回归与分类Regression & Classification

一台机械臂准备抓一个零件。它在动手之前就想知道：这次抓取会不会成功？它能测到的只有几个数，比如夹爪最大张开量比物体宽出多少毫米，夹爪方向和物体主轴偏了多少度。成败与这些数之间的关系没有现成的公式：摩擦、表面、物体重心都会影响结果。但机器人可以把过去几百次尝试记下来，每次记一行“特征 → 成败”，然后从这些数据里学出一个规则。

这就是监督学习：给定数据 \(\{(\mathbf x_i,y_i)\}_{i=1}^n\)，其中 \(\mathbf x_i\) 是输入（特征），\(y_i\) 是我们想预测的目标，找一个函数 \(\hat y=f_{\boldsymbol\theta}(\mathbf x)\) 让它在没见过的输入上也预测得准。目标 \(y\) 是连续数值时叫回归（关节摩擦力矩、电池剩余时间、物体距离）；目标是若干类别之一时叫分类（抓取成/败、地面是沥青/碎石/草地、图像里是哪个物体）。Part VI 的模仿学习本质上也是监督学习：输入是观测，目标是专家的动作（22.1 节）。

本节不需要神经网络。我们用最简单的模型（线性模型、逻辑回归、softmax、k 近邻）把监督学习的骨架讲清楚：模型 + 噪声假设 → 似然 → 损失 → 优化 → 在测试集上评估。后面的每一节（深度网络、扩散策略、强化学习里的价值网络）都只是把这里的“模型”换得更大，骨架不变。

A robot arm is about to pick up a part. Before it moves it would like to know: will this grasp succeed? All it can measure are a few numbers, for example how many millimetres the gripper's maximum opening exceeds the object's width, and how many degrees the gripper is misaligned with the object's main axis. No ready-made formula links these numbers to the outcome: friction, the surface and the object's centre of mass all play a part. But the robot can log its past few hundred attempts, one row "features → outcome" each, and learn a rule from that data.

That is supervised learning: given data \(\{(\mathbf x_i,y_i)\}_{i=1}^n\), where \(\mathbf x_i\) is the input (the features) and \(y_i\) the target we want to predict, find a function \(\hat y=f_{\boldsymbol\theta}(\mathbf x)\) that also predicts well on inputs it has not seen. When the target \(y\) is a continuous number it is regression (joint friction torque, remaining battery time, distance to an object); when it is one of several classes it is classification (grasp success/failure, ground is asphalt/gravel/grass, which object is in the image). The imitation learning of Part VI is supervised learning at heart too: the input is an observation and the target is the expert's action (Lesson 22.1).

This lesson needs no neural networks. With the simplest models (linear models, logistic regression, softmax, k nearest neighbours) we lay out the skeleton of supervised learning: model + noise assumption → likelihood → loss → optimization → evaluation on a test set. Every later lesson (deep networks, diffusion policies, value networks in reinforcement learning) only swaps in a bigger "model"; the skeleton stays the same.

1.13 节把直线拟合当作几何问题：方程比未知数多，取让残差平方和最小的解。4.6 节从概率角度给了同一个答案的理由：如果测量 = 真值 + 高斯噪声，那么最大似然估计就是最小二乘。现在把两者放在一起，用到“拟合一个函数”上。假设数据来自下面的模型，各点的噪声 \(\varepsilon_i\) 相互独立、同分布：

Lesson 1.13 treated line fitting as geometry: more equations than unknowns, so take the solution with the smallest sum of squared residuals. Lesson 4.6 gave the probabilistic reason for the same answer: if measurement = truth + Gaussian noise, the maximum-likelihood estimate is least squares. Now put the two together and apply them to fitting a function. Suppose the data come from the model below, with the noise \(\varepsilon_i\) of different points independent and identically distributed:

每一条候选直线都对应一个“生成数据的故事”：在每个 \(x_i\) 处，\(y\) 应该落在以直线为中心、宽度为 \(\sigma\) 的高斯钟形里。数据的似然是所有钟形在各自数据点处的高度之积；它越大，这条直线把数据解释得越好。取负对数，乘积变成求和，就得到负对数似然（NLL）：残差平方和除以 \(2\sigma^2\)，再加一个只与 \(\sigma\) 有关的常数。所以无论 \(\sigma\) 取多少，让 NLL 最小的直线就是最小二乘直线。下图里你可以亲手拖直线、拖数据点，看两者怎样一起变。

