一、最小二乘法

对于给定的数据集\(D = {(x_1,y_1),(x_2,y_2), ...,(x_m,y_m)}\)，其中\(x_i=(x_{i1};x_{i2}; ...;x_{id})\)。
对上述数据进行拟合：

\[f(x_i)= \hat \omega^T \hat{x_i} \]

其中：\(\hat\omega = (\omega_1;\omega_2; ..., \omega_d;b)\) , \(\hat x_i = (x_{i1};x_{i2}; ...;x_{id};1)\)
最小二乘法是使用均方误差函数进行度量，可以通过求导，令导数等于零，直接求出解析解。均方误差函数为：

\[E(\hat \omega)=\frac{1}{m}\sum_{i=1}^m(f(x_i)-y_i)^2 \]

\(X=(x_1^T;x_2^T; ...;x_m^T), Y=(y_1;y_2;...;y_m)\),则：

\[E(\hat \omega)=(X*\hat\omega)^T(X*\hat\omega) \]

上式对\(\omega\)求导，得：

\[\frac{\partial E}{\partial\hat\omega}=\frac{2}{m}X^T(X\hat\omega-Y) \]

令上述导数等于0，得：

\[\hat\omega^*=(X^TX)^{-1}X^TY \]

这就是要求的最优解
使用上述方法，随机生成三维数据集，使用最小二乘法进行线性回归

clc;
M = 50;
dim = 2;

X = 10*randn(M,dim);
Y = 10*rand(M,1);
figure(1);
scatter3(X(:,1),X(:,2),Y,'filled');

X_2 = ones(M,1);
X = [X,X_2];

omega = (X'*X)\X'*Y;
[xx,yy] = meshgrid(-20:0.2:20,-20:0.2:20);

zz = omega(1,1)*xx+omega(2,1)*yy+omega(3,1);
hold on;
surf(xx,yy,zz);

效果

二、梯度下降法

相较于均方误差函数，对\(\omega_j , j = 1, ..., d\)求导得：

\[\frac{\partial f}{\partial \omega_j}=\frac{2}{m}\sum_{i=1}^m x_{ij}(x_{ij}\omega_j-y_i) \]

使用matlab生成三维随机数，检验程序有效性

clc;
close all;
M = 50; %%50个样本
dim = 2;
N = dim+1;

X = 10*randn(M,dim);
Y = 10*rand(M,1);
figure(1);
scatter3(X(:,1),X(:,2),Y,'filled');

X_2 = ones(M,1);
X = [X,X_2];
iterate = 300;  %%迭代300次
count = 0;
omega = zeros(dim+1,1);
err = 1000;
delta_t = 0.01;
loss_data = zeros(1,iterate);
while count <= iterate && err > 0.1

    count = count+1;
    delta_omega = zeros(N,1);
    for i = 1:N 
        temp_omega = 0;
        for j = 1:M
            temp_omega = temp_omega+X(j,i)*(X(j,i)*omega(i,1)-Y(j,1));    
        end

        delta_omega(i,1) = temp_omega/M;
    end
    omega = omega - delta_t*delta_omega;

    disp(omega);
    err = (Y-X*omega)'*(Y-X*omega);

    disp(err);
    loss_data(1,count) = err;
    
end
[xx,yy] = meshgrid(-20:0.2:20,-20:0.2:20);
% 
zz = omega(1,1)*xx+omega(2,1)*yy+omega(3,1);
hold on;
surf(xx,yy,zz);
figure(2);
 x_t = linspace(0,iterate,size(loss_data,2));
 plot(x_t,loss_data);

效果
样本和分类面

损失函数随迭代次数变化

conclusion
1、对于\(\Delta \omega\),注意是军方误差，如果不除以数据集元素个数，所得梯度向量的模可能过大，程序不能收敛