Schmidt L, Santurkar S, Tsipras D, et al. Adversarially Robust Generalization Requires More Data[C]. neural information processing systems, 2018: 5014-5026.

@article{schmidt2018adversarially,

title={Adversarially Robust Generalization Requires More Data},

author={Schmidt, Ludwig and Santurkar, Shibani and Tsipras, Dimitris and Talwar, Kunal and Madry, Aleksander},

pages={5014--5026},

year={2018}}

本文在二分类高斯模型和伯努利模型上分析adversarial, 指出对抗稳定的模型需要更多的数据支撑.

主要内容

高斯模型定义: 令\(\theta^* \in \mathbb{R}^n\)为均值向量, \(\sigma >0\), 则\((\theta^*, \sigma)\)-高斯模型按照如下方式定义: 首先从等概率采样标签\(y \in \{\pm 1\}\), 再从\(\mathcal{N}(y \cdot \theta^*, \sigma^2I)\)中采样\(x \in \mathbb{R}^d\).

伯努利模型定义: 令\(\theta^* \in \{\pm1\}^d\)为均值向量, \(\tau >0\), 则\((\theta^*, \tau)\)-伯努利模型按照如下方式定义: 首先等概率采样标签\(y \in \{\pm 1\}\), 在从如下分布中采样\(x \in \{\pm 1\}^d\):

\[x_i =
\left \{
\begin{array}{rl}
y \cdot \theta_i^* & \mathrm{with} \: \mathrm{probability} \: 1/2+\tau \\
-y \cdot \theta_i^* & \mathrm{with} \: \mathrm{probability} \: 1/2-\tau
\end{array} \right.
\]

分类错误定义: 令\(\mathcal{P}: \mathbb{R}^d \times \{\pm 1\} \rightarrow \mathbb{R}\)为一分布, 则分类器\(f:\mathbb{R}^d \rightarrow \{\pm1\}\)的分类错误\(\beta\)定义为\(\beta=\mathbb{P}_{(x, y) \sim \mathcal{P}} [f(x) \not =y]\).

Robust分类错误定义: 令\(\mathcal{P}: \mathbb{R}^d \times \{\pm 1\} \rightarrow \mathbb{R}\)为一分布, \(\mathcal{B}: \mathbb{R}^d \rightarrow \mathscr{P}(\mathbb{R}^d)\)为一摄动集合. 则分类器\(f:\mathbb{R}^d \rightarrow \{\pm1\}\)的\(\mathcal{B}\)-robust 分类错误率\(\beta\)定义为\(\beta=\mathbb{P}_{(x, y) \sim \mathcal{P}} [\exist x' \in \mathcal{B}(x): f(x') \not = y]\).

注: 以\(\mathcal{B}_p^{\epsilon}(x)\)表示\(\{x' \in \mathbb{R}^d|\|x'-x\|_p \le \epsilon\}\).

高斯模型

upper bound

定理18: 令\((x_1,y_1),\ldots, (x_n,y_n) \in \mathbb{R}^d \times \{\pm 1\}\) 独立采样于同分布\((\theta^*, \sigma)\)-高斯模型, 且\(\|\theta^*\|_2=\sqrt{d}\). 令\(\hat{w}:=\bar{z}/\|\bar{z}\| \in \mathbb{R}^d\), 其中\(\bar{z}=\frac{1}{n} \sum_{i=1}^n y_ix_i\). 则至少有\(1-2\exp(-\frac{d}{8(\sigma^2+1)})\)的概率, 线性分类器\(f_{\hat{w}}\)的分类错误率至多为:

\[\exp (-\frac{(2\sqrt{n}-1)^2d}{2(2\sqrt{n}+4\sigma)^2\sigma^2}).
\]