Every candidate line comes with a "story of how the data were generated": at each \(x_i\), \(y\) should fall inside a Gaussian bell centred on the line with width \(\sigma\). The likelihood of the data is the product of the heights of the bells at their data points; the larger it is, the better the line explains the data. Taking the negative log turns the product into a sum, the negative log-likelihood (NLL): the sum of squared residuals divided by \(2\sigma^2\), plus a constant that depends only on \(\sigma\). So whatever \(\sigma\) is, the line that minimizes the NLL is the least-squares line. In the figure below you can drag the line and the data points yourself and watch the two move together.

默认数据下，最小二乘直线是 \(\hat y=0.54x+0.49\)，残差平方和 2.04，RMS 残差 0.48；你在初始的水平线 \(y=0\) 处看到的残差平方和是 17.97。第二个任务揭示了一件事：\(\sigma\) 的最大似然估计恰好是 RMS 残差 \(\hat\sigma=\sqrt{\tfrac1n\sum r_i^2}\)（第 5 节推导）。它比生成数据用的 0.7 小，一部分是只有 9 个点的抽样波动，另一部分是系统性的：直线本身是照着这 9 个点拟合的，残差天然偏小，这正是 4.6 节说 \(\hat\sigma^2_{\mathrm{ML}}\) 偏小、常把分母换成 \(n-2\)（这里减去两个参数）的原因。

With the default data the least-squares line is \(\hat y=0.54x+0.49\), with a sum of squared residuals of 2.04 and an RMS residual of 0.48; the flat starting line \(y=0\) gives 17.97. The second task shows something: the maximum-likelihood estimate of \(\sigma\) is exactly the RMS residual \(\hat\sigma=\sqrt{\tfrac1n\sum r_i^2}\) (derived in Section 5). It is smaller than the 0.7 used to generate the data, partly from the sampling luck of only 9 points and partly systematically: the line was fitted to these very 9 points, so its residuals come out small. That is why Lesson 4.6 says \(\hat\sigma^2_{\mathrm{ML}}\) is biased low and the denominator is often changed, here to \(n-2\) for the two fitted parameters.

直线太简单了。真实的关节摩擦力矩 \(\tau_f\) 和关节速度 \(v\) 的关系是弯的：静止附近有一个跳变（库仑摩擦，8.4 节），再加上随速度线性增长的粘滞摩擦。好消息是，只要把输入先变成一组特征 \(\boldsymbol\phi(v)=(\phi_1(v),\dots,\phi_d(v))\)，再让模型是特征的线性组合 \(\hat y=\boldsymbol\theta^\top\hl{phi}{\boldsymbol\phi}(v)\)，那么预测对 \(\boldsymbol\theta\) 仍是线性的，NLL 仍是 \(\boldsymbol\theta\) 的二次函数，解仍然是 1.13 节的正规方程，只是矩阵 \(A\) 的每一行换成了 \(\boldsymbol\phi(v_i)^\top\)。特征可以是多项式 \(1,v,v^2,\dots\)，可以是一排高斯“鼓包”（径向基函数，RBF），也可以是物理给出的 \(1,\operatorname{sgn}v,v\)。

A straight line is too simple. A real joint's friction torque \(\tau_f\) depends on the joint velocity \(v\) in a bent way: a jump near standstill (Coulomb friction, Lesson 8.4) plus viscous friction that grows linearly with speed. The good news: first turn the input into a set of features \(\boldsymbol\phi(v)=(\phi_1(v),\dots,\phi_d(v))\) and make the model a linear combination of the features, \(\hat y=\boldsymbol\theta^\top\hl{phi}{\boldsymbol\phi}(v)\). Then the prediction is still linear in \(\boldsymbol\theta\), the NLL is still quadratic in \(\boldsymbol\theta\), and the answer is still the normal equations of Lesson 1.13, with each row of the matrix \(A\) replaced by \(\boldsymbol\phi(v_i)^\top\). The features can be polynomials \(1,v,v^2,\dots\), a row of Gaussian "bumps" (radial basis functions, RBF), or the \(1,\operatorname{sgn}v,v\) that physics suggests.

