Gradient Descent Vectorization

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  • 我们假定 X X 为数据集,其中每一行 x ( i ) x^{(i)} 为一个样本,列数代表其特征数量;
  • Y Y 为其真实值,每行 y ( i ) y^{(i)} 与每个输入样本对应;
  • Θ \Theta 为每个特征的权重;
    X = [ x 0 ( 0 ) x 1 ( 0 ) x n ( 0 ) x 0 ( 1 ) x 1 ( 1 ) x n ( 1 ) x 0 ( m ) x 0 ( m ) x 0 ( m ) ] = [ x ( 0 ) x ( 1 ) x ( m ) ] X = \left[ \begin{matrix} x_0^{(0)} & x_1^{(0)} & \cdots & x_n^{(0)} \\ x_0^{(1)} & x_1^{(1)} & \cdots & x_n^{(1)} \\ \vdots & \vdots & \ddots & \vdots\\ x_0^{(m)} & x_0^{(m)} & \cdots & x_0^{(m)} \end{matrix} \right] = \left[ \begin{matrix} x^{(0)} \\ x^{(1)} \\ \vdots \\ x^{(m)} \end{matrix} \right]

Y = [ y ( 0 ) y ( 1 ) y ( m ) ] Y = \left[ \begin{matrix} y^{(0)} \\ y^{(1)} \\ \vdots\\ y^{(m)} \\ \end{matrix} \right]

Θ = [ θ 0 θ 1 θ n ] \Theta = \left[ \begin{matrix} \theta_0 \\ \theta_1 \\ \vdots \\ \theta_n \\ \end{matrix} \right]

神经网络结构
如图所示为一个简单的神经网络,仅包含一个输入层和一个输出层,没有任何的隐藏层;对于其中的某一个样本 x ( i ) x^{(i)} 来说,其输出的预测值 y ^ ( i ) \hat y^{(i)} 为:
(1) y ^ ( i ) = θ 0 x 0 ( i ) + θ 1 x 1 ( i ) + . . . + θ n x n ( i ) = x ( i ) θ \hat y^{(i)} = \theta_0x_0^{(i)} + \theta_1x_1^{(i)} + ... + \theta_nx_n^{(i)} = x^{(i)}\theta \tag{1}
则所有样本的预测值 Y ^ \hat Y 为:
Y ^ = [ y ^ ( 0 ) y ^ ( 1 ) y ^ ( m ) ] \hat Y = \left[ \begin{matrix} \hat y^{(0)} \\ \hat y^{(1)} \\ \vdots \\ \hat y^{(m)} \end{matrix} \right]
我们以预测值与真实值的平方误差为损失函数 L L
(2) L = 1 m i = 1 m 1 2 ( y ^ ( i ) y ( i ) ) 2 = 1 2 m i = 1 m ( x ( i ) θ y ( i ) ) 2 L = \frac{1}{m}\sum_{i=1}^m\frac{1}{2}(\hat y^{(i)} - y^{(i)})^2 \\ = \frac{1}{2m}\sum_{i=1}^m( x^{(i)}\theta - y^{(i)})^2 \tag{2}
假设我们现在要计算 θ j \theta_j 经梯度下降后的更新值,其中 α \alpha 为学习率:
(3) θ j = θ j α L θ j \theta_j = \theta_j - \alpha\frac{\partial L}{\partial \theta_j} \tag{3}
我们对损失函数,即公式(2)求 θ j \theta_j 的微分:
(4) L θ j = 1 m i = 1 m ( x ( i ) θ y ( i ) ) ( x ( i ) θ ) θ j = 1 m i = 1 m ( x ( i ) θ y ( i ) ) x j ( i ) \frac{\partial L}{\partial \theta_j} = \frac{1}{m}\sum_{i=1}^m( x^{(i)}\theta - y^{(i)})\frac{\partial( x^{(i)}\theta)}{\partial \theta_j} \\ =\frac{1}{m}\sum_{i=1}^m( x^{(i)}\theta - y^{(i)})x_j^{(i)} \tag{4}
我们记 e ( i ) = x ( i ) θ y ( i ) e^{(i)} = x^{(i)}\theta -y^{(i)} ,用 E E 表示所有的 e ( i ) e^{(i)} 有:
E = [ e ( 0 ) e ( 1 ) e ( m ) ] = [ x ( 0 ) θ y ( 0 ) x ( 1 ) θ y ( 1 ) x ( m ) θ y ( m ) ] = X Θ Y E=\left[ \begin{matrix} e^{(0)} \\ e^{(1)} \\ \vdots \\ e^{(m)} \end{matrix} \right]= \left[ \begin{matrix} x^{(0)}\theta -y^{(0)} \\ x^{(1)}\theta -y^{(1)} \\ \vdots \\ x^{(m)}\theta -y^{(m)} \end{matrix} \right] = X\Theta-Y

则公式(4)可表示为:
(5) L θ j = 1 m i = 1 m e ( i ) x j ( i ) = 1 m ( x j ( 0 ) , x j ( 1 ) , . . . , x j ( m ) ) E \frac{\partial L}{\partial \theta_j}=\frac{1}{m}\sum_{i=1}^me^{(i)}x_j^{(i)} \\ =\frac{1}{m}(x_j^{(0)},x_j^{(1)},...,x_j^{(m)})E \tag{5}
将公式(5)代入公式(3)中可以得到:
(6) θ j = θ j α 1 m ( x j ( 0 ) , x j ( 1 ) , . . . , x j ( m ) ) E \theta_j = \theta_j - \alpha\frac{1}{m}(x_j^{(0)},x_j^{(1)},...,x_j^{(m)})E \tag{6}
因此,我们可以得到所有权重的梯度更新为:
Θ = [ θ 0 θ 1 θ n ] = [ θ 0 θ 1 θ n ] α m [ x 0 ( 0 ) x 0 ( 1 ) x 0 ( m ) x 1 ( 0 ) x 1 ( 1 ) x 1 ( m ) x n ( 0 ) x n ( 1 ) x n ( m ) ] E = Θ α m X T E = Θ α m X T ( X Θ Y ) \Theta = \left[ \begin{matrix} \theta_0 \\ \theta_1 \\ \vdots \\ \theta_n \\ \end{matrix} \right] = \left[ \begin{matrix} \theta_0 \\ \theta_1 \\ \vdots \\ \theta_n \\ \end{matrix} \right] - \frac{\alpha}{m} \left[ \begin{matrix} x_0^{(0)} & x_0^{(1)} & \cdots & x_0^{(m)} \\ x_1^{(0)} & x_1^{(1)} & \cdots & x_1^{(m)} \\ \vdots & \vdots & \ddots & \vdots\\ x_n^{(0)} & x_n^{(1)} & \cdots & x_n^{(m)} \end{matrix} \right]E \\ = \Theta- \frac{\alpha}{m}X^TE=\Theta- \frac{\alpha}{m}X^T(X\Theta-Y)

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