Hilbert Space Theory

Complex random variables form a Hilbert space with inner product $\langle X, Y \rangle = \mathbb{E}\left[XY^*\right]$. If we have a random complex vector, then we can use Hilbert Theory in a more efficient manner by looking at the matrix of inner products. For simplicity, we will call this the “inner product” of two complex vectors.

Definition 1

Let the inner product between two random, complex vectors $\boldsymbol{Z_1}, \boldsymbol{Z_2}$ be defined as

$\langle \boldsymbol{Z_1}, \boldsymbol{Z_2} \rangle = \mathbb{E}\left[\boldsymbol{Z_1}\boldsymbol{Z_2}^*\right]$

The ij-th entry of the matrix is simply the scalar inner product $\mathbb{E}\left[X_iY_j^*\right]$where $X_i$ and $Y_j$ are the ith and jth entries of $\boldsymbol{X}$ and $\boldsymbol{Y}$ respectively. This means the matrix is equivalent to the cross correlation $R_{XY}$ between the two vectors. We can also specify the auto-correlation $R_X = \langle \boldsymbol{X}, \boldsymbol{X} \rangle$and auto-covariance $\Sigma_X = \langle \boldsymbol{X} - \mathbb{E}\left[\boldsymbol{X}\right] , \boldsymbol{X} - \mathbb{E}\left[\boldsymbol{X}\right] \rangle$. One reason why we can think of this matrix as the inner product is because it also satisfies the properties of inner products. In particular, it is

1. Linear: $\langle \alpha_1\boldsymbol{V_1}+\alpha_2\boldsymbol{V_2}, \boldsymbol{u} \rangle = \alpha_1\langle \boldsymbol{V_1}, \boldsymbol{u} \rangle + \alpha_2\langle \boldsymbol{V_2}, \boldsymbol{u} \rangle$.

2. Reflexive: $\langle \boldsymbol{U}, \boldsymbol{V} \rangle = \langle \boldsymbol{V}, \boldsymbol{U} \rangle ^*$.

3. Non-degeneracy: $\langle \boldsymbol{V}, \boldsymbol{V} \rangle = \boldsymbol{0} \Leftrightarrow \boldsymbol{V} = \boldsymbol{0}$.

Since we are thinking of the matrix as an inner product, we can also think of the norm as a matrix.

Definition 2

The norm of a complex random vector is given by $\|\boldsymbol{Z}\|^2 = \langle \boldsymbol{Z}, \boldsymbol{Z} \rangle$.

When thinking of inner products as matrices instead of scalars, we must rewrite the Hilbert Projection Theorem to use matrices instead.

Theorem 1 (Hilbert Projection Theorem)

The minimization problem $\min_{\hat{\boldsymbol{X}}(\boldsymbol{Y})}\|\hat{\boldsymbol{X}}(\boldsymbol{Y}) - \boldsymbol{X}\|^2$ has a unique solution which is a linear function of $\boldsymbol{Y}$. The error is orthogonal to the linear subspace of $\boldsymbol{Y}$ (i.e $\langle \boldsymbol{X} - \hat{\boldsymbol{X}}, \boldsymbol{Y} \rangle = \boldsymbol{0}$)

When we do a minimization over a matrix, we are minimizing it in a PSD sense, so for any other linear function $\boldsymbol{X}'$,

$\|\boldsymbol{X}-\hat{\boldsymbol{X}}\|^2 \preceq \|\boldsymbol{X} - \boldsymbol{X}'\|^2.$

Innovations

Suppose we have jointly distributed random variables $Y_0, Y_1,\cdots,Y_n$. Ideally, we would be able to “de-correlate” them so each new vector $E_0$ captures the new information which is orthogonal to previous random vectors in the sequence. Since vectors of a Hilbert Space operate like vectors in $\mathbb{R}^n$, we can simply do Gram-Schmidt on the $\{Y_i\}_{i=0}^n$.

Definition 3

Given jointly distributed random vectors $\{Y_i\}_{i=0}^n$ with $\mathcal{L}_i = \text{span}\{Y_j\}_{j=0}^i$, the ith innovation $E_i$ is given by

$E_i = Y_i - \text{proj}(Y_i|\mathcal{L}_{i-1}) = Y_i - \sum_{j=0}^{i-1}\frac{\langle Y_i, E_j \rangle }{\|E_j\|^2}E_j$

Innovations have two key properties.

1. $\forall i\neq j,\ \langle E_i, E_j \rangle =0$
2. $\forall i,\ \text{span}\{Y_j\}_{j=0}^i = \text{span}\{E_j\}_{j=0}^i$

We can also write innovations in terms of a matrix where $\boldsymbol{\varepsilon} = A\boldsymbol{Y}$ where $\boldsymbol{\varepsilon} = \begin{bmatrix}E_0 & E_1 & \cdots & E_n\end{bmatrix}^T$ and $\boldsymbol{Y} = \begin{bmatrix}Y_0 & Y_1 & \cdots & Y_n\end{bmatrix}^T$. Since each $E_i$ only depends on the previous $Y_i$, then A must be lower triangular, and because we need each $E_i$ to be mutually orthogonal, $R_{\varepsilon}$ should be diagonal. $R_{\varepsilon} = AR_YA^*$, so if $R_Y \succ 0$, then we can use its unique LDL decomposition $R_Y = LDL^*$ and let $A = L^{-1}$.