\documentclass[a4paper]{scrartcl}
\usepackage{enumerate}
\usepackage{amsmath,amssymb}

\renewcommand\paragraph[1]{\medskip\par\noindent\textit{#1}}

\begin{document}
\title{Processing of Stochastic Signals}
\subtitle{(Execerises 8)}
\date{December 22, 2011}
\maketitle

\textbf{8.2.a:} $\underline{z} = \underline{x} + j \underline{y}$, $\underline{\underline{C}} \underline{z},\underline{x},\underline{y}$ real-valued \\

\begin{center}
according to (3.177) \\

$\underline{\underline{C}}_{\underline{z}} = \underline{\underline{C}}_{\underline{x}} + \underline{\underline{C}}_{\underline{y}} + \left( \underline{\underline{C}}_{\underline{y,x}} - \underline{\underline{C}}^{T}_{\underline{y,x}} \right)$ \\

in case of circular symmetry about the mean we have according to (3.185) \\ 

$\underline{\underline{C}}_{\underline{x}}=\underline{\underline{C}}_{\underline{y}},\underline{\underline{C}}_{\underline{y,x}}=-\underline{\underline{C}}_{\underline{x,y}}$ \\

and therefore \\ 

$\underline{\underline{C}}_{\underline{z}} = 2 \underline{\underline{C}}_{\underline{x}} + 2j\underline{\underline{C}}_{\underline{y,x}}$
\end{center}

\textbf{8.2.b:}\\
\begin{enumerate}
\item{
diagonal elements of $\underline{\underline{C}}_{\underline{x,y}}$ zero\\
\begin{tabular}{ccc}
(3.185) and (3.186) & $\underline{\underline{C}}_{\underline{y,x}} = \underline{\underline{C}}^{T}_{\underline{x,y}}$ , $\underline{\underline{C}}_{\underline{y,x}} = -\underline{\underline{C}}_{\underline{x,y}}$ \\
\end{tabular} \\
$\Rightarrow$ Diagonal Elements: $\left( \underline{\underline{C}}_{\underline{x,y}} \right)_{i,i} = -\left( \underline{\underline{C}}_{\underline{x,y}} \right)_{i,i}$ \\
$\Rightarrow$ and this is only possible if $\left( \underline{\underline{C}}_{\underline{x,y}} \right)_{i,i} = 0$
}
\item{
$\underline{x},\underline{y}$ uncorrelated iff $\underline{\underline{C}}_{\underline{z}} \in R^{NxN}$ $\Rightarrow$ imaginary part has to become zero:\\
$\underline{\underline{C}}_{\underline{z}}=2\underline{\underline{C}}_{\underline{x}} + 2j\underbrace{\underline{\underline{C}}_{\underline{y,x}}}_{=0}$ \\
$\underline{\underline{C}}_{\underline{y,x}}=0$ gives that $\underline{x},\underline{y}$ are uncorrelated.
}
\item{
$\underline{\underline{C}}_{\underline{z}}\underline{\underline{C}}^{H}_{\underline{z}}$ real-valued iff $\underline{\underline{C}}_{\underline{x}}\underline{\underline{C}}^{T}_{\underline{x,y}}=\underline{\underline{C}}_{\underline{x,y}}\underline{\underline{C}}_{\underline{x}}$ \\
$
\begin{pmatrix}
\underline{\underline{C}}_{\underline{x}}\underline{\underline{C}}^{T}_{\underline{x,y}}=\underline{\underline{C}}_{\underline{x,y}}\underline{\underline{C}}^{T}_{\underline{x}}=\underline{\underline{C}}_{\underline{x,y}}\underline{\underline{C}}_{\underline{x}} \Rightarrow \underline{\underline{C}}_{\underline{x}}=\underline{\underline{C}}^{T}_{\underline{x}} \\
\underline{\underline{C}}^{T}_{\underline{x,y}}=\underline{\underline{C}}_{\underline{y,x}}=-\underline{\underline{C}}_{\underline{x,y}}
\end{pmatrix}
$ \\
\begin{tabular}{rl}
$\underline{\underline{C}}_{\underline{z}}\underline{\underline{C}}^{H}_{\underline{z}}$ & $ = \left( 2\underline{\underline{C}}_{\underline{x}} + 2j\underline{\underline{C}}^{T}_{\underline{x,y}}\right)\left( 2\underline{\underline{C}}_{\underline{x}} + 2j\underline{\underline{C}}^{T}_{\underline{x,y}}\right)^{H}$ \\
 & $ = \left( 2\underline{\underline{C}}_{\underline{x}} + 2j\underline{\underline{C}}^{T}_{\underline{x,y}}\right)\left( 2\underline{\underline{C}}_{\underline{x}} - 2j\underline{\underline{C}}^{T}_{\underline{x,y}}\right)$ \\
 & $ = 4\underline{\underline{C}}_{\underline{x}} + 4j\underline{\underline{C}}^{T}_{\underline{x,y}}\underline{\underline{C}}_{\underline{x}}-4j\underline{\underline{C}}_{\underline{x}}\underline{\underline{C}}_{\underline{x,y}}+4\underline{\underline{C}}^{T}_{\underline{x,y}}\underline{\underline{C}}_{\underline{x,y}}$ \\
 & $ = 4\left(\underline{\underline{C}}_{\underline{x}}\underline{\underline{C}}_{\underline{x}} + \underline{\underline{C}}^{T}_{\underline{x,y}}\underline{\underline{C}}_{\underline{x,y}} \right) +4j\left( \underline{\underline{C}}^{T}_{\underline{x,y}}\underline{\underline{C}}_{\underline{x}}+\underline{\underline{C}}_{\underline{x}}\underline{\underline{C}}^{T}_{\underline{x,y}}\right)$ \\
\end{tabular}\\
 For this to be real valued the following must hold \\
\begin{tabular}{rl} \\
$\underline{\underline{C}}_{\underline{x}} \underline{\underline{C}}^{T}_{\underline{x,y}} - \underline{\underline{C}}_{\underline{x,y}} \underline{\underline{C}}_{\underline{x}}$ & $= 0$ \\
$\underline{\underline{C}}_{\underline{x}} \underline{\underline{C}}^{T}_{\underline{x,y}}$ & $= \underline{\underline{C}}_{\underline{x,y}} \underline{\underline{C}}_{\underline{x}}$ \\
\end{tabular}\\
therefore this holds under our assumption.
}
\end{enumerate}\vspace{1cm}

