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In one of the research paper following equations are given :

$y(n) = h_0s(n)+w(n),d = 0$

$y(n) = h_1s(n)+w(n),d = 1$

where $n$ ranges from $1$ to $N$ and

$h_0,h_1$ are wireless channel assumed as $\sim\mathcal{C}{N}(0,\sigma^2)$. $s(n)\sim\mathcal{C}{N}(0,P_s)$ and it represents transmitted signal. $w(n)\sim\mathcal{C}{N}(0,N_w)$ and it represents transmitted AWGN.

In the paper it is written that distribution of received signal, $y(n)$ when $d = 0$ is $\textbf{y}(n)\sim\mathcal{C}{N}(\textbf{0},\sigma^2_0I_N)$.----(1)

I am not getting how the distribution of received signal, $y(n)$, in (1) is zero mean complex Gaussian.

Any help in this regard will be highly appreciated.

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  • Since $w$ and $h_0s$ are independent (in the statistical sense), the variance of their sum must be the sum of their variances.

  • Sums of Gaussians are Gaussians.

Therefore, in the formula $\sigma_0^2 = \sigma^2+P_s$, and the diagonal shape of the covariance implies that $s(n_1)$ is uncorrelated to $s(n_2)$ for $n_1\ne n_2$. That's it.

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  • $\begingroup$ Ok..I understood your answer to some extent. But I had a very basic doubt . In $y(n)$ expression, $h_0$, $s(n)$ and $w(n)$ all are zero mean complex Gaussian distribution. And $h_0$, $s(n)$ are in product. So how Gaussian x Gaussian is Gaussian? $\endgroup$
    – paru
    Aug 16, 2021 at 11:37
  • $\begingroup$ you're confusing what is random for how long. Assume $h_0$ and $h_1$ to be constants for the duration of your transmission (just that their value is randomly chosen before): that's why they are $h_0$ and not $h_0(n)$. $\endgroup$
    – Marcus Müller
    Aug 16, 2021 at 11:48
  • $\begingroup$ Ok.. understood perfectly...Thanks for your brilliant response.. $\endgroup$
    – paru
    Aug 16, 2021 at 11:51
  • $\begingroup$ What will be the distribution of $y(n)$ if $s(n)$ is PSK signal instead of zero mean complex Gaussian... $\endgroup$
    – paru
    Aug 16, 2021 at 11:52
  • $\begingroup$ Don't ask new questions in the comments. Also, I think you should be able to answer that yourself! $\endgroup$
    – Marcus Müller
    Aug 16, 2021 at 11:53

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