默认数据下，多项式在 7 阶时测试 RMSE 最低（0.098 N·m）；12 阶时训练 RMSE 0.040，正好是噪声水平，测试 RMSE 却到了 2.5 N·m，比摩擦本身还大。物理特征 \((1,\operatorname{sgn}v,v)\) 学到 \(\hat{\boldsymbol\theta}=(0.062,\,0.364,\,0.249)\)，与真值 \((0.05,\,0.35,\,0.25)\) 很接近，测试 RMSE 0.065：剩下的误差主要来自 \(\operatorname{sgn}\) 在零速附近与真实的平滑过渡不一样（模型误差）。这里只是先碰一下“过拟合”：训练误差低不等于模型好，一定要在没参与训练的数据上打分。为什么会这样、怎样系统地选模型复杂度（验证集、交叉验证、正则化），是 16.2 节的内容。

With the default data the polynomial's lowest test RMSE comes at degree 7 (0.098 N·m); at degree 12 the training RMSE is 0.040, exactly the noise level, while the test RMSE reaches 2.5 N·m, larger than the friction itself. The physics features \((1,\operatorname{sgn}v,v)\) learn \(\hat{\boldsymbol\theta}=(0.062,\,0.364,\,0.249)\), close to the truth \((0.05,\,0.35,\,0.25)\), with a test RMSE of 0.065; what is left comes mostly from \(\operatorname{sgn}\) differing from the true smooth transition near zero speed (model error). This is only a first touch of "overfitting": a low training error does not make a good model, and you must score on data that took no part in training. Why this happens and how to choose model complexity systematically (validation sets, cross-validation, regularization) is the subject of Lesson 16.2.

回到开头的抓取。每次尝试记两个特征：\(x_1\) = 张开余量（夹爪最大张开量减物体宽度，mm），\(x_2\) = 夹爪与物体主轴的夹角偏差（°），标签 \(y=1\) 表示成功、\(y=0\) 表示失败。能不能还用最小二乘去拟合 0 和 1？可以算，但它的输出会跑到 \([0,1]\) 之外，也不是一个概率，而且远处那些“明显成功”的点会把直线拽偏（第 7 节误区 5）。更好的做法还是先写一个噪声模型：结果是一枚有偏的硬币，成功的概率随特征平滑变化，

Back to the grasp of the opening. Each attempt records two features: \(x_1\) = opening margin (the gripper's maximum opening minus the object's width, mm) and \(x_2\) = the misalignment between the gripper and the object's main axis (°); the label \(y=1\) means success, \(y=0\) failure. Could we still fit the 0s and 1s by least squares? It computes, but its output leaves \([0,1]\), is not a probability, and far-away "obvious success" points drag the line off (Section 7, pitfall 5). The better route is again to write a noise model first: the outcome is a biased coin whose probability of success varies smoothly with the features,

这里 \(\sigma(\cdot)\) 是 logistic（sigmoid）函数，不是噪声的标准差。\(z\) 叫 logit，即对数几率 \(\log\frac{p}{1-p}\)：4.2 节的“对数几率”专栏已经说明，两类特征是等方差高斯时后验恰好是 \(z\) 线性的 sigmoid。\(L\) 是伯努利分布的平均负对数似然，也就是标签分布与预测分布之间的交叉熵（4.11 节），机器学习库里叫二元交叉熵（BCE）。它没有闭式解，但它是凸的（第 5 节），用 6.3 节的梯度下降一定能走到全局最优，而且梯度出奇地简单：\(\nabla_{\mathbf w}L=\frac1n\sum_i(\sigma(z_i)-y_i)\,\mathbf x_i\)，即“预测概率减标签”乘以输入。

Here \(\sigma(\cdot)\) is the logistic (sigmoid) function, not the noise standard deviation. \(z\) is called the logit, the log-odds \(\log\frac{p}{1-p}\): the log-odds box of Lesson 4.2 showed that when the two classes have equal-variance Gaussian features, the posterior is exactly a sigmoid of a linear \(z\). \(L\) is the mean negative log-likelihood of a Bernoulli distribution, i.e. the cross-entropy between the label distribution and the predicted distribution (Lesson 4.11); libraries call it binary cross-entropy (BCE). It has no closed form, but it is convex (Section 5), so the gradient descent of Lesson 6.3 is sure to reach the global optimum, and the gradient is surprisingly simple: \(\nabla_{\mathbf w}L=\frac1n\sum_i(\sigma(z_i)-y_i)\,\mathbf x_i\), "predicted probability minus label" times the input.

