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!


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.


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