating over $x^\prime.$
– whuber
Dec 31, 2020 at 21:30
• @whuber Ok, thanks for the clarification. Jan 1, 2021 at 6:21
• Could you please clarify your question, then? If it isn't answered by the symmetry, what are you trying to ask?
– whuber
Jan 1, 2021 at 14:37
• @whuber You indeed answered my question. Am I supposed to delete this? Or should I post a community answer? Jan 1, 2021 at 14:39
• If you think the point isn't entirely trivial then posting an answer would be good.
– whuber
Jan 1, 2021 at 14:46

As whuber stated in the comments, the statement below (1.2) states $$k$$ is symmetric, which implies integrating over $$x$$ will give the same result as integrating over $$x^\prime$$.