Calculation of the LMMSE Channel Estimator

Consider a MIMO system equipped with $$N_t$$ transmit antennas and $$N_r$$ receive antennas. The received signal over $$L$$ snapshots are given by

$$Y = H X + Z,$$

where $$X$$ is the $$N_t \times L$$ transmited signal, $$H$$ is the $$N_r \times N_t$$ channel matrix to be estimated, and $$Z$$ is the $$N_r \times L$$ Gaussian noise with zero mean and $$\sigma^2$$ variance.

When calculating the classic LMMSE estimator of $$H$$, one first set $$\hat{H} = Y A$$ and solve the following optimization problem: $$\min_{A} \quad \mathcal{E} := \operatorname{E}\left[ \operatorname{tr}\left((H - Y A) )^H (H - Y A)\right) \right],$$ where $$\mathcal{E}$$ is the MSE of the estimator $$\hat H$$ (See Section 5).

However, following the LMMSE definition from the statistics literature, the estimator $$\tilde{h} := \operatorname{vec}(\tilde{H})$$ should be something like $$\tilde{h} = B\operatorname{vec}(Y)+c$$ with parameter $$B$$ amd $$c$$. This offers more degree of freedom as compared to previously assumed $$\hat{H} = Y A$$ (as the matrix $$B$$ is larger and the introduction of the constant term $$c$$).

My question is: what is the underlining assumption when one uses $$\hat{H} = Y A$$ to calculate the LMMSE instead of the generic one, i.e., $$\tilde{h} = B\operatorname{vec}(Y)+c$$?

Some of my thoughts:

1. If we assume $$\operatorname{E}[H] = 0$$, then $$\operatorname{E}[Y] = 0$$, and hence $$c$$ must be zero to ensure that the LMMSE estimator is unbiased.

Any thoughts on this problem is appreciated!

The linear MMSE estimator is always unbiased, see Wikipedia article, $$\text{E}[\hat{x}]=\bar{x}$$.
The reasoning is similar but slightly different from what you said: if you can assume $$E[H]=0$$ (and $$Z$$ has zero mean), no need to estimate $$c=0$$.
• Thanks you for your answer. Your answer definitly helps. This solves part of my question that $c = 0$. The other part of my question is why using the small matrix $A$ instead of the big matrix $B$ to represent the linear transformation? Thanks in advance! Commented Nov 7, 2023 at 4:29
• Thank you for your response. The linked Wikipedia page shows the general framework of LMMSE when the vector $x$ is estimated by the observation $y$. If one directly uses this framework in the channel estimation, then one would like to estimate $\operatorname{vec}(H)$ based on the observation $\operatorname{vec}(Y)$, and hence the model formulation should be $\tilde{h} = B\operatorname{vec}(Y)+c$. But I found the above mentioned model different from the model that I learned from paper. Commented Nov 7, 2023 at 11:01
• Maybe I have done something wrong. This Wikipedia page shows the method for the estimation of a random vector. To apply this method to the channel estimation problem (where the channel to be estimated is a matrix), I introduced the $\operatorname{vec}(\cdot)$ operator. Commented Nov 7, 2023 at 11:47
• @maphadofan From $\hat{H} = YA$, if you vectorize it $$\textrm{vec}(\hat{H}) = \textrm{vec}(I_{N_r} YA)=(A^T\otimes I_{N_r}) \textrm{vec}(Y)$$ you get $B = A^T\otimes I_{N_r}$. Therefore, the two formulation should be equivalent. However, as I don't have real experience using the vectorization transform, you may want to ask for the confirmation elsewhere. Also, let me delete my other misleading comments. Commented Nov 7, 2023 at 13:28