The correlated Rayleigh envelopes using a set of zero-mean complex Gaussian RVs (Random Variables) is given by $$G_{k}=\sigma_{k}(\sqrt{1-\lambda_k^2}X_k+\lambda_kX_0)+i\sigma_{k}(\sqrt{1-\lambda_k^2}Y_k+\lambda_kY_0),$$ where $i=\sqrt{-1}$, $\lambda_k\in(-1,1) \setminus \{0\}$, and $X_k$, $Y_k (k=0,\cdots,N)$ are independent and follow the Gaussian distribution $\mathcal{N}(0,1/2)$. Then, for any $k,j\in \{0,\cdots,N\}$, $\mathbb{E}\{X_kY_j\}=0$, and $\mathbb{E}\{X_kX_j\}=\mathbb{E}\{Y_kY_j\}=\frac{1}{2}\delta_{k,j}$, where $\delta_{k,j}$ is defined as $\delta_{k,k}=1$ and $\delta_{k,j}=0$ for $k\neq j$.

Therefore, $G_k$ is a zero-mean complex Gaussian distribution with $\mathcal{CN}(0,\sigma_k^2)$, and $|G_k|$ is Rayleigh distributed with mean-square value $\mathbb{E}\{|G_k|^2\}=\sigma_k^2$.

Next, considering that $X_0=x_0$ and $Y_0=y_0$ are fixed. How to derive the PDF (Probability Density Function) of $|G_k|$ conditioned on $x_0$ and $y_0$, as $f_{|G_k||x_{0},y_{0}}(r_k|x_0,y_0)$ indicated?

In Appendix I of the paper Novel Simple Representations for Gaussian Class Multivariate Distributions With Generalized Correlation, it was said that the required conditional distribution is a Rician distribution, which is given as $$f_{|G_k||x_{0},y_{0}}(r_k|x_0,y_0)=\frac{r_k}{\Omega_k^2}\exp(-\frac{r_k^2+\mu_k^2}{2\Omega_k^2})I_0(\frac{r_k\mu_k}{\Omega_k^2}),$$ where $$\mu_k^2=\mu_x^2+\mu_y^2, \mu_x=\lambda_kx_0, \mu_y=\lambda_ky_0,$$ $$\Omega_k^2=\sigma_k^2(\frac{1-\lambda_k^2}{2}), k=1,\cdots,N.$$ I would like to know the detailed derivation, thanks very much indeed!


1 Answer 1


Well, now I think the above conditional Rician PDF is false, which means it's a mistake in Appendix I of the paper, and it really misleads my further calculation.

When $X_0$ and $Y_0$ are fixed, mean values should be $$\mu_x = \sigma_k\lambda_kX_0, \mu_y = \sigma_k\lambda_kY_0,$$ and the variance of the real and the imaginary part of $G_k$ are identical as given by $$\Omega_k^2=:\mathrm{Var}(\mathfrak{R}\{G_k\})=\mathrm{Var}(\mathfrak{I}\{G_k\})=\sigma_k^2(1-\lambda_k^2)/2.$$ Thus, we would get a Rician distribution with the above PDF.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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