Regularized Normal Equation for Linear Re-gression Given a data set
{ar(), y()}i=1,.-.,m with x()∈ R" and g(∈ R, the generalform of
regularized linear regression is as follows n (he(zr)- g)3+入>0号 (1) ”
2m i=1 j=1 Derive the normal equation.

设

$X=\left[\begin{array}{c}(x^{(1)}{)}^T\\ (x^{(2)}{)}^T\\ ...\\ (x^{(m)}{)}^T\end{array}\right]$ ， $Y=\left[\begin{array}{c}{y}^{(1)}\\ y^{(2)}\\ ...\\ y^{(m)}\end{array}\right]$，

因此，$X\theta -Y=\left[\begin{array}{c}(x^{(1)}{)}^T\theta \\ (x^{(2)}{)}^T\theta \\ ...\\ (x^{(m)}{)}^T\theta \end{array}\right]-\left[\begin{array}{c}{y}^{(1)}\\ y^{(2)}\\ ...\\ y^{(m)}\end{array}\right]=\left[\begin{array}{c}{h}_{\theta }(x^{(1)})-y^{(1)}\\ h_{\theta }(x^{(2)})-y^{(2)}\\ ...\\ h_{\theta }(x^{(m)})-y^{(m)}\end{array}\right]$，

损失函数可以表达为$J(\theta )=\frac{1}{2m}[(X\theta -Y{)}^T(X\theta -Y)+\lambda {\theta }^T\theta ]$，

${\nabla }_{\theta }J(\theta )={\nabla }_{\theta }\frac{1}{2m}[(X\theta -Y{)}^T(X\theta -Y)+\lambda {\theta }^T\theta ]$
$=\frac{1}{2m}[{\nabla }_{\theta }(X\theta -Y{)}^T(X\theta -Y)+{\nabla }_{\theta }\lambda {\theta }^T\theta ]$

${\nabla }_{\theta }\lambda {\theta }^T\theta =\lambda {\nabla }_{\theta }{\theta }^T\theta =\lambda {\nabla }_{\theta }tr(\theta {\theta }^T)=\lambda L\theta$

因此，${\nabla }_{\theta }J(\theta )=\frac{1}{2m}(X^{T}X\theta -X^{T}Y+\lambda L\theta )$

令${\nabla }_{\theta }J(\theta )=0$，当$X$矩阵各列向量线性独立时，$X^{T}X$矩阵可逆，存在唯一解$\theta =(X^{T}X+\lambda L{)}^{-1}{X}^{T}Y$.

aussian Discriminant Analysis Model Given m training data {x() ,
g)}i=1,… ,m，assume that y ~ Bernoulli(b)，ay =0～N(uo,2),x \ y = 1
～N(u1,>).Hence, we have p(y)= ”(1 一 )1一u .p(zl y =0)=(2z7)"/72 3(1/
exp(一士(a一 uo)T>-1(a 一o))op(al y= 1)=(2n)n/l2(1/a exp (一是(a
一u1)TE-1(a一ui))The log-likelihood function is m l(, /Lo,41,>)= log ][
[p(r(), g); o, uo,41,) i二1 m
=logp(x()| g() ; ，uo,41,2)p(g() ; ) i—1 Solve p，o，u1 and 2 by maximizing l(, Lo，u1，>). Hint: xtr(AX-1B)=一(X-1BAX-1)T，VA|A=|A|(A-1)T

这里 高斯判别分析（GDA）公式推导

3MLE for Naive Bayes Consider the following definition of MLE problem
for multinomials. Theinput to the problem is a finite set J,and a
weight cg > 0 for each gy ∈ y. The output from the problem is the
distribution p* that solves the followingmaximization problem. p*= arg
max > c y log py y∈ (i) Prove that,the vector p* has components p,-Cy
for Vy ∈ y,where N = >ucycy.(Hint: Use the theory of
Lagrangemultiplier) (1i) Using the above consequence，prove that,the
maximum-likelihood esti- mates for Naive Bayes model are as follows
p)=之1 1(y()=gy) m and Ps(a l y)=>E1 1(g=y Aa,=z) 〉岩11(g(阈)= g)

(i)设拉格朗日函数为$L(\Omega ,\alpha )={\sum }_{y\in Y}{c}_{y}log{p}_y-\alpha ({\sum }_{y\in Y}{p}_y-1)$，其中$\alpha$为拉格朗日乘子，

对$p_y$求偏导，令$\frac{\partial }{\partial p_y}L(\Omega ,\alpha )=0$，

求得$p_y^{\ast }=\frac{c_y}{\alpha }$，代入${\sum }_{y\in Y}{p}_y^{\ast }=1$得$\frac{{\sum }_{y\in Y}{c}_y}{\alpha }=1$，

而$N={\sum }_{y\in Y}{c}_y$，因此$\alpha =N$，进而$p_y^{\ast }=\frac{c_y}{N}$

(ii)贝叶斯的最大似然模型的目标函数为

$max{\sum }_{i=1}^{m}logp(y^{(i)})+{\sum }_{i=1}^m{\sum }_{j=1}^{n}log{p}_j(x_j^{(i)}|y^{(i)})$

设标签种类数为$k$，则$p(y)$满足约束${\sum }_{i=1}^{k}p(y)=1$，以及$p(x_j|y)$满足约束${\sum }_{j=1}^{n}p(x_j|y)=1$，且所有概率均是非负的。

注意到加号两边可以分开独立进行优化，对于加号左边考虑优化模型：

$max{\sum }_{i=1}^{m}logp(y^{(i)})$

$s.t.{\sum }_{i=1}^{k}p(y)=1$

将标签$y$在训练集中的出现次数$cnt(y)$视为权重$c_y$，其中$cnt(y)={\sum }_{i=1}^{m}1(y^{(i)}=y)$，因此

$max{\sum }_{i=1}^{m}logp(y^{(i)})=max{\sum }_{i=1}^{k}cnt(y)logp(y)$，根据第一问的结论有$p^{\ast }(y)=\frac{cnt(y)}{m}=\frac{{\sum }_{i=1}^{m}1(y^{(i)}=y)}{m}$.

同理，将特征$x_j$在训练集标签为$y$的样本中的出现次数$cnt(x_j|y)$视为权重$c_y$，其中$cnt(x_j|y)={\sum }_{i=1}^{m}1(y^{(i)}=y\wedge x_j^{(i)}=x)$，因此

$$\begin{gathered} max{\sum }_{i=1}^m{\sum }_{j=1}^{n}log{p}_j(x_j^{(i)}|y^{(i)}) \\ =max{\sum }_{j=1}^n{\sum }_{i=1}^{m}log{p}_j(x_j^{(i)}|y^{(i)}) \\ =max{\sum }_{j=1}^{n}cnt(x_j|y)log{p}_j(x_j|y) \end{gathered}$$

根据第一问的结论有$p_j^{\ast }(x_j|y)=\frac{cnt(x_j|y)}{cnt(y)}=\frac{{\sum }_{i=1}^{m}1(y^{(i)}=y\wedge x_j^{(i)}=x)}{{\sum }_{i=1}^{m}1(y^{(i)}=y)}$，证毕。