2
$\begingroup$

Let $\mathbf{U}$ be a random unitary matrix and $\mathbf{z}$ be a random i.i.d complex Gaussian vector (unitary invariant). Assume that the following relation is satisfied: \begin{align} \mathbf{y}=\mathbf{U}\mathbf{s}+\mathbf{z}. \end{align} Are the following relations hold for the mutual information between $\mathbf{s}$ and $\mathbf{y}$? \begin{align} I(\mathbf{s};\mathbf{y})&=I(\mathbf{s};\mathbf{U}^{\mathrm{H}}\mathbf{y})\\ &=I(\mathbf{s};\mathbf{s}+\mathbf{U}^{\mathrm{H}}\mathbf{z})\\ &=I(\mathbf{s};\mathbf{s}+\mathbf{z})\\ &=h(\mathbf{s}+\mathbf{z})-h(\mathbf{z}), \end{align} where $(.)^{\mathrm{H}}$ is a hermitian opertor.

$\endgroup$
1
  • $\begingroup$ This relationship is clearly true for deterministic $\mathbf{U}$. $\endgroup$
    – Math_Y
    May 13 at 23:40
2
+50
$\begingroup$

No, this is not correct. Consider as a counterexample the case that $s$ can take only two values, a unit vector $e$ or minus $e$. Since $s$ is rotated randomly to construct $y=Us+z$, knowledge of $s$ gives you no information on $y$, so the mutual information $I(s,y)=0$.

On the other hand, knowledge of $s$ does give you information on the sum $s+z=\pm e+z$, so $I(s,s+z)\neq 0$, contradicting the third equality in the OP.

The error appears already in the first equality: $I(s,y)=I(s,U_0y)$ for any given unitary $U_0$, but this is not the same unitary as in the construction $y=Us+z$, since that $U$ is unknown.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.