y[n]=hx[n]+w[n]
h=0.1,x是要估计的,已知x只可能发+10或-10及概率,\(w[n]\sim N(0,1)\),接收到的信号是y[n]=0.5
问题:如何估计x[n]?
方法:ML、MAP
1、先验概率不同,p(x=+10)=0.2,p(x=-10)=0.8
ML:
\[\text { 似然函数: } p(y(n) \mid x(n))=\left\{\begin{array}{l} \frac{1}{\sqrt{2 \pi}} e^{-\frac{(y(n)-1)^{2}}{2}}=0.35, x(n)=+10 \\ \frac{1}{\sqrt{2 \pi}} e^{-\frac{(y(n)+1)^{2}}{2}}=0.13, x(n)=-10 \end{array}\right.\]判x[n]为+10
MAP
\[\begin{aligned} &\text { 后验概率 } \\ &p(x(n) \mid y(n))=p(x(n)) f(y(n) \mid x(n))= \begin{cases}0.2 \times \frac{1}{\sqrt{2 \pi}} e^{-\frac{(y(n)-1)^{2}}{2}}=0.07, & x(n)=+10 \\ 0.8 \times \frac{1}{\sqrt{2 \pi}} e^{-\frac{(y(n)+1)^{2}}{2}}=0.10, & x(n)=-10\end{cases} \end{aligned}\]判x[n]为-10
2、先验概率相同,p(x=+10)=0.5,p(x=-10)=0.5
ML与MAP等价
判x[n]为+10