考虑MIMO信道的一般表达式：

$$Y= HX +N \qquad (1)$$

其中

$$X-H_t \times 1 $$

$$Y-H_r \times 1  $$

$$ X,N- i.i.d Gaussian Distribution$$

同时定义

$$ R_{xx}= E[XX^H] $$

$$ R_{nn}= E[NN^H] = \sigma ^2 I_{Nr}$$

$$ R_{yy}= E[YY^H]= HR_{xx}H^H+ \sigma ^2 I_{Nr}$$ 

那么 $P=tr(R_{xx})$ 为发射信号总功率，$H(Y|X)= H(HX+N|X) = H(N|X)= H(N)$ 并且

$$C= \arg max \quad I(X,Y)= H(Y)-H(N)$$ 

当信道AWGN信道时，信源X服从高斯分布时$I(X,Y)$取得最大值，此时接收信号Y也服从高斯分布。

(1) 考虑一维连续高斯分布,此时需要计算连续信息熵 (微分熵), 假设 $X \sim \phi (0, \sigma) = (1/\sqrt {2 \pi \sigma ^2}) e^{-x^2/2 \sigma ^2}$

可计算得到微分熵 

$$ h(X) = - \int x \ln x dx $$

$$\qquad = - \int \phi [- \frac {-x^2}{2\sigma ^2}-ln \sqrt {2\pi \sigma ^2}]dx  $$

$$\qquad = \frac {E[X^2]}{2 \sigma ^2} + \frac {1}{2} \ln 2\pi \sigma ^2 $$

$$\qquad = \frac {1}{2} + \frac {1}{2}\ln 2 \pi \sigma ^2$$

$$\qquad = \frac {1}{2} \ln \text {e} + \frac {1}{2} 2 \pi \sigma ^2$$

$$\qquad = \frac {1}{2} \ln 2 \pi \text {e} \sigma ^2$$

(2) 对于N维高斯信道，其中概率密度函数表达式如下

$$ f_X(x_1,,,x_N) = \frac {1}{\sqrt {(2\pi)^N|\sum|}} \text {exp} [-\frac {1}{2}(X-\mu )^T \Sigma ^{-1}(X-\mu)] \qquad (2)$$

假设$x_1,x_2,,,x_N$ 为$i.i.d$分布，即 $\sum = \sigma_X ^2 I_{N}$ 并且 $|\sum| = \prod _{i=1}^{N} \sigma ^2 $. (2)式可进一步简化为

$$f_X(x_1,,,x_N) = \prod _{i=1}^{N} \frac {1}{\sqrt {2\pi \sigma ^2}}\text {exp} ^{-\frac {(x_i-\mu _i)^2}{2\sigma ^2}} = \prod _{i=1}^{N} f(x_i)$$

重新计算微分熵可得

$$ h(X)= - \int f_X(x_1,,,x_N) \ln ^{f_X(x_1,,,x_N)} dx_1...dx_N$$

$$ \qquad = - \int f_X(x_1,,,x_N) [-\ln ^{\sqrt {(2\pi)^N|\Sigma|}}- \frac {1}{2}(X-\mu )^T \Sigma ^{-1}(X-\mu)]dx_1...dx_N$$

$$ \qquad = \frac {1}{2}\ln ^{(2\pi)^N|\Sigma|} + \frac {1}{2} \int _{x_1}f(x_1)[\int _{x_2}f(x_2)[.....[\int _{x_N}f(x_N) \sum _{i=1}^{N} \frac {(x_i-\mu _{i})^2}{\sigma ^2}dx_N]dx_{N-1}]....]dx_1 $$

$$\qquad = \frac {1}{2}\ln ^{(2\pi)^N|\Sigma|}+ \frac {N}{2} $$

$$\qquad = \frac {1}{2} \ln ^{(2\pi)^N|\Sigma|\text{e}^N} $$

$$\qquad = \frac {1}{2} \ln ^{|\Sigma|(2\pi \text{e})^N}$$

将上述结果代入 $C= \arg max \quad I(X,Y)= H(Y)-H(N)$ 中可得

$$ C = 1/2 \log _{2} {(2\pi \text{e})^{N}|\Sigma _Y|} -1/2 \log _{2} {(2\pi \text{e})^{N}|\Sigma _N|} $$

$$ \qquad =1/2 \log _2 {\frac {|HR_{xx}H^T+\sigma ^{2}I_{N_r}|}{|\sigma ^{2}I_{N_r}|}}$$

$$\qquad = 1/2 \log _2 {| I_{N_r} + \frac {HR_{xx}H^T}{\sigma ^2}|}$$

证明完毕