朴素贝叶斯法是基于贝叶斯定理与特征条件独立假设的分类方法

对于给定的训练数据集，首先基于特征条件独立假设学习输入/输出的联合概率分布 p ( x , y ) p(x,y) p(x,y)；然后基于此模型，对给定的输入 x x x,利用贝叶斯定理求出后验概率 p ( y ∣ x ) p(y|x) p(y∣x)最大的输出 y y y
利用训练数据学习 p ( x ∣ y ) p(x|y) p(x∣y)和 p ( y ) p(y) p(y)的估计，得到联合概率分布： p ( x , y ) = p ( y ) p ( x ∣ y ) p(x,y)=p(y)p(x|y) p(x,y)=p(y)p(x∣y)
概率估计可以使极大似然估计或贝叶斯估计

朴素贝叶斯法的基本假设是条件独立性，
P ( X = x ∣ Y = c k ) = P ( X ( 1 ) = x ( 1 ) , X ( 2 ) = x ( 2 ) , . . . , X ( n ) = x ( n ) ∣ Y = c k ) = ∏ j = 1 n P ( X ( j ) = x ( j ) ∣ Y = c k ) \begin{aligned} P(X=x|Y=c_{k})&=P(X^{(1)}=x^{(1)},X^{(2)}=x^{(2)},...,X^{(n)}=x^{(n)}|Y=c_{k})\\ &=\prod \limits_{j=1}^{n}P(X^{(j)}=x^{(j)}|Y=c_{k}) \end{aligned} P(X=x∣Y=ck​)​=P(X(1)=x(1),X(2)=x(2),...,X(n)=x(n)∣Y=ck​)=j=1∏n​P(X(j)=x(j)∣Y=ck​)​
这是一个较强的假设，由于这一假设，模型包含的条件概率的数量大为减少，朴素贝叶斯法的学习与预测大为简化，高效易于实现，然而分类的性能不一定很高

P ( Y ∣ X ) = P ( X , Y ) P ( X ) = P ( Y ) P ( X ∣ Y ) ∑ Y P ( Y ) P ( X ∣ Y ) P(Y|X)=\frac {P(X,Y)}{P(X)}=\frac {P(Y)P(X|Y)}{\sum \limits_{Y}P(Y)P(X|Y)} P(Y∣X)=P(X)P(X,Y)​=Y∑​P(Y)P(X∣Y)P(Y)P(X∣Y)​
将输入 x x x分到后验概率最大的类 y y y
y = a r g max ⁡ c k P ( Y = c k ) ∏ j = 1 n P ( X ( j ) = x ( j ) ∣ Y = c k ) y=arg\max \limits{_{c_{k}}P(Y=c_{k})}\prod \limits_{j=1}^{n}P(X^{(j)}=x^{(j)}|Y=c_{k}) y=argmaxck​​P(Y=ck​)j=1∏n​P(X(j)=x(j)∣Y=ck​)
后验概率最大等价于0-1损失函数时的期望风险最小化
朴素贝叶斯法实际上学习到的生成数据的机制，所以属于生成模型
条件独立假设等于说用于分类的特征在类确定的条件下都是独立的，这一假设使朴素贝叶斯法变得简单，但有时会牺牲一定的分类准确率。

先验概率 P ( Y = c k ) P(Y=c_{k}) P(Y=ck​)的极大似然估计 P ( Y = c k ) = ∑ i = 1 N I ( y i = c k ) N , k = 1 , 2 , . . . , K P(Y=c_{k})=\frac{\sum \limits_{i=1}^{N}I(y_{i}=c_{k})}{N},k=1,2,...,K P(Y=ck​)=Ni=1∑N​I(yi​=ck​)​,k=1,2,...,K
设第 j j j个特征 x ( j ) x^{(j)} x(j)可能取值的集合为 { a j 1 , a j 2 , . . . , a j S j } \{a_{j1},a_{j2},...,a_{jS_j}\} {aj1​,aj2​,...,ajSj​​}，
条件概率 P ( X ( j ) = a j l ∣ Y = c k ) P(X^{(j)}=a_{jl}|Y=c_{k}) P(X(j)=ajl​∣Y=ck​)的极大似然估计 P ( X ( j ) = a j l ∣ Y = c k ) = ∑ i = 1 N I ( x i ( j ) = a j l , y i = c k ) ∑ i = 1 N I ( y i = c k ) P(X^{(j)}=a_{jl}|Y=c_{k})=\frac {\sum \limits_{i=1}^{N}I(x_{i}^{(j)}=a_{jl},y_{i}=c_{k})}{\sum \limits_{i=1}^{N}I(y_{i}=c_{k})} P(X(j)=ajl​∣Y=ck​)=i=1∑N​I(yi​=ck​)i=1∑N​I(xi(j)​=ajl​,yi​=ck​)​ j = 1 , 2 , . . . , n ; l = 1 , 2 , . . . , S j ; k = 1 , 2 , . . . , K j=1,2,...,n;l=1,2,...,S_{j};k=1,2,...,K j=1,2,...,n;l=1,2,...,Sj​;k=1,2,...,K
x i ( j ) x_{i}^{(j)} xi(j)​是第 i i i个样本的第 j j j个特征； a j l a_{jl} ajl​是第 j j j个特征可能取的第 l l l个值; I I I为指示函数

