From f1bbcc259b2eb13574ddccbe7cd40d79aa7f4ae8 Mon Sep 17 00:00:00 2001 From: Shine wOng <1551885@tongji.edu.cn> Date: Thu, 9 Jan 2020 09:21:51 +0800 Subject: [PATCH] modify some latex formulas. --- ml/linear regression/linear regression.md | 70 +++++++++++------------ 1 file changed, 35 insertions(+), 35 deletions(-) diff --git a/ml/linear regression/linear regression.md b/ml/linear regression/linear regression.md index 433fbb1..2f9c162 100644 --- a/ml/linear regression/linear regression.md +++ b/ml/linear regression/linear regression.md @@ -79,7 +79,7 @@ $$ 因此可以写出似然函数$L(\theta)$ $$ -L(\theta) = \Pi_{i = 1}^m f(y^{(i)}|x^{(i)}) = (\frac{1}{\sqrt{2\pi}\sigma})^m\cdot e^{-\frac{1}{2\sigma^2}\Sigma_{i = 1}^m (y^{(i)} - \theta^Tx^{(i)})^2}\\\ +L(\theta) = \Pi_{i = 1}^m f(y^{(i)}|x^{(i)}) = (\frac{1}{\sqrt{2\pi}\sigma})^m\cdot e^{-\frac{1}{2\sigma^2}\Sigma_{i = 1}^m (y^{(i)} - \theta^Tx^{(i)})^2}\\\\ lnL(\theta) = -mln(\sqrt{2\pi}\sigma) - \frac{1}{2\sigma^2}\Sigma_{i = 1}^m(y^{(i)} - \theta^Tx^{(i)})^2 $$ @@ -135,7 +135,7 @@ $$ 学习率的选择在梯度下降法中至关重要。如果学习率太小，则需要更多次迭代才能找到最优解，需要较长的学习时间；而如果学习率太大，也有可能导致收敛速度慢，甚至可能会发散。这是因为学习率太大时，不再满足全微分方程中微小变化量的条件，因此每次迭代后被优化函数都未必可以取到比上一次迭代更小的值。学习率的选择对$J(\theta)$收敛速度的影响如下图所示： -![select_alpha](images/select_alpha.png) +[外链图片转存失败,源站可能有防盗链机制,建议将图片保存下来直接上传(img-RqKOi1Aq-1578532827322)(images/select_alpha.png)] 可以看到，当$\alpha$很小时（如图中红色曲线所示），损失函数$J(\theta)$收敛速度较慢。随着$\alpha$逐渐增大，曲线逐渐变得陡峭，收敛速度越来越快。可是当$\alpha$很大时（$\alpha = 1.3$，图中黑色曲线）收敛速度反而相对此前变慢了；$\alpha$继续增大将导致代价函数发散（图中洋红色曲线）。 @@ -145,7 +145,7 @@ $$ 从上面的讨论中可以看出，梯度下降法最终将收敛到某个局部最优点，而无法保证收敛到全局最优点。实际上，当选择不同的初始值时，使用梯度下降法往往会收敛到不同的极值点，如下图所示： -![gd_diff_path](images/gd_diff_path.jpg) +[外链图片转存失败,源站可能有防盗链机制,建议将图片保存下来直接上传(img-epUtOmFv-1578532827324)(images/gd_diff_path.jpg)] 因此，使用梯度下降法无法保证得到的结果一定是全局最优解。然而，需要指出，对于线性回归与平方损失函数，则不存在这个问题，因为平方损失函数是**凸函数**。证明如下： @@ -158,7 +158,7 @@ $$ 所以 $$ -\frac{\partial}{\partial \theta_j}J(\theta) = \frac{1}{m}\Sigma_{i = 1}^m(h_\theta(x^{(i)}) - y^{(i)})x_j^{(i)}\\ +\frac{\partial}{\partial \theta_j}J(\theta) = \frac{1}{m}\Sigma_{i = 1}^m(h_\theta(x^{(i)}) - y^{(i)})x_j^{(i)}\\\\ \frac{\partial^2}{\partial \theta_j\partial \theta_k}J(\theta) = \frac{1}{m}\Sigma_{i = 1}^mx_j^{(i)}x_k^{(i)} $$ @@ -167,10 +167,10 @@ $$ $$ H = \frac{1}{m}\left[ \begin{matrix} - \Sigma_{i = 1}^mx_0^{(i)^2} & \Sigma_{i = 1}^mx_0^{(i)}x_1^{(i)} & \cdots & \Sigma_{i = 1}^mx_0^{(i)}x_n^{(i)}\\ - \Sigma_{i = 1}^mx_1^{(i)}x_0^{(i)} & \Sigma_{i = 1}^mx_1^{(i)^2} & \cdots & \Sigma_{i = 1}^mx_1^{(i)}x_n^{(i)}\\ - \vdots & \vdots & & \vdots&\\ - \Sigma_{i = 1}^mx_n^{(i)}x_0^{(i)} & \Sigma_{i = 1}^mx_n^{(i)}x_1^{(i)} & \cdots & \Sigma_{i = 1}^mx_n^{(i)^2}\\ + \Sigma_{i = 