可分数据为什么停不下来？当一条直线能把两类完全分开时，把 \(\mathbf w\) 和 \(b\) 一起放大 2 倍，边界不动，但每个点的 \(|z|\) 都变大，每个点的概率都更接近它的标签，交叉熵就更小。所以损失没有最小值，只有下确界 0，最大似然解在无穷远处：梯度下降会让 \(\|\mathbf w\|\) 大致按 \(\log t\) 一直增长，模型变得无比自信（3000 步时 9.4，20000 步时 13.6）。实践中的解药是正则化（给 \(\mathbf w\) 一个先验，4.6 节的 MAP）或提前停止，见 16.2 节。

Why does separable data never stop? When a line can split the two classes perfectly, scaling \(\mathbf w\) and \(b\) together by 2 leaves the boundary in place but makes every \(|z|\) bigger, every point's probability closer to its label, and the cross-entropy smaller. So the loss has no minimum, only an infimum of 0, and the maximum-likelihood solution sits at infinity: gradient descent makes \(\|\mathbf w\|\) grow roughly like \(\log t\) for ever and the model becomes boundlessly confident (9.4 at 3000 steps, 13.6 at 20000). The cures used in practice are regularization (a prior on \(\mathbf w\), the MAP of Lesson 4.6) or early stopping, see Lesson 16.2.

一台轮式机器人想知道自己正在什么地面上走，好调整速度和牵引控制。它有两个便宜的特征：车身竖直方向振动的 RMS（m/s²）和车轮打滑率（%）。类别有三个：沥青、碎石、草地。逻辑回归的推广是给每一类 \(k\) 一个 logit \(z_k=\mathbf w_k^\top\mathbf x+b_k\)，再用 softmax 把 \(K\) 个 logit 变成 \(K\) 个加起来为 1 的概率：

A wheeled robot wants to know what ground it is on, to adapt its speed and traction control. It has two cheap features: the RMS of the body's vertical vibration (m/s²) and the wheel slip ratio (%). There are three classes: asphalt, gravel, grass. The generalization of logistic regression gives every class \(k\) its own logit \(z_k=\mathbf w_k^\top\mathbf x+b_k\) and turns the \(K\) logits into \(K\) probabilities summing to 1 with the softmax:

噪声模型是“掷一个 \(K\) 面的有偏骰子”（类别分布），损失仍是负对数似然 = 交叉熵。类别 \(k\) 与 \(j\) 之间的边界是 \(z_k=z_j\)，即 \((\mathbf w_k-\mathbf w_j)^\top\mathbf x+(b_k-b_j)=0\)：仍是直线，所以 softmax 回归把平面切成若干个凸的多边形区域。想要弯曲的边界，就像 FIG 2 一样先做特征映射：加入 \(x_1^2,x_2^2,x_1x_2\) 之后，边界在原平面里变成二次曲线（椭圆、双曲线），但模型对参数仍是线性的，损失仍是凸的。

The noise model is "roll a biased \(K\)-sided die" (a categorical distribution), and the loss is again the negative log-likelihood = cross-entropy. The boundary between classes \(k\) and \(j\) is \(z_k=z_j\), i.e. \((\mathbf w_k-\mathbf w_j)^\top\mathbf x+(b_k-b_j)=0\): still a straight line, so softmax regression cuts the plane into convex polygonal regions. For curved boundaries, map the features first as in FIG 2: after adding \(x_1^2,x_2^2,x_1x_2\) the boundaries in the original plane become conics (ellipses, hyperbolas), yet the model is still linear in its parameters and the loss still convex.

“地形”数据上线性 softmax 训练 200 步后测试准确率 94.0%。继续训练，训练准确率还会升到 98.7%，测试交叉熵却在约 200 步后从 0.194 慢慢回升（5000 步时 0.227）：这又是一次过拟合的预告。“圆环”上线性模型的测试准确率只有 49%，因为没有任何一组直线能围出一个圆；二次特征 200 步就到 97.3%，1000 步到 99.3%。

On "terrain", linear softmax reaches 94.0% test accuracy after 200 steps. Keep training and training accuracy climbs to 98.7%, yet the test cross-entropy slowly rises again after about 200 steps (from 0.194 to 0.227 at 5000 steps): another preview of overfitting. On "ring" the linear model's test accuracy is only 49%, because no set of straight lines can enclose a circle; quadratic features reach 97.3% in 200 steps and 99.3% in 1000.

