两视角分析

考虑下面两个视角：
${\mathcal{R}}_1={\mathcal{H}}_1{\mathcal{W}}_1+{\mathcal{Z}}_1$
${\mathcal{R}}_2={\mathcal{H}}_2{\mathcal{W}}_2+{\mathcal{Z}}_2$

• ${\mathcal{W}}_1$ has no direct prior, but is conditionally dependent on ${\mathcal{W}}_2$ through their correlation.
• ${\mathcal{W}}_2\sim \mathcal{N}({\mu }_{w_2},{\sigma }_{w_2}^2)$ and the correlation with ${\mathcal{W}}_1$ is defined by:
  $\mathrm{Cov}({\mathcal{W}}_1,{\mathcal{W}}_2)=K{\sigma }_{w_1}{\sigma }_{w_2}$
• ${\mathcal{H}}_1,{\mathcal{H}}_2$ are independent Rayleigh-distributed:
  ${\mathcal{H}}_i\sim \text{Rayleigh}({\sigma }_{h_i})i=1,2$
• ${\mathcal{Z}}_1,{\mathcal{Z}}_2$ are independent Gaussian noises:
  ${\mathcal{Z}}_i\sim \mathcal{N}(0,{\sigma }_z^2),i=1,2$
  解码端接收到${\mathcal{R}}_1$与${\mathcal{R}}_2$对于${\mathcal{W}}_1$的恢复，${\mathcal{R}}_2$相当于是side information，由于${\mathcal{W}}_1$与${\mathcal{W}}_2$存在相关性，所以${\mathcal{R}}_2$对于${\mathcal{W}}_1$的恢复相当于一种有损的先验，为什么是有损的先验，因为${\mathcal{R}}_2$的传输过程有噪声。现在考虑对于${\mathcal{W}}_1$的恢复过程没有完美的CSI，只有估计的CSI，求${\mathcal{R}}_1$与${\mathcal{R}}_2$下的贝叶斯分布

Joint Likelihood Function
$p({\mathcal{R}}_1,{\mathcal{R}}_2\mid {\mathcal{W}}_1,{\mathcal{W}}_2)=\int p({\mathcal{R}}_1\mid {\mathcal{W}}_1,{\mathcal{H}}_1)p({\mathcal{R}}_2\mid {\mathcal{W}}_2,{\mathcal{H}}_2)p({\mathcal{H}}_1)p({\mathcal{H}}_2)d{\mathcal{H}}_{1}d{\mathcal{H}}_2$
Marginalizing Rayleigh Distributions
$p({\mathcal{R}}_i\mid {\mathcal{W}}_i)=\int \frac{h_i}{{\sigma }_{h_i}^2}\exp \left(-\frac{h_i^2}{2{\sigma }_{h_i}^2}\right)\frac{1}{\sqrt{2\pi {\sigma }_z^2}}\exp \left\{-\frac{({\mathcal{R}}_i-h_i{\mathcal{W}}_i{)}^2}{2{\sigma }_z^2}\right\}d{h}_i$
Since ${\mathcal{W}}_1$ has no direct prior, we model it conditionally based on ${\mathcal{W}}_2$ as 【PRML,p87】
${\mathcal{W}}_1\mid {\mathcal{W}}_2\sim \mathcal{N}\left(K\frac{{\sigma }_{w_1}}{{\sigma }_{w_2}}({\mathcal{W}}_2-{\mu }_{w_2}),(1-K^2){\sigma }_{w_1}^2\right)$

$\left[\begin{array}{c}{\mathcal{W}}_1\\ {\mathcal{W}}_2\end{array}\right]\sim \mathcal{N}\left(\left[\begin{array}{c}0\\ {\mu }_{w_2}\end{array}\right],\left[\begin{array}{cc}{\sigma }_{w_1}^2& K{\sigma }_{w_1}{\sigma }_{w_2}\\ K{\sigma }_{w_1}{\sigma }_{w_2}& {\sigma }_{w_2}^2\end{array}\right]\right)$
我们想要得到 ${\mathcal{W}}_1\mid {\mathcal{W}}_2$ 的条件分布。
根据多元高斯分布的性质，条件分布仍然是高斯分布，其均值和协方差如下：
$\mathbb{E}[{\mathcal{W}}_1\mid {\mathcal{W}}_2]={\mu }_1+{\Sigma }_{12}{\Sigma }_{22}^{-1}({\mathcal{W}}_2-{\mu }_2)$其中：${\mu }_1=0$, ${\mu }_2={\mu }_{w_2}$${\Sigma }_{12}=K{\sigma }_{w_1}{\sigma }_{w_2}$${\Sigma }_{22}={\sigma }_{w_2}^2$
条件协方差公式$\mathrm{Var}[{\mathcal{W}}_1\mid {\mathcal{W}}_2]={\Sigma }_{11}-{\Sigma }_{12}{\Sigma }_{22}^{-1}{\Sigma }_{21}$
${\Sigma }_{11}={\sigma }_{w_1}^2,{\Sigma }_{12}=K{\sigma }_{w_1}{\sigma }_{w_2},{\Sigma }_{22}={\sigma }_{w_2}^2$
$\mathrm{Var}[{\mathcal{W}}_1\mid {\mathcal{W}}_2]={\sigma }_{w_1}^2-\frac{(K{\sigma }_{w_1}{\sigma }_{w_2}{)}^2}{{\sigma }_{w_2}^2}={\sigma }_{w_1}^2(1-K^2)$

$\mathbb{E}[{\mathcal{W}}_1\mid {\mathcal{W}}_2]=K\frac{{\sigma }_{w_1}}{{\sigma }_{w_2}}({\mathcal{W}}_2-{\mu }_{w_2})$

Using Bayes’ theorem, we derive the posterior of ${\mathcal{W}}_1$:

• $p({\mathcal{W}}_1$