用极大似然估计可能会出现所要估计的概率值为0的情况，采用贝叶斯估计来解决这一问题
条件概率的贝叶斯估计是 P λ ( X ( j ) = a j l ∣ Y = c k ) = ∑ i = 1 N I ( x i ( j ) = a j l , y i = c k ) + λ ∑ i = 1 N I ( y i = c k ) + S i λ P_{\lambda}(X^{(j)}=a_{jl}|Y=c_{k})=\frac {\sum \limits_{i=1}^{N}I(x_{i}^{(j)}=a_{jl},y_{i}=c_{k})+\lambda}{\sum \limits_{i=1}^{N}I(y_{i}=c_{k})+S_{i}\lambda} Pλ​(X(j)=ajl​∣Y=ck​)=i=1∑N​I(yi​=ck​)+Si​λi=1∑N​I(xi(j)​=ajl​,yi​=ck​)+λ​
式中 λ > 0 \lambda>0 λ>0,常取 λ = 1 \lambda=1 λ=1，这时称为拉普拉斯平滑,显然有
P λ ( X ( j ) = a j l ∣ Y = c k ) > 0 P_{\lambda}(X^{(j)}=a_{jl}|Y=c_{k})>0 Pλ​(X(j)=ajl​∣Y=ck​)>0
∑ l = 1 S j P λ ( X ( j ) = a j l ∣ Y = c k ) = 1 \sum \limits_{l=1}^{S_{j}}P_{\lambda}(X^{(j)}=a_{jl}|Y=c_{k})=1 l=1∑Sj​​Pλ​(X(j)=ajl​∣Y=ck​)=1
l = 1 , 2 , . . . , S j , k = 1 , 2 , . . . , K l=1,2,...,S_{j},k=1,2,...,K l=1,2,...,Sj​,k=1,2,...,K
表明贝叶斯估计是一种概率分布。同理，先验概率的贝叶斯估计是 P λ ( Y = c k ) = ∑ i = 1 N I ( y i = c k ) + λ N + K λ , k = 1 , 2 , . . . , K P_{\lambda}(Y=c_{k})=\frac{\sum \limits_{i=1}^{N}I(y_{i}=c_{k})+\lambda}{N+K\lambda},k=1,2,...,K Pλ​(Y=ck​)=N+Kλi=1∑N​I(yi​=ck​)+λ​,k=1,2,...,K

输入：训练数据 T = { ( x 1 , y 1 ) , ( x 2 , y 2 ) , . . . , ( x N , y N ) } T=\{(x_1,y_1),(x_2,y_2),...,(x_N,y_N)\} T={(x1​,y1​),(x2​,y2​),...,(xN​,yN​)},其中 x i = ( x i ( 1 ) , x i ( 2 ) , . . . , x i ( N ) ) x_{i}=(x_{i}^{(1)},x_{i}^{(2)},...,x_{i}^{(N)}) xi​=(xi(1)​,xi(2)​,...,xi(N)​), x i ( j ) x_{i}^{(j)} xi(j)​是第 i i i个样本的第 j j j个特征， x i ( j ) ∈ { a j 1 , a j 2 , . . . , a j S j } x_{i}^{(j)}\in \{a_{j1},a_{j2},...,a_{jS_{j}}\} xi(j)​∈{aj1​,aj2​,...,ajSj​​}, a j l a_{jl} ajl​是第 j j j个特征可能取的第 l l l个值， j = 1 , 2 , . . . , n , l = 1 , 2 , . . . , S j , y i ∈ { c 1 , c 2 , . . . , c K } j=1,2,...,n,l=1,2,...,S_{j},y_{i}\in\{c_{1},c_2,...,c_K\} j=1,2,...,n,l=1,2,...,Sj​,yi​∈{c1​,c2​,...,cK​};实例 x x x;