1}^mx_0^{(i)^2} & \Sigma_{i = 1}^mx_0^{(i)}x_1^{(i)} & \cdots & \Sigma_{i = 1}^mx_0^{(i)}x_n^{(i)}\\\\ + \Sigma_{i = 1}^mx_1^{(i)}x_0^{(i)} & \Sigma_{i = 1}^mx_1^{(i)^2} & \cdots & \Sigma_{i = 1}^mx_1^{(i)}x_n^{(i)}\\\\ + \vdots & \vdots & & \vdots&\\\\ + \Sigma_{i = 1}^mx_n^{(i)}x_0^{(i)} & \Sigma_{i = 1}^mx_n^{(i)}x_1^{(i)} & \cdots & \Sigma_{i = 1}^mx_n^{(i)^2}\\\\ \end{matrix} \right] $$ @@ -229,25 +229,25 @@ $$ $$ X = \left[ \begin{matrix} - x^{(1)^T}\\ - x^{(2)^T}\\ - \vdots\\ + x^{(1)^T}\\\\ + x^{(2)^T}\\\\ + \vdots\\\\ x^{(m)^T} \end{matrix} \right],\ \ Y = \left[ \begin{matrix} - y^{(1)}\\ - y^{(2)}\\ - \vdots\\ + y^{(1)}\\\\ + y^{(2)}\\\\ + \vdots\\\\ y^{(m)} \end{matrix} \right],\ \ \theta = \left[ \begin{matrix} - \theta_0\\ - \theta_1\\ - \vdots\\ + \theta_0\\\\ + \theta_1\\\\ + \vdots\\\\ \theta_n \end{matrix} \right] @@ -290,7 +290,7 @@ $$ $$ \begin{aligned} -J(\theta) = &\frac{1}{2m}[(X\theta - Y)^T(X\theta - Y) + \lambda(I\theta)^TI\theta]\\ +J(\theta) = &\frac{1}{2m}[(X\theta - Y)^T(X\theta - Y) + \lambda(I\theta)^TI\theta]\\\\ = &\frac{1}{2m}(\theta^TX^TX\theta - 2Y^TX\theta + Y^TY + \lambda(I\theta)^TI\theta] \end{aligned} $$ @@ -299,9 +299,9 @@ $$ $$ I = \left[ \begin{matrix} -0 & 0 & \cdots & 0\\ -0 & 1 & \cdots & 0\\ -\vdots & \vdots & & 0\\ +0 & 0 & \cdots & 0\\\\ +0 & 1 & \cdots & 0\\\\ +\vdots & \vdots & & \vdots\\\\ 0 & 0 & \cdots & 1 \end{matrix} \right]. @@ -318,9 +318,9 @@ $$ $$ \theta = (X^TX + \lambda I)^{-1}X^TY = (X^TX + \lambda \left[ \begin{matrix} -0 & 0 & \cdots & 0\\ -0 & 1 & \cdots & 0\\ -\vdots & \vdots & & 0\\ +0 & 0 & \cdots & 0\\\\ +0 & 1 & \cdots & 0\\\\ +\vdots & \vdots & & \vdots\\\\ 0 & 0 & \cdots & 1 \end{matrix} \right])^{-1}X^TY @@ -343,17 +343,17 @@ $$ $$ X = \left[ \begin{matrix} - x^{(1)^T}\\ - x^{(2)^T}\\ - \vdots\\ + x^{(1)^T}\\\\ + x^{(2)^T}\\\\ + \vdots\\\\ x^{(m)^T} \end{matrix} \right] = \left[ \begin{matrix} -x_0^{(1)} & x_1^{(1)} & \cdots & x_n^{(1)}\\ -x_0^{(2)} & x_1^{(2)} & \cdots & x_n^{(2)}\\ -\vdots & \vdots & & \vdots\\ -x_0^{(m)} & x_1^{(m)} & \cdots & x_n^{(m)}\\ +x_0^{(1)} & x_1^{(1)} & \cdots & x_n^{(1)}\\\\ +x_0^{(2)} & x_1^{(2)} & \cdots & x_n^{(2)}\\\\ +\vdots & \vdots & & \vdots\\\\ +x_0^{(m)} & x_1^{(m)} & \cdots & x_n^{(m)}\\\\ \end{matrix} \right] $$ @@ -363,10 +363,10 @@ $$ $$ X^T = \left[ \begin{matrix} -x_0^{(1)} & x_0^{(2)} & \cdots & x_0^{(m)}\\ -x_1^{(1)} & x_1^{(2)} & \cdots & x_1^{(m)}\\ -\vdots & \vdots & & \vdots\\ -x_n^{(1)} & x_n^{(2)} & \cdots & x_n^{(m)}\\ +x_0^{(1)} & x_0^{(2)} & \cdots & x_0^{(m)}\\\\ +x_1^{(1)} & x_1^{(2)} & \cdots & x_1^{(m)}\\\\ +\vdots & \vdots & & \vdots\\\\ +x_n^{(1)} & x_n^{(2)} & \cdots & x_n^{(m)}\\\\ \end{matrix} \right] $$