最小均方误差推导（RPML ）

推导如下

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令 $A=t-\Phi w$
$dA=t-\Phi dw=-\Phi dw$
$$f=A^{T}RA$$
$$df=(dA{)}^{T}RA+A^{T}RdA$$$$df=tr((dA{)}^{T}RA+A^{T}RdA)$$$$df=tr(A^T{R}^{T}dA+A^{T}RdA)$$$$=tr(A^T(R^T+R)dA)$$$$=tr(2A^{T}RdA)$$$$df=-tr(2A^{T}R\Phi dw)$$
由于$df=（(\frac{\partial f}{\partial w}{)}^{T}dw）$
$$\frac{\partial f}{\partial w}={\Phi }^T{R}^{T}A$$$$={\Phi }^T{R}^T(t-\Phi w)$$$$={\Phi }^{T}R(t-\Phi w)$$
令$\frac{\partial f}{\partial w}==0$$${\Phi }^{T}Rt={\Phi }^{T}R\Phi w$$$$w^{\ast }=({\Phi }^{T}R\Phi {)}^{-1}{\Phi }^{T}Rt$$