输出：实例 x x x的分类

（1）计算先验概率及条件概率
P ( Y = c k ) = ∑ i = 1 N I ( y i = c k ) N , k = 1 , 2 , . . . , K P(Y=c_{k})=\frac{\sum \limits_{i=1}^{N}I(y_{i}=c_{k})}{N},k=1,2,...,K P(Y=ck​)=Ni=1∑N​I(yi​=ck​)​,k=1,2,...,K
P ( X ( j ) = a j l ∣ Y = c k ) = ∑ i = 1 N I ( x i ( j ) = a j l , y i = c k ) ∑ i = 1 N I ( y i = c k ) P(X^{(j)}=a_{jl}|Y=c_{k})=\frac {\sum \limits_{i=1}^{N}I(x_{i}^{(j)}=a_{jl},y_{i}=c_{k})}{\sum \limits_{i=1}^{N}I(y_{i}=c_{k})} P(X(j)=ajl​∣Y=ck​)=i=1∑N​I(yi​=ck​)i=1∑N​I(xi(j)​=ajl​,yi​=ck​)​
j = 1 , 2 , . . . , n ; l = 1 , 2 , . . . , S j ; k = 1 , 2 , . . . , K j=1,2,...,n;l=1,2,...,S_{j};k=1,2,...,K j=1,2,...,n;l=1,2,...,Sj​;k=1,2,...,K
（2）对于给定的实例 x = ( x ( 1 ) , x ( 2 ) , . . . , x ( N ) ) x=(x^{(1)},x^{(2)},...,x^{(N)}) x=(x(1),x(2),...,x(N)),计算 P ( Y = c k ) ∏ j = 1 n P ( X ( j ) = x ( j ) ∣ Y = c k ) , k = 1 , 2 , . . . , K P(Y=c_{k})\prod \limits_{j=1}^{n}P(X^{(j)}=x^{(j)}|Y=c_{k}),k=1,2,...,K P(Y=ck​)j=1∏n​P(X(j)=x(j)∣Y=ck​),k=1,2,...,K
（3）确定实例 x x x的类 y = a r g max ⁡ c k P ( Y = c k ) ∏ j = 1 n P ( X ( j ) = x ( j ) ∣ Y = c k ) y=arg\max \limits{_{c_{k}}P(Y=c_{k})}\prod \limits_{j=1}^{n}P(X^{(j)}=x^{(j)}|Y=c_{k}) y=argmaxck​​P(Y=ck​)j=1∏n​P(X(j)=x(j)∣Y=ck​)

习题：用极大似然估计法推出朴素贝叶斯中的概率估计公式
P ( Y = c k ) = ∑ i = 1 N I ( y i = c k ) N , k = 1 , 2 , . . . , K P(Y=c_{k})=\frac{\sum \limits_{i=1}^{N}I(y_{i}=c_{k})}{N},k=1,2,...,K P(Y=ck​)=Ni=1∑N​I(yi​=ck​)​,k=1,2,...,K
P ( X ( j ) = a j l ∣ Y = c k ) = ∑ i = 1 N I ( x i ( j ) = a j l , y i = c k ) ∑ i = 1 N I ( y i = c k ) P(X^{(j)}=a_{jl}|Y=c_{k})=\frac {\sum \limits_{i=1}^{N}I(x_{i}^{(j)}=a_{jl},y_{i}=c_{k})}{\sum \limits_{i=1}^{N}I(y_{i}=c_{k})} P(X(j)=ajl​∣Y=ck​)=i=1∑N​I(yi​=ck​)i=1∑N​I(xi(j)​=ajl​,yi​=ck​)​

