稳健秩一信号检测器

在假设检验 \(H_1\) 下，联合概率密度函数为：

$$\begin{array}{rl}f(x,X_L)& =\frac{\exp [-(x-s_t{)}^H{R}^{-1}(x-s_t)-tr(R^{-1}S)]}{{\pi }^{N(L+1)}{\left|\begin{array}{c}R\end{array}\right|}^{L+1}}\end{array}$$

\(R\) 的最大似然估计为：
$${\hat{R}}_1=\frac{1}{L+1}{T}_1$$
其中
$$T_1=S+(x-s_t)(x-s_t{)}^H$$
\(R\) 的最大似然估计推导过程：
$$f(x,X_L)=\frac{\exp [-tr\left\{R^{-1}[S+(x-s_t)(x-s_t{)}^H]\right\}]}{{\pi }^{N(L+1)}{\left|\begin{array}{c}R\end{array}\right|}^{L+1}}$$

$$\ln f(x,X_L)=-tr\left\{R^{-1}[S+(x-s_t)(x-s_t{)}^H]\right\}-N(L+1)\ln \pi -(L+1)\ln \left|R\right|$$

$$d\ln f(x,X_L)=tr\{R^{-1}(dR)R^{-1}[S+(x-s_t)(x-s_t{)}^H]\}-(L+1)tr\{R^{-1}(dR)\}$$

令上式等于零，可解得
$$\begin{array}{rl}{\hat{R}}_1& =\frac{S+(x-s_t)(x-s_t{)}^H}{L+1}\end{array}$$

再将 \(\hat R_1\) 代入联合概率密度函数得

$$\begin{array}{rl}f(x,X_L;{\hat{R}}_1)& =\frac{\exp [-tr\left\{\frac{L+1}{T_1}[S+(x-s_t)(x-s_t{)}^H]\right\}]}{{\pi }^{N(L+1)}{\left|\begin{array}{c}\frac{T_1}{L+1}\end{array}\right|}^{L+1}}\\ & =\frac{\exp [-tr\{(L+1)I\}]}{{\left(\frac{\pi }{L+1}\right)}^{N(L+1)}{\left|\begin{array}{c}{T}_1\end{array}\right|}^{L+1}}\\ & =c{\left|\begin{array}{c}{T}_1\end{array}\right|}^{-(L+1)}\end{array}$$

其中，\(c=\left [ \frac{(L+1)}{e\pi} \right ] ^ {N(L+1)}\)