机器学习笔记——百度云盘资料

最新推荐文章于 2021-04-22 20:01:51 发布

炸天豪

最新推荐文章于 2021-04-22 20:01:51 发布

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本文以银行贷款预测为例，深入浅出地介绍了线性回归的基本概念。通过将顾客的工资和年龄作为特征，贷款值作为标签，阐述了回归问题的本质及线性回归的目标：即寻找一条最佳拟合线来预测连续值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

$\begin{align} \log L\left ( \boldsymbol{\theta } \right )&=\sum_{i=1}^{n}\log \frac{1}{\sqrt{2\pi }\sigma }exp\left ( -\frac{\left ( y^{\left ( i \right )}-\boldsymbol{\theta }^{T} x^{\left ( i \right )}\right )^{2}}{2\sigma ^{2}} \right )\\&=m\log \frac{1}{\sqrt{2\pi }\sigma }-\frac{1}{\sigma ^{2}}\cdot \frac{1}{2}\sum_{i=1}^{n}\left ( y^{\left ( i \right )}-\boldsymbol{\theta ^{T}} x^{\left ( i \right )}\right )^{2} \end{align}$

$J\left ( \theta \right )=\frac{1}{2}\sum_{i=1}^{n}\left ( y^{\left ( i \right )}-\boldsymbol{\theta ^{T}}x^{i} \right )^{2}$

$J\left ( \theta \right )=\frac{1}{2}\sum_{i=1}^{n}\left ( h_{\theta } \left ( x^{\left ( i \right )} \right )-y^{\left ( i \right )}\right )^{2}=\frac{1}{2}\left ( \boldsymbol{X} \boldsymbol{\theta }-y\right )^{T}\left ( \boldsymbol{X} \boldsymbol{\theta }-y \right )$

$\nabla J\left ( \theta \right )=\nabla \left ( \frac{1}{2}\left ( \boldsymbol{X} \boldsymbol{\theta }-y \right )^{T} \left ( \boldsymbol{X} \boldsymbol{\theta }-y \right )\right )=\nabla \left ( \frac{1}{2} \left ( \boldsymbol{\theta }^{T} \boldsymbol{X}^{T}-y^{T}\right )\left ( \boldsymbol{X} \boldsymbol{\theta }-y \right )\right )\\=\nabla \left ( \frac{1}{2} \left ( \boldsymbol{\theta } ^{T}\boldsymbol{X}^{T}\boldsymbol{X}\boldsymbol{\theta }-\boldsymbol{\theta }^{T}\boldsymbol{X}^{T}y-y^{T}\boldsymbol{X}\boldsymbol{\theta }+y^{T}y\right )\right )\\=\frac{1}{2}\left ( 2\boldsymbol{X}^{T} \boldsymbol{X}\boldsymbol{\theta }-\boldsymbol{X}y-\left ( y^{T}\boldsymbol{X} \right )^{T}\right )=\boldsymbol{X}^{T}\boldsymbol{X}\boldsymbol{\theta }-\boldsymbol{X}^{T}y$

$\boldsymbol{\theta }=\left ( \boldsymbol{X}^{T} \boldsymbol{X}\right )^{-1}\boldsymbol{X}^{T}y$

$g\left ( z \right )\frac{1}{1+e^{-z}}$

$h_{\theta }\left ( x \right )=g\left ( \boldsymbol{\theta } ^{T}x\right )=\frac{1}{1+e^{-\boldsymbol{\theta } ^{T}x}}$

$\theta _{0}+\theta _{1}x_{1}+\cdots +\theta _{i}x_{i}=\sum_{i=1}^{m}\theta _{i}x_{i}=\boldsymbol{\theta }^{T}\boldsymbol{X}$

$P\left ( y=1|x;\theta \right )=h_{\theta }\left ( x \right )\\ P\left ( y=0 |x;\theta\right )=1-h_{\theta }\left ( x \right )\\ \Rightarrow P\left ( y|x;\theta\right )=\left( h_{\theta }\left ( x \right ) \right)^{y}\left ( 1-h_{\theta }\left ( x \right ) \right )^{1-y}$

$L\left ( \theta \right )=\prod_{i=1}^{m}P\left ( y_{i}|x;\theta \right )=\prod_{i=1}^{m}\left( h_{\theta }\left ( x_{i} \right ) \right)^{y_{i}}\left ( 1-h_{\theta }\left ( x_{i} \right ) \right )^{1-y_{i}}$

$l\left (\theta \right )=\log L\left ( \theta \right )=\prod_{i=1}^{m}\left ( y_{i}\log h_{\theta }\left ( x_{i} \right ) +\left ( 1-y_{i} \right )\log \left ( 1-h_{\theta }\left ( x_{i} \right ) \right )\right )$

$l\left ( \theta \right )=\log \left ( y_{i}\log h_{\theta }\left ( x_{i} \right )+\left ( 1-y_{i} \right )\log \left ( 1-h_{\theta } \left ( x_{i} \right )\right ) \right )$