解答：把 P ( Y = c k ) ， P ( X ( j ) = a j l ∣ Y = c k ) P(Y=c_{k})，P(X^{(j)}=a_{jl}|Y=c_{k}) P(Y=ck​)，P(X(j)=ajl​∣Y=ck​)当做参数, ∑ k = 1 K P ( y = c k ) = 1 \sum \limits_{k=1}^{K}P(y=c_k)=1 k=1∑K​P(y=ck​)=1作为约束条件来求解参数值

由假设可知： P ( y ) = ∏ k = 1 K P ( y = c k ) I ( y = c k ) P(y)=\prod \limits_{k=1}^{K}P(y=c_{k})^{I(y=c_{k})} P(y)=k=1∏K​P(y=ck​)I(y=ck​) ，
P ( x ∣ y = c k ) = ∏ j = 1 n P ( x ( j ) ∣ y = c k ) = ∏ j = 1 n ∏ l = 1 S j P ( x ( j ) = a j l ∣ y = c k ) I ( x ( j ) = a j l , y = c k ) P(x|y=c_k)=\prod \limits_{j=1}^{n}P(x^{(j)}|y=c_{k})=\prod \limits_{j=1}^{n} \prod \limits_{l=1}^{S_j}P(x^{(j)}=a_{jl}|y=c_{k})^{I(x^{(j)}=a_{jl},y=c_k)} P(x∣y=ck​)=j=1∏n​P(x(j)∣y=ck​)=j=1∏n​l=1∏Sj​​P(x(j)=ajl​∣y=ck​)I(x(j)=ajl​,y=ck​)
令 φ = { P ( Y = c k ) ， P ( X ( j ) = a j l ∣ Y = c k ) } \varphi = \{P(Y=c_{k})，P(X^{(j)}=a_{jl}|Y=c_{k})\} φ={P(Y=ck​)，P(X(j)=ajl​∣Y=ck​)},对数似然函数为：
L ( φ ) = l o g ∏ i = 1 N P ( x i , y i ; φ ) = l o g ∏ i = 1 N P ( x i ∣ y i ; φ ) P ( y i ; φ ) = l o g ∏ i = 1 N ∏ j = 1 n P ( x i ( j ) ∣ y i ; φ ) P ( y i ; φ ) = ∑ i = 1 N ( P ( y i ; φ ) + ∑ j = 1 n P ( x i ( j ) ∣ y i ; φ ) ) = ∑ i = 1 N [ ∑ k = 1 K l o g P ( y = c k ) I ( y i = c k ) + ∑ j = 1 n ∑ l = 1 S j ∑ k = 1 K l o g P ( x i ( j ) = a j l ∣ y i = c k ) I ( x i ( j ) = a j l , y i = c k ) ] = ∑ i = 1 N [ ∑ k = 1 K I ( y i = c k ) l o g P ( y = c k ) + ∑ j = 1 n ∑ l = 1 S j ∑ k = 1 K I ( x i ( j ) = a j l , y i = c k ) l o g P ( x i ( j ) = a j l ∣ y i = c k ) ] \begin{aligned} L(\varphi)&=log\prod \limits_{i=1}^{N}P(x_i,y_i;\varphi)=log\prod \limits_{i=1}^{N}P(x_i|y_i;\varphi)P(y_{i};\varphi)\\ &=log\prod \limits_{i=1}^{N} \prod \limits_{j=1}^{n}P(x_i^{(j)}|y_i;\varphi)P(y_{i};\varphi)\\ &=\sum \limits_{i=1}^{N} (P(y_{i};\varphi) + \sum \limits_{j=1}^{n}P(x_i^{(j)}|y_i;\varphi))\\ &=\sum \limits_{i=1}^{N} [\sum \limits_{k=1}^{K}logP(y=c_k)^{I(y_i=c_k)} + \sum \limits_{j=1}^{n} \sum \limits_{l=1}^{S_j}\sum \limits_{k=1}^{K}log P(x_i^{(j)}=a_{jl}|y_i=c_k)^{I(x_i^{(j)}=a_{jl},y_i=c_k)}]\\ &=\sum \limits_{i=1}^{N} [\sum \limits_{k=1}^{K}{I(y_i=c_k)}logP(y=c_k) + \sum \limits_{j=1}^{n} \sum \limits_{l=1}^{S_j}\sum \limits_{k=1}^{K}{I(x_i^{(j)}=a_{jl},y_i=c_k)}logP(x_i^{(j)}=a_{jl}|y_i=c_k)] \end{aligned} L(φ)​=logi=1∏N​P(xi​,yi​;φ)=logi=1∏N​P(xi​∣yi​;φ)P(yi​;φ)=logi=1∏N​j=1∏n​P(xi(j)​∣yi​;φ)P(yi​;φ)=i=1∑N​(P(yi​;φ)+j=1∑n​P(xi(j)​∣yi​;φ))=i=1∑N​[k=1∑K​logP(y=ck​)I(yi​=ck​)+j=1∑n​l=1∑Sj​​k=1∑K​logP(xi(j)​=ajl​∣yi​=ck​)I(xi(j)​=ajl​,yi​=ck​)]=i=1∑N​[k=1∑K​I(yi​=ck​)logP(y=ck​)+j=1∑n​l=1∑Sj​​k=1∑K​I(xi(j)​=ajl​,yi​=ck​)logP(xi(j)​=ajl​∣yi​=ck​)]​
