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Let $\mathbf{x},\mathbf{y} \in \mathbb{C}^{N}$ be two independent vectors, with elements distributed independently as $\mathcal{C}\mathcal{N}(0,\sigma_{\mathbf{x}}^{2})$ and $\mathcal{C}\mathcal{N}(0,\sigma_{\mathbf{y}}^{2})$, respectively.

I need to derive the PDF of $Z = |\mathbf{x}^{\mathrm{H}} \mathbf{y}|^{2}$, denoted $f_{Z}(z)$.

Tentative solution:

  • I know that when $\sigma_{\mathbf{x}}^{2} = \sigma_{\mathbf{x}}^{2} = 1$ and one of the two vectors is normalized (e.g., $\|\mathbf{x}\|_{2}=1$), we have $f_{Z}(z) = \exp(-z)$, but I can't derive the distribution for the general case.
  • I can write $|\mathbf{x}^{\mathrm{H}} \mathbf{y}|^{2} = \mathbf{x}^{\mathrm{H}} \mathbf{y} \mathbf{y}^{\mathrm{H}} \mathbf{x} = \big| \sum_{n=1}^{N} x_{n}^{*} y_{n} \big|^{2}$, where $x_{n}$ and $y_{n}$ are the elements of $\mathbf{x}$ and $\mathbf{y}$, respectively, but here I get stuck.
TheDon
  • 919

1 Answers1

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This paper "Distribution of Inner Product of Two Complex Gaussian Vectors and its Application to MPSK Performance" provides the pdf of the real and imaginary parts of the inner product resul $z=z_1+jz_2$. From there you can find the distribution of the absolute $|z|$ which is Rayleigh distribution. The square of the Rayleigh distributed random variable is exponentially distributed.

Ahmed Abouelfadl
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