https://youtu.be/zV5D6EtkLzA

Simulation Solutions

• $H H^{\dagger} = K = E \Lambda E^{\dagger}$
  • $\Rightarrow I = H^{-1} E \Lambda E^{\dagger} H^{-1 \dagger}$
  • $U^{\dagger} \triangleq H^{-1}E \Lambda^{1 \over 2}$는 unitary
  • $\Rightarrow H = E \Lambda^{1 \over 2} U$
• Simulation Solutions
  • $z_{0}(u) = E \Lambda^{1 \over 2} U w(u)$
• orthogonal matrix $U$ 는 항상 diagonal 위에는 모두 $0$이 나오는 $E \Lambda^{1 \over 2}U$를 선택할 수 있다.
  • $E \Lambda^{1 \over 2}U$는 causal operator가 된다.

Example 4.1

• Eigenvector들을 이용한 Covariance matrix의 Causal Factorization
• random vector $y(u)$에 대하여
  • Covariance matrix $K_{y} = \left[ \begin{array}{rrr} 1 & -{1 \over 2} & -{1 \over 2} \\ -{1 \over 2} & 1 & -{1 \over 2} \\ -{1 \over 2} & - {1 \over 2} & 1 \end{array} \right]$
  • $HH^{\dagger} = K_{y}$ 가 되는 matrix $H$ 를 찾으면
  • $y(u)$는 $y(u) = Hw(u)$를 통해 simulate 될 수 있다.
• $H$를 구하는 방법
  • $(K_{y} - \lambda I)e = 0$
  • $det(K_{y} - \lambda I) = 0$
  • $det \left[ \begin{array}{rrr} 1 - \lambda & -{1 \over 2} & -{1 \over 2} \\ -{1 \over 2} & 1 - \lambda & -{1 \over 2} \\ -{1 \over 2} & - {1 \over 2} & 1 - \lambda \end{array} \right] = -{\lambda \over 4} (2 \lambda - 3)^{2} = 0$
  • $K_{y}e_{1} = 0 \Rightarrow e_{1}^{t} = {1 \over \sqrt{3}}(1, 1, 1)$
  • $(K_{y} - {3 \over 2} I)e_{2} = 0 \Rightarrow e_{2}^{t} = {1 \over \sqrt{2}}(1, -1, 0)$
  • $(K_{y} - {3 \over 2} I)e_{3} = 0 \Rightarrow e_{3}^{t} = \sqrt{{2 \over 3}}({1 \over 2}, {1 \over 2}, -1)$
  • $E \Lambda^{1 \over 2} = \left[ \begin{array}{rrr} 0 & {\sqrt{3} \over 2} & {1 \over 2} \\ 0 & -{\sqrt{3} \over 2} & {1 \over 2} \\ 0 & 0 & -1 \end{array} \right]$
  • $y(u) = \left[ \begin{array}{rrr} {\sqrt{3} \over 2} & {1 \over 2} \\ -{\sqrt{3} \over 2} & {1 \over 2} \\ 0 & -1 \end{array} \right] w(u)$
  • By choosing $U = \left[ \begin{array}{rrr} 0 & 0 & 1 \\ {\sqrt{3} \over 2} & -{1 \over 2} & 0 \\ {1 \over 2} & {\sqrt{3} \over 2} & 0 \end{array} \right]$
  • $E \Lambda^{1 \over 2} U = \left[ \begin{array}{rrr} 1 & 0 & 0 \\ -{1 \over 2} & {\sqrt{3} \over 2} & 0 \\ -{1 \over 2} & -{\sqrt{3} \over 2} & 0 \end{array} \right]$
  • $z(u) = \left[ \begin{array}{rrr} 1 & 0 \\ -{1 \over 2} & {\sqrt{3} \over 2} \\ -{1 \over 2} & -{\sqrt{3} \over 2} \end{array} \right] w(u)$

Brute Force Factorization

• $K = HH^{\dagger}$
  • $\left[ \begin{array}{rrr} 1 & -{1 \over 2} & -{1 \over 2} \\ -{1 \over 2} & 1 & -{1 \over 2} \\ -{1 \over 2} & - {1 \over 2} & 1 \end{array} \right] = \left[ \begin{array}{rrr} h_{11} & 0 & 0 \\ h_{21} & h_{22} & 0 \\ h_{31} & h_{32} & h_{33} \end{array} \right] \left[ \begin{array}{rrr} h_{11}^{} & h_{21}^{} & h_{31}^{} \\ 0 & h_{22}^{} & h_{32}^{} \\ 0 & 0 & h_{33}^{} \end{array} \right]$
  • $1 = |h_{11}|^{2} \Leftarrow h_{11} = i$
  • $-{1 \over 2} = h_{21} h_{11}^{*} = -h_{21}i \Leftrightarrow h_{21} = - {i \over 2}$
  • $-{1 \over 2} = h_{31} h_{11}^{*} = -h_{31}i \Leftrightarrow h_{31} = - {i \over 2}$
  • $1 = |h_{21}|^{2} + |h_{22}|^{2} = {1 \over 4} + |h_{22}|^{2} \Leftarrow h_{22} = - {\sqrt{3} \over 2}$
  • $H = \left[ \begin{array}{rrr} i & 0 \\ -{i \over 2} & -{\sqrt{3} \over 2} \\ -{i \over 2} & {\sqrt{3} \over 2} \end{array} \right]$
• $K = E \Lambda E^{\dagger}$
  • $= \left[ \begin{array}{rrrr} \lambda_{1} e_{1} & \lambda_{2} e_{2} & ... & \lambda_{n} e_{n} \end{array} \right] \left[ \begin{array}{rrrr} e_{1}^{\dagger} \\ e_{2}^{\dagger} \\ ... \\ e_{n}^{\dagger} \end{array} \right] = \sum_{i=1}^{n} \lambda_{i} e_{i} e_{i}^{\dagger}$
  • $K$의 eigenvalue들은 $K$의 spectrum이라 부른다.
• $x = \sum_{j = 1}^{n} a_{j} e_{j}$
  • $y = Kx = (\sum_{i = 1}^{n} \lambda_{i} e_{i} e_{i}^{\dagger})(\sum_{j = 1}^{n} a_{j} e_{j})$
  • $e_{i}$의 orthonormaliity를 이용해서
  • $y = \sum_{i = 1}^{n} \lambda_{i} a_{i} e_{i}$

Random Vectors의 Eigen-Representation

• simulation problem
  • $z(u) = E \Lambda^{1 \over 2} w(u) + m_{z}$
  • $z(u) = m_{z} + \sum_{j=1}^{n} x_{j}(u) e_{j} = m_{z} + \sum_{j = 1}^{n} \sqrt{\lambda_{j}} w_{j}(u) e_{j}$
• uncorrelated coefficients와 함께 알려진 벡터들의 random 선형 결합 $z(u)$의 representation를 Karhuenen-Loeve expansion의 finite-dimensional analog라고 한다.

Average Length Measures for Random Vectors

• $\mathbb{E} \{|z(u)|^{2}\} = \mathbb{E} \{ z^{\dagger}(u) z(u) \} = \sum_{t=1}^{n} R_{z}(t, t) = TR(R_{z})$
• $\mathbb{E} \{ |z(u)|^{2} \} = \sum_{i = 1}^{n} \lambda_{i}$

Directional Preference