PNN点击率预估模型公式推导

1.

$l_{p}^{n}=oldsymbol{W}_{p}^{n} odot oldsymbol{p}=sum_{i=1}^{N} sum_{j=1}^{N} heta_{i}^{n} heta_{j}^{n}leftlangleoldsymbol{f}_{i}, oldsymbol{f}_{j} ight angle=leftlanglesum_{i=1}^{N} oldsymbol{delta}_{i}^{n}, sum_{i=1}^{N} oldsymbol{delta}_{i}^{n} ight angle$

$delta_{i}^{n}= heta_{i}^{n}oldsymbol{f}_{i}$

所以

$l_{p}^{n}=||sum_{i=1}^{n}delta_{i}^{n}||=||sum_{i=1}^{n} heta_{i}^{n}oldsymbol{f}_{i}||$

有两个疑问：

1.WHY？$sum_{i=1}^{N} sum_{j=1}^{N} heta_{i}^{n} heta_{j}^{n}leftlangleoldsymbol{f}_{i}, oldsymbol{f}_{j} ight angle=leftlanglesum_{i=1}^{N} oldsymbol{delta}_{i}^{n}, sum_{i=1}^{N} oldsymbol{delta}_{i}^{n} ight angle$

2.文中说$leftlangle sumlimits_{i=1}^{N}{mathbf{delta }_{i}^{n}},sumlimits_{i=1}^{N}{mathbf{delta }_{i}^{n}} ight angle = left| sumlimits_{i=1}^{N}{mathbf{delta }_{i}^{n}} ight|$

但是计算出来$leftlangle sumlimits_{i=1}^{N}{mathbf{delta }_{i}^{n}},sumlimits_{i=1}^{N}{mathbf{delta }_{i}^{n}} ight angle e left| sumlimits_{i=1}^{N}{mathbf{delta }_{i}^{n}} ight|$