前面的模型都是参数化的：训练把数据“压缩”进 \(\mathbf w,b\)，之后数据可以扔掉。k 近邻（k-NN）走另一个极端：根本不训练，把所有训练样本都存着；要分类一个新点，就找离它最近的 \(k\) 个训练样本，让它们投票。它是非参数方法：模型的复杂度随数据量增长，任何形状的边界都能表示，所以常被用作“不做任何假设时能做到多好”的基线。它的代价是：预测时要和所有样本比距离；“距离”完全取决于你怎样度量各个特征，单位不同的特征必须先标准化；维数高时“最近的邻居”也离得很远（维数灾难）。

The models so far are parametric: training "compresses" the data into \(\mathbf w,b\) and the data can then be thrown away. k nearest neighbours (k-NN) goes to the other extreme: it does not train at all and simply stores every training sample; to classify a new point it finds the \(k\) training samples closest to it and lets them vote. It is a non-parametric method: its complexity grows with the data and it can represent boundaries of any shape, which is why it is often used as a baseline for "how well can you do with no assumptions". The price: every prediction compares distances to all samples; the "distance" depends entirely on how you measure each feature, so features with different units must be standardized first; and in high dimensions even the "nearest" neighbours are far away (the curse of dimensionality).

标准化开启时，测试准确率在 \(k=17\) 和 \(k=23\) 到 \(35\) 之间达到 94.0%，与 FIG 4 的线性 softmax 一样；关掉标准化，\(k=25\) 时只有 83.3%，\(k=41\) 时 80.0%。小 \(k\) 的边界跟着每一个噪声点走（方差大），大 \(k\) 的边界被远处的点抹平（偏差大），这又是 16.2 节偏差–方差权衡的一个例子。在“圆环”上，k-NN 不需要任何特征工程，\(k=3\) 就有 96.7%：这正是非参数方法的长处。

With standardization on, test accuracy reaches 94.0% at \(k=17\) and for \(k=23\) to \(35\), the same as FIG 4's linear softmax; with it off, \(k=25\) gives only 83.3% and \(k=41\) gives 80.0%. A small \(k\) makes the boundary follow every noisy point (high variance); a large \(k\) smears it with far-away points (high bias), another example of the bias–variance trade-off of Lesson 16.2. On "ring", k-NN needs no feature engineering at all and gets 96.7% with \(k=3\): exactly the strength of a non-parametric method.

模型 \(y_i=\boldsymbol\theta^\top\boldsymbol\phi(\mathbf x_i)+\varepsilon_i\)，\(\varepsilon_i\sim\mathcal N(0,\sigma^2)\) 独立同分布。把 \(\boldsymbol\phi(\mathbf x_i)^\top\) 排成设计矩阵 \(\Phi\in\R^{n\times d}\) 的第 \(i\) 行，\(\mathbf y=(y_1,\dots,y_n)\)。负对数似然是

The model is \(y_i=\boldsymbol\theta^\top\boldsymbol\phi(\mathbf x_i)+\varepsilon_i\) with \(\varepsilon_i\sim\mathcal N(0,\sigma^2)\) i.i.d. Stack \(\boldsymbol\phi(\mathbf x_i)^\top\) as row \(i\) of the design matrix \(\Phi\in\R^{n\times d}\), and let \(\mathbf y=(y_1,\dots,y_n)\). The negative log-likelihood is

对 \(\boldsymbol\theta\) 求最小时第一项与 \(\boldsymbol\theta\) 无关，第二项正是 1.13 节的 \(\|A\mathbf x-\mathbf b\|^2\)（\(A=\Phi\)，\(\mathbf b=\mathbf y\)），所以 \(\hat{\boldsymbol\theta}\) 满足正规方程 \(\Phi^\top\Phi\,\hat{\boldsymbol\theta}=\Phi^\top\mathbf y\)，与 \(\sigma\) 无关（FIG 1 的第二个任务）。再对 \(\sigma\) 求导并令其为零：\(-n/\sigma+\|\mathbf r\|^2/\sigma^3=0\)，得 \(\hat\sigma^2=\tfrac1n\|\mathbf y-\Phi\hat{\boldsymbol\theta}\|^2\)，即残差的均方。数值上不要真的去求 \((\Phi^\top\Phi)^{-1}\)：多项式特征的 \(\Phi\) 条件数很大，\(\Phi^\top\Phi\) 把它平方，FIG 2 用的是 QR 分解（1.13 节的伪逆也可以）。每个噪声不同的点可以给权重 \(1/\sigma_i^2\)，就是 1.13 节的加权最小二乘。