\textbf{8.3.a:}\\
$\underline{s}=
\begin{pmatrix}
s_1 \\ s_2
\end{pmatrix}, \mu_s=
\begin{pmatrix}
1/2 \\ 1/2
\end{pmatrix}, \sigma^2_{s_1}=\sigma^2_{s_2}=7/4, R_{s_1,s_2}=1/2$ \\
$\underline{n}=
\begin{pmatrix}
n_1 \\ n_2
\end{pmatrix},\underline{\mu_n}=\underline{0}, \sigma^2_{n_1}=\sigma^2_{n_2}=1, R_{n_1,n_2}=-1/4$ \\
$\underline{y}=\underline{n}+\underline{s}$ \\
according to (3.272) \\
$A_{LMMSE} = \underline{\underline{R}}_s \left( \underline{\underline{R}}_s + \underline{\underline{R}}_n \right) ^{-1} = \underline{\underline{R}}_s\underline{\underline{R}}^{-1}_y$ \\
$\underline{\underline{R}}_{\underline{s}}=
\begin{pmatrix}
\sigma^{2}_{s_1}+\mu^{2}_{s_1} & 1/2 \\
1/2 & \sigma^{2}_{s_2}+\mu^{2}_{s_2} \\
\end{pmatrix}
=
\begin{pmatrix}
2 & 1/2 \\
1/2 & 2 \\
\end{pmatrix}
$ \\
\begin{tabular}{cc}
$\underline{\underline{R}}_{\underline{n}}=
\begin{pmatrix}
1 & -1/4 \\
-1/4 & 1 \\
\end{pmatrix}$
&
$\underline{\underline{R}}_{\underline{y}}=
\begin{pmatrix}
3 & 1/4 \\
1/4 & 3 \\
\end{pmatrix}$ \\
\end{tabular} \\

\begin{tabular}{lccc}
$\underline{\underline{R}}^{-1}_{\underline{y}}:$
&
(
	\begin{tabular}{cc|cc} \\
		3 & 1/4 & 1 & 0 \\
		1/4 & 3 & 0 & 1 \\
	\end{tabular}
)
&
$=^{{Z_{1}-\frac{Z_{2}}{12}}}$
&
(
	\begin{tabular}{cc|cc} \\
		$\frac{143}{48}$ & 0 & 1 & $-\frac{1}{12}$ \\
		1/4 & 3 & 0 & 1 \\
	\end{tabular}
)
\\
$=^{\frac{143}{12}Z_2-Z_1}$
&
(
	\begin{tabular}{cc|cc} \\
		143/48 & 0 & 1 & -1/12 \\
		0 & 143/4 & -1 & 12 \\
	\end{tabular}
)
&
$=^{12Z_1}$
&
(
	\begin{tabular}{cc|cc} \\
		143/4 & 0 & 12 & -1 \\
		0 & 143/4 & -1 & 12 \\
	\end{tabular}
) \\
\end{tabular} \\
$\underline{\underline{R}}^{-1}_{\underline{y}}=\frac{4}{143}
\begin{pmatrix}
12 & -1 \\
-1 & 12 \\
\end{pmatrix}
=
\begin{pmatrix}
48/143 & -4/143 \\
-4/143 & 48/143 \\
\end{pmatrix} $\\
$A_{LMMSE}=\frac{4}{143}
\begin{pmatrix}
47/2 & 4 \\
4 & 47/2 \\
\end{pmatrix}=
\begin{pmatrix}
94/143 & 16/143 \\
16/143 & 94/143 \\
\end{pmatrix} $\\
$\hat{s}_{LMMSE}=
\begin{pmatrix}
94/143 & 16/143 \\
16/143 & 94/143 \\
\end{pmatrix}
\begin{pmatrix}
y_1 \\
y_2 \\
\end{pmatrix}
$