关于第一个参数 P ( Y = c k ) P(Y=c_{k}) P(Y=ck​)求导： ∂ L ( φ ) ∂ P ( y = c k ) = ∂ ∂ P ( y = c k ) ∑ i = 1 N ∑ k = 1 K I ( y i = c k ) l o g P ( y = c k ) \frac {\partial {L(\varphi)}}{\partial P(y=c_k)}=\frac {\partial}{\partial P(y=c_k)}\sum \limits_{i=1}^{N}\sum \limits_{k=1}^{K}{I(y_i=c_k)}logP(y=c_k) ∂P(y=ck​)∂L(φ)​=∂P(y=ck​)∂​i=1∑N​k=1∑K​I(yi​=ck​)logP(y=ck​)
由约束条件可知： P ( y = c K ) = 1 − ∑ k = 1 K − 1 P ( y = c k ) P(y=c_K)=1-\sum \limits_{k=1}^{K-1}P(y=c_k) P(y=cK​)=1−k=1∑K−1​P(y=ck​)
⇒ ∂ L ( φ ) ∂ P ( y = c k ) = ∂ ∂ P ( y = c k ) ∑ i = 1 N [ ∑ k = 1 K − 1 I ( y i = c k ) l o g P ( y = c k ) + I ( y i = c K ) l o g P ( y = c K ) ] = ∂ ∂ P ( y = c k ) ∑ i = 1 N [ ∑ k = 1 K − 1 I ( y i = c k ) l o g P ( y = c k ) + I ( y i = c K ) l o g ( 1 − ∑ k = 1 K − 1 P ( y = c k ) ) ] \Rightarrow\frac {\partial {L(\varphi)}}{\partial P(y=c_k)}=\frac {\partial}{\partial P(y=c_k)}\sum \limits_{i=1}^{N}[\sum \limits_{k=1}^{K-1}{I(y_i=c_k)}logP(y=c_k)+I(y_i=c_K)logP(y=c_K)]\\ =\frac {\partial}{\partial P(y=c_k)}\sum \limits_{i=1}^{N}[\sum \limits_{k=1}^{K-1}{I(y_i=c_k)}logP(y=c_k)+I(y_i=c_K)log(1-\sum \limits_{k=1}^{K-1}P(y=c_k))] ⇒∂P(y=ck​)∂L(φ)​=∂P(y=ck​)∂​i=1∑N​[k=1∑K−1​I(yi​=ck​)logP(y=ck​)+I(yi​=cK​)logP(y=cK​)]=∂P(y=ck​)∂​i=1∑N​[k=1∑K−1​I(yi​=ck​)logP(y=ck​)+I(yi​=cK​)log(1−k=1∑K−1​P(y=ck​))]
先来求 P ( y = c 1 ) P(y=c_1) P(y=c1​)的估计值：
0 = ∂ ∂ P ( y = c 1 ) ∑ i = 1 N [ ∑ k = 1 K − 1 I ( y i = c k ) l o g P ( y = c k ) + I ( y i = c K ) l o g ( 1 − ∑ k = 1 K − 1 P ( y = c k ) ) ] = ∑ i = 1 N [ I ( y i = c 1 ) P ( y = c 1 ) − I ( y i = c K ) 1 − ∑ a = 1 K − 1 P ( y = c a ) ] = ∑ i = 1 N [ I ( y i = c 1 ) P ( y = c 1 ) − I ( y i = c K ) P ( y = c K ) ] \begin{aligned} 0&=\frac {\partial}{\partial P(y=c_1)}\sum \limits_{i=1}^{N}[\sum \limits_{k=1}^{K-1}{I(y_i=c_k)}logP(y=c_k)+I(y_i=c_K)log(1-\sum \limits_{k=1}^{K-1}P(y=c_k))]\\ &=\sum \limits_{i=1}^{N}[\frac{I(y_i=c_1)}{P(y=c_1)}-\frac{I(y_i=c_K)}{1-\sum\limits_{a=1}^{K-1}P(y=c_a)}]\\ &=\sum \limits_{i=1}^{N}[\frac{I(y_i=c_1)}{P(y=c_1)}-\frac{I(y_i=c_K)}{P(y=c_K)}] \end{aligned} 0​=∂P(y=c1​)∂​i=1∑N​[k=1∑K−1​I(yi​=ck​)logP(y=ck​)+I(yi​=cK​)log(1−k=1∑K−1​P(y=ck​))]=i=1∑N​[P(y=c1​)I(yi​=c1​)​−1−a=1∑K−1​P(y=ca​)I(yi​=cK​)​]=i=1∑N​[P(y=c1​)I(yi​=c1​)​−P(y=cK​)I(yi​=cK​)​]​