When minimizing over \(\boldsymbol\theta\) the first term does not depend on \(\boldsymbol\theta\), and the second is exactly the \(\|A\mathbf x-\mathbf b\|^2\) of Lesson 1.13 (\(A=\Phi\), \(\mathbf b=\mathbf y\)). So \(\hat{\boldsymbol\theta}\) solves the normal equations \(\Phi^\top\Phi\,\hat{\boldsymbol\theta}=\Phi^\top\mathbf y\), independently of \(\sigma\) (the second task of FIG 1). Differentiating with respect to \(\sigma\) and setting to zero, \(-n/\sigma+\|\mathbf r\|^2/\sigma^3=0\), gives \(\hat\sigma^2=\tfrac1n\|\mathbf y-\Phi\hat{\boldsymbol\theta}\|^2\), the mean squared residual. Numerically, do not actually form \((\Phi^\top\Phi)^{-1}\): with polynomial features \(\Phi\) is badly conditioned and \(\Phi^\top\Phi\) squares that; FIG 2 uses a QR factorization (the pseudo-inverse of Lesson 1.13 also works). Points with different noise levels can be weighted by \(1/\sigma_i^2\), which is the weighted least squares of Lesson 1.13.

\(y_i\in\{0,1\}\)，\(p_i=\sigma(z_i)\)，\(z_i=\mathbf w^\top\mathbf x_i+b\)。伯努利分布 \(p(y\mid\mathbf x)=p^{y}(1-p)^{1-y}\) 的平均负对数似然就是二元交叉熵。用 \(\sigma'(z)=\sigma(z)(1-\sigma(z))\) 求导，几项会神奇地消掉：

\(y_i\in\{0,1\}\), \(p_i=\sigma(z_i)\), \(z_i=\mathbf w^\top\mathbf x_i+b\). The mean negative log-likelihood of the Bernoulli distribution \(p(y\mid\mathbf x)=p^{y}(1-p)^{1-y}\) is the binary cross-entropy. Differentiating with \(\sigma'(z)=\sigma(z)(1-\sigma(z))\), terms cancel almost magically:

三点值得记住。梯度的形状：“预测减目标”乘输入，和线性回归的 \(\nabla_{\boldsymbol\theta}=\frac1n\sum(\hat y_i-y_i)\boldsymbol\phi_i\)（差一个常数因子）一模一样；softmax 也一样，\(\partial L/\partial\mathbf z_i=\frac1n(\mathbf p_i-\mathbf e_{y_i})\)，\(\mathbf e_{y_i}\) 是独热向量。这不是巧合：三者都是“指数族分布 + 典范连接函数”的广义线性模型。到了 16.3 节，这个 \(\mathbf p-\mathbf e_y\) 就是反向传播从输出层送回去的第一个信号。凸性：Hessian 是一组半正定矩阵的加权和，所以 \(L\) 是凸的，没有坏的局部极小，梯度下降、牛顿法（逻辑回归里叫 IRLS，迭代重加权最小二乘）都收敛到全局最优（数据可分时除外：最优在无穷远）。边界的几何：

Three things are worth remembering. The shape of the gradient: "prediction minus target" times the input, the same as linear regression's \(\nabla_{\boldsymbol\theta}=\frac1n\sum(\hat y_i-y_i)\boldsymbol\phi_i\) (up to a constant factor); softmax is the same again, \(\partial L/\partial\mathbf z_i=\frac1n(\mathbf p_i-\mathbf e_{y_i})\) with \(\mathbf e_{y_i}\) the one-hot vector. This is no coincidence: all three are generalized linear models, "an exponential-family distribution + its canonical link". In Lesson 16.3 this \(\mathbf p-\mathbf e_y\) is the first signal that backpropagation sends back from the output layer. Convexity: the Hessian is a weighted sum of positive semidefinite matrices, so \(L\) is convex with no bad local minima; gradient descent and Newton's method (called IRLS, iteratively reweighted least squares, for logistic regression) both reach the global optimum (except on separable data, where the optimum is at infinity). The geometry of the boundary:

决策边界是一个超平面，\(\mathbf w\) 是它的法向量（1.7 节点积的几何）。\(p=0.1\) 和 \(p=0.9\) 两条等高线到边界的距离是 \(2.197/\|\mathbf w\|\)：\(\|\mathbf w\|\) 越大，过渡带越窄、模型越自信，FIG 3 的可分数据里你看到的就是它。上式里的 \(d_\perp(\mathbf x)\) 是点 \(\mathbf x\) 到边界的欧氏距离（在标准化后的坐标里）。阈值不一定取 0.5：把判成功的阈值改成 \(t\)，边界就平移到 \(z=\log\frac{t}{1-t}\)，第 6 节会利用这一点。

The decision boundary is a hyperplane and \(\mathbf w\) is its normal vector (the geometry of the dot product, Lesson 1.7). The \(p=0.1\) and \(p=0.9\) contours lie \(2.197/\|\mathbf w\|\) from the boundary: the larger \(\|\mathbf w\|\), the narrower the transition band and the more confident the model, which is what the separable data of FIG 3 shows. \(d_\perp(\mathbf x)\) above is the Euclidean distance from the point \(\mathbf x\) to the boundary (in standardized coordinates). The threshold need not be 0.5: deciding "success" at a threshold \(t\) shifts the boundary to \(z=\log\frac{t}{1-t}\), which Section 6 exploits.

\(\mathbf z=W\mathbf x+\mathbf b\)，\(W\in\R^{K\times d}\)，\(p_k=e^{z_k}/\sum_je^{z_j}\)。给所有 logit 加同一个常数，概率不变，所以 \(K\) 组参数里只有 \(K-1\) 组是可辨识的；\(K=2\) 时 \(p_1=\sigma(z_1-z_0)\)，正是逻辑回归，\(\mathbf w=\mathbf w_1-\mathbf w_0\)。数值上要先减去最大的 logit 再取指数（否则 \(e^{z}\) 会溢出），库里的交叉熵函数都直接吃 logit 而不是概率，就是为了把 log 和 softmax 合在一起稳定地算。

\(\mathbf z=W\mathbf x+\mathbf b\) with \(W\in\R^{K\times d}\), \(p_k=e^{z_k}/\sum_je^{z_j}\). Adding the same constant to every logit leaves the probabilities unchanged, so only \(K-1\) of the \(K\) parameter sets are identifiable; with \(K=2\), \(p_1=\sigma(z_1-z_0)\), which is logistic regression with \(\mathbf w=\mathbf w_1-\mathbf w_0\). Numerically, subtract the largest logit before exponentiating (or \(e^{z}\) overflows); library cross-entropy functions take logits rather than probabilities precisely so that the log and the softmax can be combined and computed stably.

\(\hat y(\mathbf x)=\arg\max_k\big|\{i\in\hl{nbr}{\mathcal N_k(\mathbf x)}:y_i=k\}\big|\)，\(\mathcal N_k(\mathbf x)\) 是离 \(\mathbf x\) 最近的 \(k\) 个训练样本（这里的距离是标准化后的欧氏距离）。平票时本页取票数并列的类里离得最近的那个样本的类别。Cover 与 Hart（1967）证明，样本数趋于无穷时，1-NN 的错误率不超过最优（贝叶斯）错误率的两倍。预测一次要 \(O(nd)\) 次运算；机器人上数据量大时用 k-d 树或近似最近邻搜索（与 12.1 节点云最近邻搜索用的 k-d 树相同）。

\(\hat y(\mathbf x)=\arg\max_k\big|\{i\in\hl{nbr}{\mathcal N_k(\mathbf x)}:y_i=k\}\big|\), where \(\mathcal N_k(\mathbf x)\) is the set of the \(k\) training samples closest to \(\mathbf x\) (here in standardized Euclidean distance). On a tie this page takes, among the tied classes, the class of the nearest sample. Cover and Hart (1967) proved that as the number of samples goes to infinity, the error rate of 1-NN is at most twice the optimal (Bayes) error rate. One prediction costs \(O(nd)\) operations; with large data on a robot, k-d trees or approximate nearest-neighbour search are used (the same k-d trees that Lesson 12.1 uses for nearest neighbours in point clouds).

