For the following, I do not understand how to solve for $d$
Problem Statement:
Choose $\mathbf{c}$ and $d$ to minimize $MSE=E(|y_k-a_{k-d}|^2)= E(|\mathbf{c}^T\mathbf{r}_k-a_{k-d}|^2)$
where $d\in\mathbb Z^+_0$ and $c_i\in\mathbb{R}$. Also $a_k$ are iid that means $E(a_ka_i)=0\forall i\neq k$ and $E(a_k^2)=E_a$
Solution: $\mathbf{c}=\mathbf{R}^{-1}\mathbf{p}$
is obtained for a given $d$
\begin{align} MSE &= E(|y_k-a_{k-d}|^2) \\ &= E(|\mathbf{c}^T\mathbf{r}_k-a_{k-d}|^2)\\ &= \mathbf{c}^TE(\mathbf{r}_k^{}\mathbf{r}_k^T)\mathbf{c} + E(|a_{k-d}|^2) - 2\mathtt{Re}\{\mathbf{c}^TE(\mathbf{r}_ka_{k-d})\} \\ &= \mathbf{c}^T\mathbf{R}\mathbf{c} + E_a - 2\mathtt{Re}\{\mathbf{c}^T\mathbf{p})\}\\ &= (\mathbf{c}-\mathbf{R}^{-1}\mathbf{p})^T\mathbf{R}(\mathbf{c}-\mathbf{R}^{-1}\mathbf{p})+E_a-\mathbf{p}^{T}\mathbf{R}\mathbf{p}\tag{1} \end{align}
where
- $\mathbf{R}=E(\mathbf{r}_k\mathbf{r}_k^T)$
- $\mathbf{p}=E(\mathbf{r}_ka_{k-d})$
- $\mathbf{r}_k = \mathbf{H}\mathbf{a}_k+\mathbf{n}_k$
- $\mathbf{H} = [\mathbf{h}_0 \mathbf{h}_1 \cdots \mathbf{h}_d \cdots \mathbf{h}_{N+\mu}]$ where $\mathbf{h}_d$ is $(d+1)^{th}$ column of $\mathbf{H}$
- $\mathbf{a}_k = [a_k a_{k-1} \cdots a_d \cdots a_{k-(N+\mu)}]^T$ and $a_i$ is iid
- $n_i\sim N(0,\sigma^2)$
- $\mathbf{c} = [c_0 c_1 \cdots c_{N}]$
My guess is Since $\mathbf{H},\sigma^2$ is known, therefore I compute $\mathbf{R}^{-1}$ only once and then sweep $d\in[0,1,2,...,N+\mu]$, and compute $\mathbf{c}$ and $MSE$ for each value of $d$ and then choose one $d$ which yield minimum MSE.
But this is an exhaustive search and doesnt sound like a good idea.