Koh P W, Liang P. Understanding black-box predictions via influence functions[C]. international conference on machine learning, 2017: 1885-1894.

@article{koh2017understanding,

title={Understanding black-box predictions via influence functions},

author={Koh, Pang Wei and Liang, Percy},

pages={1885--1894},

year={2017}}

本文介绍了如果计算(估计)损失关于样本的一些影响因子, 并介绍了一些应用范围.

主要内容

假设样本\(z_1,\ldots, z_n\), \(z_i = (x_i,y_i) \in \mathcal{X} \times \mathcal{Y}\), 通过最小化经验损失

\[\hat{\theta} := \arg \min_{\theta \in \Theta} \frac{1}{n} \sum_{i=1}^n L(z_i, \theta),
\]

找到最优解.

且假设\(L\)关于样本和参数都是二阶可导且强凸的.

样本重要性分析

显然, 此时给定一个测试样本\(z_{test}\), 其对应的损失为\(L(z_{test},\hat{\theta})\), 那么衡量一个样本重要性的一个重要指标便是, 倘若在移除样本\(z\)的情况下重新训练模型, 对应的参数和损失的变化.

假设在移除样本\(z\)的情况下训练得到的最优参数为\(\hat{\theta}_{-z}\), 并引入

\[\hat{\theta}_{\epsilon, z} := \arg \min_{\theta \in \Theta} \frac{1}{n}\sum_{i=1}^n L(z_i, \theta)+\epsilon L(z,\theta),
\]

易得\(\hat{\theta}_{-z} = \hat{\theta}_{-\frac{1}{n},z}\).

\[\tag{1}
\mathcal{I}_{up, params} (z) := \frac{d \hat{\theta}_{\epsilon, z}}{d \epsilon}|_{\epsilon=0} = -H_{\hat{\theta}}^{-1} \nabla_{\theta} L(z, \hat{\theta}),
\]

其中\(H_{\theta}:= \frac{1}{n} \sum_{i=1}^n \nabla_{\theta}^2 L(z_i, \hat{\theta})\).

我们可以得到, 参数的变化量的一阶近似

\[\hat{\theta}_{-z} - \hat{\theta} \approx -\frac{1}{n} \mathcal{I}_{up,params} (z).
\]

进一步, 我们定义损失的变化量

\[\tag{2}
\begin{array}{ll}
\mathcal{I}_{up, loss} (z, z_{test})
& := \frac{dL(z_{test}, \hat{\theta}_{\epsilon, z})}{d \epsilon} |_{\epsilon = 0} \\
& = \nabla_{\theta} L(z_{test}, \hat{\theta})^T \frac{d \hat{\theta}_{\epsilon, z}}{d \epsilon} |_{\epsilon = 0} \\
& = -\nabla_{\theta} L(z_{test}, \hat{\theta})^T H_{\hat{\theta}}^{-1} \nabla_{\theta} L(z, \hat{\theta}).
\end{array}
\]

样本摄动对损失的影响

倘若我们对其中一个样本\(z\)添加一个扰动\(\delta\), 并在新的数据\(z_{\delta}:=(x+\delta,y)\)上训练, 得到模型, 其参数和损失会如何变化?

我们定义

\[\hat{\theta}_{\epsilon, z_{\delta}, -z}:= \arg \min_{\theta \in \Theta} \frac{1}{n} \sum_{i=1}^n L(z_i, \theta) + \epsilon L(z_{\delta},\theta)-\epsilon L(z,\theta),
\]

并令\(\hat{\theta}_{z_{\delta},-z}:= \hat{\theta}_{\frac{1}{n}, z_{\delta}, -z}\)

同样可以证明

\[\tag{3}
\frac{d \hat{\theta}_{\epsilon, z_{\delta}, -z}}{d \epsilon} |_{\epsilon=0} = -H_{\hat{\theta}}^{-1} (\nabla_{\theta} L(z_{\delta}, \hat{\theta}) -\nabla_{\theta} L(z, \hat{\theta})).
\]

\[\hat{\theta}_{z_{\delta}, -z}-\hat{\theta} \approx -\frac{1}{n}H_{\hat{\theta}}^{-1} (\nabla_{\theta} L(z_{\delta}, \hat{\theta}) -\nabla_{\theta} L(z, \hat{\theta})) \approx -\frac{1}{n} H_{\hat{\theta}}^{-1} \nabla_x \nabla_{\theta} L(z_{\delta}, \hat{\theta}) \delta.
\]

\[\tag{5}
\mathcal{I}_{pert, loss}(z, z_{test})^T:= \nabla_{\delta} L(z_{test}, \hat{\theta}_{z_{\delta},-z})^T \approx -\frac{1}{n} \nabla_{\theta} L(z_{test}, \hat{\theta})^T H_{\hat{\theta}}^{-1} \nabla_x \nabla_{\theta} L(z, \hat{\theta}).
\]

注:文章这里没有\(\frac{1}{n}\)且是等号(我卡在这个地方了, 推不出来).