P ( y = c K ) P(y=c_K) P(y=cK​)在此为由 P ( y = c 1 ) ， P ( y = c 2 ) , . . . , P ( y = c K − 1 ) P(y=c_1)，P(y=c_2),...,P(y=c_{K-1}) P(y=c1​)，P(y=c2​),...,P(y=cK−1​)决定的一个值
∑ i = 1 N [ I ( y i = c 1 ) P ( y = c 1 ) − I ( y i = c K ) P ( y = c K ) ] = 0 \begin{aligned} \sum \limits_{i=1}^{N}[\frac{I(y_i=c_1)}{P(y=c_1)}-\frac{I(y_i=c_K)}{P(y=c_K)}]=0 \\ \end{aligned} i=1∑N​[P(y=c1​)I(yi​=c1​)​−P(y=cK​)I(yi​=cK​)​]=0​ ⇒ P ( y = c K ) ∑ i = 1 N I ( y i = c 1 ) − P ( y = c 1 ) ∑ i = 1 N I ( y i = c K ) = 0 \begin{aligned} \Rightarrow P(y=c_K)\sum \limits_{i=1}^{N}I(y_i=c_1)-P(y=c_1)\sum \limits_{i=1}^{N}I(y_i=c_K)=0\\ \end{aligned} ⇒P(y=cK​)i=1∑N​I(yi​=c1​)−P(y=c1​)i=1∑N​I(yi​=cK​)=0​ P ( y = c 1 ) = ∑ i = 1 N I ( y i = c 1 ) ∑ i = 1 N I ( y i = c K ) P ( y = c K ) P ( y = c 2 ) = ∑ i = 1 N I ( y i = c 2 ) ∑ i = 1 N I ( y i = c K ) P ( y = c K ) . . . . . . P ( y = c K ) = ∑ i = 1 N I ( y i = c K ) ∑ i = 1 N I ( y i = c K ) P ( y = c K ) \begin{aligned} P(y=c_1) &= \frac {\sum \limits_{i=1}^{N}I(y_i=c_1)}{\sum \limits_{i=1}^{N}I(y_i=c_K)} P(y=c_K)\\ P(y=c_2) &= \frac {\sum \limits_{i=1}^{N}I(y_i=c_2)}{\sum \limits_{i=1}^{N}I(y_i=c_K)} P(y=c_K)\\ &...... \\ P(y=c_K) &= \frac {\sum \limits_{i=1}^{N}I(y_i=c_K)}{\sum \limits_{i=1}^{N}I(y_i=c_K)} P(y=c_K) \end{aligned} P(y=c1​)P(y=c2​)P(y=cK​)​=i=1∑N​I(yi​=cK​)i=1∑N​I(yi​=c1​)​P(y=cK​)=i=1∑N​I(yi​=cK​)i=1∑N​I(yi​=c2​)​P(y=cK​)......=i=1∑N​I(yi​=cK​)i=1∑N​I(yi​=cK​)​P(y=cK​)​