尧图建网站 尧图建网站 YAOTU WEB BUILD 免费咨询
ARTICLE DETAIL

资讯详情

深耕网站建设与建站编程的一线实战洞察。

贝叶斯准则——多元假设检验

贝叶斯准则——多元假设检验 多元假设检验当判决结果有KKK种可能时,称为KKK元假设检验问题。通信中经常需要检测KKK个信号中哪个出现,模式识别问题中也经常遇到区分KKK种模式的问题。对于多元假设检验问题,通常采用最小错误概率准则或者贝叶斯准则,尽管纽曼 - 皮尔逊准则同样可应用,但实际中很少采用。风险函数Lij=L(Ci,Dj)L_{ij}= L \left(C_i , D_j \right)Lij​=L(Ci​,Dj​)假定对KKK种可能假设{ C1,C2,⋯ ,CK}\{C_1,C_2,\cdots,C_{K}\}{C1​,C2​,⋯,CK​}进行判决,CjC_jCj​为真判CiC_iCi​成立的代价用LijL_{ij}Lij​表示,则平均代价为R=∑i=1K∑j=1KLijP(Ci,Dj)=∑i=1K∑j=1KLijP(Dj∣Ci)P(Ci)条件概率的乘积公式 \begin{aligned} R=\sum\limits_{i = 1}^{K}\sum\limits_{j = 1}^{K}L_{ij}P\left({C}_i,D_j\right) =\sum\limits_{i = 1}^{K}\sum\limits_{j = 1}^{K}L_{ij}P\left({D}_j\mid C_i\right)P\left(C_i\right) \text{条件概率的乘积公式}\\ \end{aligned}R=i=1∑K​j=1∑K​Lij​P(Ci​,Dj​)​=i=1∑K​j=1∑K​Lij​P(Dj​∣Ci​)P(Ci​)条件概率的乘积公式​由于P(Dj∣Ci)=P(X∈Dj∣Ci)=∫Xjp(x∣Ci) dx \begin{aligned} P\left({D}_j \mid {C}_i\right) =P\left( {\boldsymbol X}\in { {\mathcal D}_j} \mid C_i \right)= \int\nolimits_{ {\mathscr X} _j} p\left({\boldsymbol x}\mid {C}_i\right) \, {\rm d}{\boldsymbol x} \end{aligned}P(Dj​∣Ci​)=P(X∈Dj​∣Ci​)=∫Xj​​p(x∣Ci​)dx​代入贝叶斯风险RRRR=∑i=1K∑j=1KLijP(X∈Xj∣Ci)P(Ci)=∑i=1K∑j=1KLij∫Xjp(x∣Ci)dxP(Ci)=∑j=1K∫Xj∑i=1KLijp(x∣Ci)P(Ci)dx(交换i和j求和的顺序)=∑j=1K∫Xj[∑i=1KLijP(Ci∣x)]p(x)dx(p(x∣Ci)P(Ci)=P(Ci∣x)p(x)) \begin{aligned} R=\sum\limits_{i = 1}^{K}\sum\limits_{j = 1}^{K}L_{ij}P\left(\boldsymbol{X}\in{ {\mathscr X} _j} \mid C_i\right)P\left(C_i\right)\\ =\sum\limits_{i = 1}^{K}\sum\limits_{j = 1}^{K}L_{ij} \int\nolimits_{ {\mathscr X} _j}p\left(\boldsymbol{x}\mid C_i\right){\rm d}\boldsymbol{x}P\left(C_i\right)\\ =\sum\limits_{j = 1}^{K}\int\nolimits_{ {\mathscr X} _j}\sum\limits_{i = 1}^{K}L_{ij}p\left(\boldsymbol{x}\mid C_i\right)P\left(C_i\right){\rm d}\boldsymbol{x} {\text{(交换}i\text{和}j\text{求和的顺序)}}\\ =\sum\limits_{j = 1}^{K}\int\nolimits_{ {\mathscr X} _j} \left[ \sum\limits_{i = 1}^{K}L_{ij}P\left(C_i\mid \boldsymbol{x}\right) \right]p\left(\boldsymbol{x}\right){\rm d}\boldsymbol{x} (p\left(\boldsymbol{x}\mid C_i\right)P\left(C_i\right)=P\left(C_i\mid \boldsymbol{x}\right)p\left(\boldsymbol{x}\right)) \end{aligned}R​=i=1∑K​j=1∑K​Lij​P(X∈Xj​∣Ci​)P(Ci​)=i=1∑K​j=1∑K​Lij​∫Xj​​
返回列表