定理21: 令\((x_1,y_1),\ldots, (x_n,y_n) \in \mathbb{R}^d \times \{\pm 1\}\) 独立采样于同分布\((\theta^*, \sigma)\)-高斯模型, 且\(\|\theta^*\|_2=\sqrt{d}\). 令\(\hat{w}:=\bar{z}/\|\bar{z}\| \in \mathbb{R}^d\), 其中\(\bar{z}=\frac{1}{n} \sum_{i=1}^n y_ix_i\). 如果

\[\epsilon \le \frac{2\sqrt{n}-1}{2\sqrt{n}+4\sigma} - \frac{\sigma\sqrt{2\log 1/\beta}}{\sqrt{d}},
\]

则至少有\(1-2\exp(-\frac{d}{8(\sigma^2+1)})\)的概率, 线性分类器\(f_{\hat{w}}\)的\(\ell_{\infty}^{\epsilon}\)-robust 分类错误率至多为\(\beta\).

lower bound

定理11: 令\(g_n\)为任意的学习算法, 并且, \(\sigma > 0, \epsilon \ge 0\), 设\(\theta \in \mathbb{R}^d\)从\(\mathcal{N}(0,I)\)中采样. 并从\((\theta,\sigma)\)-高斯模型中采样\(n\)个样本, 由此可得到分类器\(f_n: \mathbb{R}^d \rightarrow \{\pm 1\}\). 则分类器关于\(\theta, (y_1,\ldots, y_n), (x_1,\ldots, x_n)\)的\(\ell_{\infty}^{\epsilon}\)-robust 分类错误率至少

\[\frac{1}{2} \mathbb{P}_{v\sim \mathcal{N}(0, I)} [\sqrt{\frac{n}{\sigma^2+n}} \|v\|_{\infty} \le \epsilon ].
\]

伯努利模型

upper bound

令\((x, y) \in \mathbb{R}^d \times \{\pm1\}\)从一\((\theta^*, \tau)\)-伯努利模型中采样得到. 令\(\hat{w}=z / \|z\|_2\), 其中\(z=yx\). 则至少有\(1- \exp (-\frac{\tau^2d}{2})\)的概率, 线性分类器\(f_{\hat{w}}\)的分类错误率至多为\(\exp (-2\tau^4d)\).

lower bound

引理30: 令\(\theta^* \in \{\pm1\}^d\) 并且关于\((\theta^*, \tau)-伯努利模型\)考虑线性分类器\(f_{\theta^*}\),

\(\ell_{\infty}^{\tau}\)-robustness: \(f_{\theta^*}\)的\(\ell_{\infty}^{\tau}\)-robust分类误差率至多为\(2\exp (-\tau^2d/2)\).

\(\ell_{\infty}^{3\tau}\)-nonrobustness: \(f_{\theta^*}\)的\(\ell_{\infty}^{3\tau}\)-robust分类误差率至少为\(1-2\exp (-\tau^2d/2)\).

Near-optimality of \(\theta^*\): 对于任意线性分类器, \(\ell_{\infty}^{3\tau}\)-robust 分类误差率至少为\(\frac{1}{6}\).

定理31: 令\(g_n\)为任一线性分类器学习算法. 假设\(\theta^*\)均匀采样自\(\{\pm1\}^d\), 并从\((\theta^*, \tau)\)-伯努利分布(\(\tau \le 1/4\))中采样\(n\)个样本, 并借由\(g_n\)得到线性分类器\(f_{w}\).同时\(\epsilon < 3\tau\)且\(0 < \gamma < 1/2\), 则当

\[n \le \frac{\epsilon^2\gamma^2}{5000 \cdot \tau^4 \log (4d/\gamma)},
\]

\(f_w\)关于\(\theta^*, (y_1,\ldots, y_n), (x_1,\ldots, x_n)\)的期望\(\ell_{\infty}^{\epsilon}\)-robust 分类误差至少为\(\frac{1}{2}-\gamma\).

