In the simplest scenario of MIMO spatial multiplexing: $$\mathbf{y} = H\mathbf{s} + \mathbf{n}$$ where: $\mathbf{s}=[s_0,s_1,...s_{M-1}] \\\mathbf{y}=[y_0,y_1,...y_{N-1}]$

$\mathbf{n}=[n_0,n_1,...n_{N-1}]$ is zero mean complex normal independent Random vector

$M$ transmit antennas, $N$ receive antennas where $N\geq M$

The maximum likelihood solution is: $$\tilde{\mathbf{s}} = \arg\min_{\mathbf{s} \in QPSK^M} ||\mathbf{y}-H\mathbf{s}||^2$$

Now, my textbook says that due to the fact that in this scenario $H^{*}H$ is not a diagonal matrix, the least squares solution followed by proccesing is not optimal, but I can't see why this is true. To my understanding, denoting the LS solution as $\hat{s}=(H^{*}H)^{-1}H^{*}\mathbf{y}$
I can write: $$ ||\mathbf{y}-H\mathbf{s}||^2 = ||\mathbf{y}-H\hat{s}+H\hat{s}-H\mathbf{s}||^2 = \\||y-H\hat{s}||^2 + ||H(\hat{s}-s)||^2 + 2Re[(y-H\hat{s})^{*}H(\hat{s}-s)]$$

Now, by substituing the LS solution the $Re$ part vanishes. And the problem reduces to $$\arg\min_{s\in QPSK^{M}}|\hat{s}-s|^2$$ So I don't get why the ML is not optimal is this case. Can anyone please clarify? Thanks

edit: screen shot from the book enter image description here

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    $\begingroup$ Could you tell the noise properties? $\endgroup$ – Royi Jul 28 '18 at 15:40
  • $\begingroup$ Regular zero mean complex normal independent random vector $\endgroup$ – user3921 Jul 28 '18 at 15:47
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    $\begingroup$ Could you share the book name and a screen shot of the page you're talking about? I think I know what's the catch, yet just to be sure. $\endgroup$ – Royi Jul 28 '18 at 15:53
  • $\begingroup$ Thanks, I added, it from a course thought by Dr Doron Ezri - OFDM-MIMO course $\endgroup$ – user3921 Jul 28 '18 at 16:03

You must attention to the written text.
It doesn't say the ML isn't optimal, what it says is that the problem isn't regular LS problem but Least squares problem with Constraints.

The constraints make analytic solution infeasible and hence in order to solve it one must go through any signal in the space of valid signals and mark the one which minimizes the LS problem.

What they mean that the regular LS solution result doesn't necessarily yield a solution in the space of valid space of signals and even taking the closes one to the solution isn't optimal.

So we have the problem the constrained problem:

$$\begin{align*} \arg \min_{s} \quad & \frac{1}{2} {\left\| H s - y \right\|}_{2}^{2} \\ \text{subject to} \quad & s \in {QPSK}^{M} \end{align*}$$

Which its solution is optimal.

What they say isn't optimal is solving:

$$\begin{align*} \arg \min_{s} \quad & \frac{1}{2} {\left\| H s - y \right\|}_{2}^{2} \\ \end{align*}$$

And then find $ s \in {QPSK}^{M} $ as you'd do in MIMO.
Namely, the greedy method (Do LS while ignoring the constraint and then apply the constraint) isn't optimal.

Yet the problem is that there is no analytic or efficient way to solve the Constraint LS problem but doing brute force search which isn't feasible.

I agree the text isn't clear about making this observation.
Yet it well known in the optimization field that solving in this greedy 2 step method isn't guaranteed to be optimal.

I found a link to the PDF - Doron Ezri - MIMO OFDM Lecture Notes. What I talked about can be seen in part 5.4.

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  • $\begingroup$ I don't get it, basically your'e saying that using regular LS yields post processing which is far to computationally heavy and thus not optimal? $\endgroup$ – user3921 Jul 28 '18 at 16:13
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    $\begingroup$ I tried making it clearer. Read the answer again. By the way, what's the name of the book? $\endgroup$ – Royi Jul 28 '18 at 16:20
  • $\begingroup$ Theoretically, If ignoring computational complexity, the greedy way using the LS would be optimal, right? Its from a course by Dr Doron Ezri - OFDM-MIMO $\endgroup$ – user3921 Jul 28 '18 at 16:26
  • $\begingroup$ Yes, the constraint problem is optimal given AWGN. Is there a site for the course? $\endgroup$ – Royi Jul 28 '18 at 16:33
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    $\begingroup$ I found a link to the PDF - Doron Ezri - MIMO OFDM Lecture Notes. What I talked about can be seen in part 5.4. $\endgroup$ – Royi Jul 28 '18 at 17:06

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