\textbf{8.3.b:}\\
(3.265)\\
\begin{tabular}{rl}
$\epsilon_{LMMSE}$ & $= tr\left\{ \underline{\underline{R}}_{\underline{s}} - \underline{\underline{R}}^{T}_{\underline{y,s}} \underline{\underline{R}}^{-1}_{\underline{y}}\underline{\underline{R}}_{\underline{y,s}} \right\}$ \\
$\underline{\underline{R}}_{\underline{y,s}}$ & $=E\left\{(\underline{n}+\underline{s}\underline{s}^T\right\}=E\left\{\underline{n}\underline{s}^T+\underline{s}\underline{s}^T\right\}=$ \\
 & $= E\left\{\underline{n}\underline{s}^{T}\right\}+E\left\{\underline{s}\underline{s}^{T}\right\}=\underline{\underline{R}}_{\underline{s}}$ \\
$\epsilon_{LMMSE}$ & $= tr\left\{ \begin{pmatrix} 2 & 1/2 \\ 1/2 & 2 \\ \end{pmatrix} - \begin{pmatrix} 2 & 1/2 \\ 1/2 & 2 \\ \end{pmatrix}^{T}\begin{pmatrix} 48/143 & -4/143 \\ -4/143 & 48/143 \\ \end{pmatrix} \begin{pmatrix} 2 & 1/2 \\ 1/2 & 2 \\ \end{pmatrix} \right\}$ \\
 & $= tr\left\{ \begin{pmatrix} 2 & 1/2 \\ 1/2 & 2 \\ \end{pmatrix} - \begin{pmatrix} 94/143 & 16/143 \\ 16/143 & 94/143 \\ \end{pmatrix}\begin{pmatrix} 2 & 1/2 \\ 1/2 & 2 \\ \end{pmatrix}\right\}$ \\
 & $= tr\left\{\begin{pmatrix} 2 & 1/2 \\ 1/2 & 2 \\ \end{pmatrix} - \begin{pmatrix} 196/143 & 79/143 \\ 79/143 & 196/143 \\ \end{pmatrix}\right\}$ \\
 & $= tr\left\{\begin{pmatrix} 90/143 & -15/286 \\ -15/286 & 90/143 \\ \end{pmatrix}\right\}$ \\
 & $= \frac{180}{143} = \epsilon_{LMMSE}$ \\
\end{tabular} 

\textbf{8.3.c:}\\
Because the signal has non-zero mean it would be better to use the inhomogeneous LMMSE estimator instead of the homogeneous LMMSE estimator.

\textbf{8.4.a:}\\
from $\underline{\underline{W}} \in R^MxN$ and definition of $\underline{y}$\\
$\Rightarrow \underline{x} \in R^N, \underline{y} \in R^M, \underline{v} \in R^M$ \\
\textbf{8.4.b:}\\
\begin{tabular}{rll}\\
(page 101 ff.) & $\underline{\underline{\hat{x}}}_{LMMSE}$ & $= \underline{\underline{A}}_{LMMSE} \underline{y}=\underline{\underline{R}}^{T}_{\underline{x,y}}$ $\underline{\underline{R}}^{-1}_{\underline{y}}$ $\underline{y}$ \\
(3.270) & & $\left( \underline{\underline{W}} \underline{\underline{R}}_{\underline{x}} \right)^{T} \left(\underline{\underline{W}}\underline{\underline{R}}_{\underline{x}}\underline{\underline{W}}^{T} + \underline{\underline{R}}_{\underline{y}}\right)^{-1}$ $\underline{y}$ \\
(3.262) & & $ \underline{\underline{R}}_{\underline{x}} \underline{\underline{W}}^{T} \left(\underline{\underline{W}}\underline{\underline{R}}_{\underline{x}}\underline{\underline{W}}^{T} + \underline{\underline{R}}_{\underline{y}}\right)^{-1}$ $\underline{y}$ \\  
\end{tabular}\\

\textbf{8.4.c:}\\
$\epsilon_{LMMSE}= tr\left\{ \underline{\underline{R}}_{x} - \underline{\underline{R}}^{T}_{\underline{y,x}}\underline{\underline{R}}^{-1}_{\underline{y}}\underline{\underline{R}}_{\underline{y,x}} \right\}$ \\
$\underline{\underline{R}}_{\underline{y,x}} = \underline{\underline{W}}\underline{\underline{R}}_{x}$ \\
$\underline{\underline{R}}^{-1}_{\underline{y}}=\left(\underline{\underline{W}}\underline{\underline{R}}_{\underline{x}}\underline{\underline{W}}^{T} + \underline{\underline{R}}_{\underline{y}}\right)^{-1}$ \\

\textbf{8.4.d:}\\
straight forward, just apply the hint given on the sheet

\end{document}
