# understanding the distribution of received signal

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.

• 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.

• 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?
– paru
Aug 16, 2021 at 11:37
• 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 $h_0(n)$. Aug 16, 2021 at 11:48
• Ok.. understood perfectly...Thanks for your brilliant response..
– paru
Aug 16, 2021 at 11:51
• What will be the distribution of $y(n)$ if $s(n)$ is PSK signal instead of zero mean complex Gaussian...
– paru
Aug 16, 2021 at 11:52
• Don't ask new questions in the comments. Also, I think you should be able to answer that yourself! Aug 16, 2021 at 11:53