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!


1 Answer 1


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 $c=0$.

  • $\begingroup$ 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! $\endgroup$ Commented Nov 7, 2023 at 4:29
  • $\begingroup$ 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. $\endgroup$ Commented Nov 7, 2023 at 11:01
  • $\begingroup$ 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. $\endgroup$ Commented Nov 7, 2023 at 11:47
  • $\begingroup$ @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. $\endgroup$
    – AlexTP
    Commented Nov 7, 2023 at 13:28

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