2

I have a complex Gaussian vector $\mathbf{x} \triangleq (x_1; x_2; ...; x_n) \in \mathbb{C}^{n \times 1}$, where $x_i \sim \mathbb{CN}(0,1)$ i.i.d., and $\mathbf{y} \triangleq (y_1;y_2;..;y_n)\in \mathbb{C}^{n \times 1}$, where $y_i = \frac{x_i}{||x||}$. What is the expectation of the product of $\mathbf{y}$ and $\mathbf{y}^{H}$, i.e., $\mathbb{E}\{ \mathbf{y}\mathbf{y}^{H} \}$? Here, $\mathbf{y}^{H}$ denotes the hermitian transpose of $\mathbf{y}$, and $\mathbb{E}\{ \cdot \}$ denotes the expectation matrix which contains the expectation for each element in $\mathbf{y}\mathbf{y}^{H}$. Or what is the expectation of the product of ${y}_i$ and ${y}^{H}_j$, i.e., $\mathbb{E}\{{y}_i{y}^{H}_j\}$?

According to this question, I know that $\mathbf{y}$ is uniformly distributed on a unit sphere, but I have no idea what $\mathbb{E}\{ \mathbf{y}\mathbf{y}^{H} \}$ and $\mathbb{E}\{{y}_i{y}^{H}_j\}$ ($i = j $ or $i \neq j$) is, or how to calculate them. Could someone show me the exact computations?

Thanks!

humpbackwhale
  • 43

1 Answers1

0

$\mathbf A=\mathbf{yy}^H$ is an outer product, a random $n×n$ complex matrix, so $E(\mathbf{yy}^H)$ is a corresponding $n×n$ matrix of expectations. Each element of $\mathbf y$ is a unit complex number.

The diagonal elements of $\mathbf A$ are always 1, since the product of a unit complex number with its conjugate is always 1, so the expectation matrix is also 1 on the diagonal. As for $\mathbf A$'s off-diagonal elements, it is easy to see that they are uniformly distributed on the unit circle like $\mathbf y$'s entries (the product of two iid uniform unit complex numbers is likewise uniform), so their expectation is 0.

Thus $E(\mathbf{yy}^H)=\mathbf I_n$.

Parcly Taxel
  • 105,904
  • Thank you for your answer! It helps me a lot! But I have a little question that in this question, it suggests that $\mathbb{E}{\mathbf{y}\mathbf{y}^{H}} = \frac{1}{n} \mathbf{I}_n$. Why it has a factor $ \frac{1}{n}$? From my point, that question is the same as mine. Maybe I made some mistakes. Could you tell me the reasons? Thank you very much! – humpbackwhale Jan 13 '19 at 15:30
  • @humpbackwhale That's the covariance matrix, which is a different thing from the matrix here. – Parcly Taxel Jan 13 '19 at 15:33
  • Sorry to bother you again, but I think the covariance matrix in this question, i.e., $\mathbb{E}{\mathbf{Y}\mathbf{Y}^{T}}$, is the same as mine, i.e., $\mathbb{E}{ \mathbf{y} \mathbf{y}^{H} }$. Is there any difference? Thanks! – humpbackwhale Jan 13 '19 at 15:53
  • @humpbackwhale This is expectation. That is variance. You are getting the two mixed up. – Parcly Taxel Jan 13 '19 at 15:57
  • Maybe I didn't make the explanation of $\mathbf{y}$ clear. Not every element in $\mathbf{y}$ is uniformly distributed on a unit sphere, but the whole $\mathbf{y}$ is uniformly distributed on a unit sphere... – humpbackwhale Jan 13 '19 at 16:10