高效计算\(H^{-1}\)

共轭梯度

此时我们不是计算\(H_{\hat{\theta}}^{-1}\), 而是计算\(s:=H_{\hat{\theta}}^{-1}v\), 比如在计算\(\mathcal{I}_{up, loss}\)的时候, \(v=\nabla_{\theta} L(z_{test}, \hat{\theta})\), 则对于固定的\(z_{test}\)想要知道不同的\(z_i\)的影响可以直接用\(s^T \nabla_{\theta} L(z_i, \hat{\theta})\), 避免了重复运算.

即求解

\[\arg \min_t \quad \frac{1}{2} t^TH_{\hat{\theta}}t - v^Tt,
\]

假设第\(k\)步为

\[t=t_k,
\]

则利用精确直线搜索

\[\arg \min_{p} \frac{1}{2} t_{k+1}^T H_{\hat{\theta}}t_{k+1}-v^Tt_{k+1}, \: \mathrm{s.t.} \: t_{k+1}=t_k + p(H_{\theta}t_k -v),
\]

\[p= -\frac{\Delta^T H_{\hat{\theta}}t_k-v^T\Delta}{\Delta^T H_{\hat{\theta}} \Delta}, \Delta=H_{\hat{\theta}}t_k-v.
\]

随机估计

这里是估计\(H_{\hat{\theta}}^{-1}\), 为了符号简便省略下表, 因为\(H^{-1}=\sum_{i=0}^{+\infty}(I-H)^i\), 用\(H_j^{-1}= \sum_{i=0}^j (I - H)^i\)表示前\(j+1\)项的和, 易知

\[H_j^{-1} = I + (I -H)H_{j-1}^{-1}, H_j^{-1} \rightarrow H^{-1}.
\]

我们从样本中均匀挑选, 计算\(\nabla_{\theta}^2 L(z_i, \hat{\theta})\) 作为\(H\)的替代, 则

\[\tilde{H}_j^{-1}=I+(I-\nabla_{\theta}^2 L(z_{s_j}, \hat{\theta}))\tilde{H}_{j-1}^{-1}.
\]

当然, 处于稳定性的考虑, 我们可以一次性采样多个来作为\(H\)的替代.

一些应用

  1. 探索模型关于样本的内在解释, 即什么样的样本模型会更加偏好之类的;
  2. 生成对抗样本;
  3. 检测目标数据分布和训练分布是否一致;
  4. 检测训练数据的标签是否正确.

附录

(1)的证明

定义\(\Delta_{\epsilon}:= \hat{\theta}_{\epsilon, z}-\hat{\theta}\), 则

\[\mathcal{I}_{up, params} (z) = \frac{d \Delta_{\epsilon}}{d \epsilon} |_{\epsilon =0}.
\]

由一阶最优条件可知

\[0= \frac{1}{n} \sum_{i=1}^n \nabla_{\theta} L(z_i, \hat{\theta}) := R(\hat{\theta}), \\
0= R(\hat{\theta}_{\epsilon, z}) + \epsilon \nabla_{\theta} L(z, \hat{\theta}_{\epsilon, z}),
\]

把\(\hat{\theta}_{\epsilon, \theta}\)看成自变量(固定\(\epsilon\)), 第二个等式右边是关于这个变量的一个函数, 则其在\(\hat{\theta}_{\epsilon, \theta}=\hat{\theta}\)处的泰勒展式为

\[R(\hat{\theta}) + \epsilon \nabla_{\theta} L(z, \hat{\theta})+ [\nabla_{\theta} R(\hat{\theta}) + \epsilon \nabla_{\theta}^2 L(z, \hat{\theta})] \Delta_{\epsilon} + o(\Delta_{\epsilon})=0
\]

\[\Delta_{\epsilon} = -[\nabla_{\theta} R(\hat{\theta}) + \epsilon \nabla_{\theta}^2 L(z, \hat{\theta})] ^{-1} [0 + \epsilon \nabla_{\theta} L(z, \hat{\theta})+o(\Delta_{\epsilon})],
\]

因为\(\epsilon \rightarrow 0\), \(\Delta_{\epsilon} \rightarrow0\) 易知,

\[\frac{d \Delta_{\epsilon}}{d \epsilon |_{\epsilon =0}} = -H_{\hat{\theta}}^{-1} \nabla_{\theta} L(z, \hat{\theta}).
\]

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