On the expectation of a persistence diagram by the persistence weighted kernel

Abstract

In topological data analysis, persistent homology characterizes robust topological features in data and it has a summary representation, called a persistence diagram. Statistical research for persistence diagrams have been actively developed, and the persistence weighted kernel shows several advantages over other statistical methods for persistence diagrams. If data is drawn from some probability distribution, the corresponding persistence diagram have randomness. Then, the expectation of the persistence diagram by the persistence weighted kernel is well-defined. In this paper, we study relationships between a probability distribution and the persistence weighted kernel in the viewpoint of (1) the strong law of large numbers and the central limit theorem, (2) a confidence interval to estimate the expectation of the persistence weighted kernel numerically, and (3) the stability theorem to ensure the continuity of the map from a probability distribution to the expectation. In numerical experiments, we demonstrate our method gives an interesting counterexample to a common view in topological data analysis.

Appendix A: Kernel two sample test

In this section, we briefly review the kernel two sample test, following [18, 19].

1.1 Gereral framework of the two sample test

Let $(\mathcal {X}, \mathcal {B}_{\mathcal {X}})$ be a topological space, P and Q be probability distributions on $\mathcal {X}$, $X_{1},\ldots ,X_{m}, {\mathrm {i.i.d.}}\sim P$, $Y_{1},\ldots , Y_{n}, {\mathrm {i.i.d.}}\sim Q$, and $\theta (\varvec{X}_{m}, \varvec{Y}_{n})$ be a statistics. If $H_{0}$ is true, the statistics $\theta (\varvec{X}_{m}, \varvec{Y}_{n})$ depends only on P because $Y_{1},\ldots , Y_{n}, {\mathrm {i.i.d.}}\sim Q=P$. Here, we assume that the upper $\alpha$-quantile ${\hat{\xi }}_{m,n,\alpha }$ which satisfies ${\mathrm {Pr}}(\theta (\varvec{X}_{m}, \varvec{Y}_{n}) \le {\hat{\xi }}_{m,n,\alpha })=1-\alpha$ is computable when $H_{0}$ is true. When $\alpha$ is set a small value, if a pair of realizations $(\varvec{x}_{m}, \varvec{y}_{n})$ of $(\varvec{X}_{m},\varvec{Y}_{n})$ satisfies ${\hat{\xi }}_{m,n, 1-\alpha /2} \le \theta (\varvec{x}_{m}, \varvec{y}_{n}) \le {\hat{\xi }}_{m,n,\alpha /2}$, we conclude that the hypothesis $H_{0}$ is accepted under $H_{0}$. Otherwise, we conclude that $H_{0}$ is rejected. The threshold $\alpha$ is called the significance level and $\alpha =0.01$ or 0.05 is often used.

1.2 Kernel method for the two sample problem

Let k be a measurable positive definite kernel on $\mathcal {X}$ satisfying $\int _{\mathcal {X}}\int _{\mathcal {X}} k(x,y)^{2}dP(x)dQ(y) < \infty$. In [18, 19], a statistics

$$\begin{aligned}&\mathrm {MMD}_{u}(\varvec{X}_{m},\varvec{Y}_{n};k)^{2} \nonumber \\&\quad := \frac{1}{m(m-1)}\sum _{i =1}^{m} \sum _{j \ne i}^{m} k(X_{i},X_{j}) + \frac{1}{n(n-1)}\sum _{a=1}^{n}\sum _{b \ne a}^{n} k(Y_{a},Y_{b}) \nonumber \\&\qquad - \frac{2}{mn}\sum _{i=1}^{m} \sum _{a=1}^{n}k(X_{i},Y_{a}) \end{aligned}$$

(20)

is used to the two sample test and the distribution function is given as follows:

Theorem 12

(Theorem 8 in [18], Theorem 12 in [19]) Under the null hypothesis $H_{0}$, $n\mathrm {MMD}_{u}(\varvec{X}_{n},\varvec{Y}_{n};k)^{2} \rightarrow _{\mathrm {d}}\sum _{i=1}^{\infty }\lambda _{i}(z_{i}^{2}-2)$ where $z_{1},\ldots , {\mathrm {i.i.d.}}\sim \mathcal {N}(0,2)$, $\{\lambda _{i}\}_{i=1}^{\infty }$ are the solutions to the eigenvalue equation

$$\begin{aligned} \int _{\mathcal {X}} {\tilde{k}}(x, x')\psi _{i}(x)dP(x) = \lambda _{i} \psi _{i}(x'), \end{aligned}$$

(21)

and ${\tilde{k}}(x_{i},x_{j})=k(x_{i},x_{j})-\int _{\mathcal {X}}k(x_{i},x)dP(x)-\int _{\mathcal {X}}k(x,x_{j})dP(x)-\int _{\mathcal {X}}\int _{\mathcal {X}}k(x,x')dP(x)dP(x')$.

In order to obtain the distribution of $\sum _{i=1}^{\infty }\lambda _{i}(z_{i}^{2}-2)$ numerically, we approximate the eigenvalues $\{\lambda _{i}\}_{i=1}^{\infty }$ in Eq. (21). Let $\tilde{\varvec{k}}$ denote the centered Gram matrix of $\{x_{1},\ldots ,x_{n}\}$ whose (i, j) component is given by $(\tilde{\varvec{k}})_{i,j}=k(x_{i},x_{j})- n^{-1} \sum _{b=1}^{n}k(x_{i},x_{b})-n^{-1} \sum _{a=1}^{n}k(x_{a},x_{j}) + n^{-2} \sum _{a,b=1}^{n}k(x_{a},x_{b})$ and $\{{\hat{\mu }}_{i}\}_{i=1}^{n}$ be the set of the eigenvalues of $\tilde{\varvec{k}}$. Then, it is shown from Theorem 1 in [20] that $\sum _{i=1}^{n}{\hat{\lambda }}_{i}(z_{i}^{2}-2) \rightarrow _{\mathrm {d}}\sum _{i=1}^{\infty }\lambda _{i}(z_{i}^{2}-2)$ where ${\hat{\lambda }}_{i}=n^{-1}{\hat{\mu }}_{i}$. Therefore, the upper $\alpha$-quantile of $n\mathrm {MMD}_{u}(\varvec{X}_{n}, \varvec{Y}_{n})^{2}$ is numerically obtained from the histogram of $\sum _{i=1}^{n}{\hat{\lambda }}_{i}(z_{i}^{2}-2)$. Since the current null hypothesis is $P=Q$, we have $Y_{i} \sim P$ and we can approximate the eigenvalues on the aggregated data, that is, the eigenvalues are approximated by the centered Gram matrix of $\{x_{1},\ldots ,x_{n},y_{1},\ldots ,y_{n}\}$. We estimate the quantile of $\sum _{i=1}^{2n}{\hat{\lambda }}_{i}(z_{i}^{2}-2)$ by the (standard) bootstrap method. To sum up, the algorithm of the kernel two sample problem is given as follows:

Since the output ${\hat{p}}$ of Algorithm 3 is the acceptance ratio of $H_{0}$, $1-{\hat{p}}$ is the type I error when $H_{0}$ is true.