Adversarially Robust Generalization Requires More Data的更多相关文章

  1. Exploring Architectural Ingredients of Adversarially Robust Deep Neural Networks

    目录 概 主要内容 深度 宽度 代码 Huang H., Wang Y., Erfani S., Gu Q., Bailey J. and Ma X. Exploring architectural ...

  2. 自定义 ASP.NET Identity Data Model with EF

    One of the first issues you will likely encounter when getting started with ASP.NET Identity centers ...

  3. ExtJs Ext.data.Model 学习笔记

    Using a Proxy Ext.define('User', { extend: 'Ext.data.Model', fields: ['id', 'name', 'email'], proxy: ...

  4. Buffer Data

    waylau/netty-4-user-guide: Chinese translation of Netty 4.x User Guide. 中文翻译<Netty 4.x 用户指南> h ...

  5. Buffer Data RDMA 零拷贝 直接内存访问

    waylau/netty-4-user-guide: Chinese translation of Netty 4.x User Guide. 中文翻译<Netty 4.x 用户指南> h ...

  6. A Complete Tutorial on Tree Based Modeling from Scratch (in R & Python)

    A Complete Tutorial on Tree Based Modeling from Scratch (in R & Python) MACHINE LEARNING PYTHON  ...

  7. Wide and Deep Learning Model

    https://blog.csdn.net/starzhou/article/details/78845931 The Wide and Deep Learning Model(译文+Tensorlf ...

  8. Android开发训练之第五章——Building Apps with Connectivity & the Cloud

    Building Apps with Connectivity & the Cloud These classes teach you how to connect your app to t ...

  9. C# Interview Questions:C#-English Questions

    This is a list of questions I have gathered from other sources and created myself over a period of t ...

随机推荐

  1. Vue2全家桶+Element搭建的PC端在线音乐网站

    目录 1,前言 2,已有功能 3,使用 4,目录结构 5,页面效果 登录页 首页 排行榜 歌单列表 歌单详情 歌手列表 歌手详情 MV列表 MV详情 搜索页 播放器 1,前言 项目基于Vue2全家桶及 ...

  2. CPU如何同时运行多个进程?

    1 # -*- coding: utf-8 -*- 2 import re 3 mem = [x for x in re.split('[\r|\n]', ''' 4 store a 1 5 add ...

  3. Linux下强制踢掉登陆用户

    1.pkill -kill -t   tty 例:pkill -kill -t tty1

  4. Output of C++ Program | Set 14

    Predict the output of following C++ program. Difficulty Level: Rookie Question 1 1 #include <iost ...

  5. 数据库ER图基础概念

    ER图分为实体.属性.关系三个核心部分.实体是长方形体现,而属性则是椭圆形,关系为菱形. ER图的实体(entity)即数据模型中的数据对象,例如人.学生.音乐都可以作为一个数据对象,用长方体来表示, ...

  6. Can namespaces be nested in C++?

    In C++, namespaces can be nested, and resolution of namespace variables is hierarchical. For example ...

  7. java中super的几种用法,与this的区别

    1. 子类的构造函数如果要引用super的话,必须把super放在函数的首位. class Base { Base() { System.out.println("Base"); ...

  8. 【Spring Framework】Spring入门教程(五)AOP思想和动态代理

    本文主要讲解内容如下: Spring的核心之一 - AOP思想 (1) 代理模式- 动态代理 ① JDK的动态代理 (Java官方) ② CGLIB 第三方代理 AOP概述 什么是AOP(面向切面编程 ...

  9. Python 中更安全的 eval

    问题 想要将一段列表形式的字符串转为 list,但是担心这个动态的字符串可能是恶意的代码?使用 eval 将带来安全隐患.比如: # 期望是 eval('[1, 2, 3]') # 实际上是 eval ...

  10. CF135A Replacement 题解

    Content 有 \(n\) 个数 \(a_1,a_2,a_3,...,a_n\),试用 \(1\) ~ \(10^9\) 之间的数(除了本身)代替其中的一个数,使得这 \(n\) 个数的总和最小, ...