把“判为成功”当作正类。在测试集上数四个数：TP（判成功，真成功）、FP（判成功，实际失败）、FN（判失败，实际成功）、TN（判失败，真失败）。于是

Take "predicted success" as the positive class. Count four numbers on the test set: TP (predicted success, was success), FP (predicted success, actually failed), FN (predicted failure, actually succeeded), TN (predicted failure, was failure). Then

这里 Acc 是准确率，Prec 是精确率（precision），Rec 是召回率（recall），FPR 是假阳性率。精确率问的是“我说会成功的里面，有多少真成功了”，召回率（recall，也叫真阳性率 TPR）问的是“真能成功的里面，我抓住了多少”。阈值 \(t\) 从 0 扫到 1，召回率单调下降，精确率大体上升，画出来就是 PR 曲线；以 FPR 为横轴、TPR 为纵轴画就是 ROC 曲线，它下方的面积 AUC 等于“随机取一个成功样本和一个失败样本，前者得分更高”的概率。ROC 不受类别比例影响，PR 曲线和精确率则直接受它影响：失败很少时，ROC 可能很好看，但精确率仍可能不够。

Here Acc is accuracy, Prec precision, Rec recall and FPR the false-positive rate. Precision asks "of the grasps I said would succeed, how many did", recall (also the true-positive rate, TPR) asks "of the grasps that could succeed, how many did I catch". Sweeping the threshold \(t\) from 0 to 1, recall falls monotonically and precision roughly rises; plotting one against the other gives the PR curve. Plotting TPR against FPR gives the ROC curve, and the area under it, AUC, equals the probability that a randomly chosen success scores higher than a randomly chosen failure. ROC does not depend on the class ratio; the PR curve and precision depend on it directly: with few failures, the ROC curve can look excellent while precision is still not good enough.

训练准确率说明不了什么（FIG 2 和 FIG 5 已经看到），所以要留出一份测试集：训练、调参时都不碰它，最后只用一次。可即便在测试集上，“准确率 90%”也可能毫无用处。下图是一个抓取分类器在 400 次测试抓取上给出的分数（预测的成功概率）。机器人只在分数超过阈值 \(t\) 时才执行抓取。执行了却失败（FP）的代价很高：零件掉落、产线停下；放弃了本能成功的抓取（FN）只是浪费一点时间，换一个候选抓取就行。真实的拣选场景里，候选抓取大多是能成功的，失败是少数：把“类别比例”切到“现实（90% 成功）”看看会发生什么。

Training accuracy says little (FIG 2 and FIG 5 showed it), so hold out a test set: never touch it while training or tuning, and use it once at the end. Yet even on a test set, "90% accuracy" can be useless. The figure below shows the scores (predicted success probabilities) that a grasp classifier gives on 400 test grasps. The robot executes a grasp only when its score exceeds a threshold \(t\). Executing a grasp that then fails (FP) is expensive: a dropped part, a stopped line; skipping a grasp that would have worked (FN) only wastes a little time, since another candidate can be tried. In real picking most candidate grasps succeed and failures are the minority: set the class ratio to "realistic (90% success)" and see what happens.

“均衡”数据下阈值 0.5 时准确率 92.5%、精确率 0.934、召回率 0.915。“现实”数据下，让准确率最高的阈值移到了 0.22 附近（95.75%），因为失败很少，多判成功几乎不吃亏；而满足 FP = 0 且召回率 ≥ 0.85 的阈值只有 0.58 到 0.61 这一小段。别把“0 个失败”看得太绝对：测试集里一共只有 40 次失败，0/40 只能说明“失败的候选被执行”的真实比例（FPR）大概率低于 7.5% 左右（统计里的“三分之法则”：0 次发生时，95% 置信上界约为 \(3/n\)）。要宣称这个比例低于 1%，需要几百次失败样本。

On "balanced" data a threshold of 0.5 gives 92.5% accuracy, precision 0.934 and recall 0.915. On "realistic" data the threshold with the best accuracy moves to about 0.22 (95.75%), because failures are rare and predicting success costs almost nothing; while the thresholds satisfying FP = 0 with recall ≥ 0.85 form only the narrow band from 0.58 to 0.61. Do not read "0 failures" too literally: the test set holds only 40 failures in total, and 0/40 only says that the true fraction of failing candidates that get executed (the FPR) is probably below about 7.5% (the "rule of three" in statistics: with zero events in \(n\) trials, the 95% upper confidence bound is about \(3/n\)). Claiming that this fraction is below 1% needs failure samples in the hundreds.