4.1 用極大似然估計法推導樸素貝葉斯法中的先驗概率估計公式(4.8)和條件概率估計公式(4.9)

首先是(4.8) $P({Y=c_k})=\frac {\sum_{i=1}^NI(y_i=c_k)} {N}$ ###################下面開始證明############################### $P(x^{(j)}=a_{jl}|Y=c_k)$ $=\frac{\sum _{1=1}^N(x_i^{\left(j\right)}=a_{jl},y_i=c_k)}{\sum _{i=1}^{N}I(}$

似然概率 $P(y_1,y_2,...,y_n)$ $=p^{ \sum_{i=1}^N·I(y_i=c_k) }· (1-p)^{\sum_{i=1}^NI(y_i≠c_k)}$

然後求解最大似然概率： $\frac {dP(y_1,y_2,...,y_n)} {dp}$ $= \sum_{i=1}^NI(y_i=c_k)p^{\sum_{i=1}^NI(y_i=c_k)-1}·(1-p)^{\sum_{i=1}^NI(y_i≠c_k)}$ $-\sum_{i=1}^NI(y_i≠c_k)(1-p)^{\sum_{i=1}^NI(y_i≠c_k)-1}·p^{\sum_{i=1}^NI(y_i=c_k) }$

$= p^{[\sum_{i=1}^NI(y_i=c_k)]-1}·(1-p)^{[\sum_{i=1}^NI(y_i≠c_k)]-1}$ $·[(1-p)\sum_{i=1}^NI(y_i=c_k)-p\sum_{i=1}^NI(y_i≠c_k)]=0$

$∴[(1-p)\sum_{i=1}^NI(y_i=c_k)-p\sum_{i=1}^NI(y_i≠c_k)]=0$ $又∵Σ_i^NI(y_i=c_k)=p(\sum_{i=1}^NI(y_i=c_k)+\sum_{i=1}^NI(y_i≠c_k))=pN$