Would the ratio of Autocorrelation value and autocorrelation of noise for different data types be different? in Matlab Considering a model of the form:

$y[n] = x[n] + noise[n]$ and 

$y[n] = u[n] + noise[n]$ 

where the term noise represents Additive White Gaussian Noise of variance $\sigma^2_{noise}$.
I want to understand if there is any difference in the value of the terms: $\sigma^2_x/\sigma^2_{noise}$ and $\sigma^2_u/\sigma^2_{noise}$ by varying the noise level. 

$x$ represents a signal generated from Gaussian distribution whose variance is $\sigma^2_x$. In Matlab, I have used the command rand() to generate $x$.

$u$ represents another signal taking values zero or 1 having variance $\sigma^2_u$. In Matlab, I have used rand()&gt;0.5 to generate thie binary valued data.

Based on my understanding the numerator in the ratio is the autocorrelation of the data mathematically expressed as $E[\mathbf{x}[n]\mathbf{x}[n]^T]$ and  $E[\mathbf{u}[n]\mathbf{u}[n]^T]$. Why are these terms giving same result?

In most papers, the variance of the data irrespective of whether it takes numeric or symbols is considered to be 1. 

Question1) Is my approach of implementing the ratio correct? I cannot understand what the value of $\sigma^2_x$ and $\sigma^2_u$ should be in practice. Implementing this way would give the same value for the ratio for both the different kinds of data $x$ and $u$. Would the auto correlation for numeric values x and binary data u be the same value?

Question2) If the ratio is the same then what is the need for using binary signal?

N = 128;
var_x =1;
var_u = 1;
signal_x = rand(1,N);
signal_u = rand(1,N)&gt;=0.5; % this creates 0/1 data
noise = 0;
index = 1;
for noise = 0:5:30
noise_var  = 10^(-noise/10);
ratio_x(index)= 10*log10(var_x/noise_var); %in dB
ratio_u(index)= 10*log10(var_u/noise_var);
index = index+1;
end
I am not really sure what you want to do so please comment and I will modify my answer.

The case that $x$ and $u$ seems statistically identical is that $x$ is standard Gaussian random variable and $u$ has $\pm 1$ values.

Use your notations, $x$ is Gaussian random variable with zero mean and unit variance (deducted from the fact that you used the standard rand function, even though the randn() function must be used for Gaussian pdf).

Let's stay with real $x$ for the sake of simplicity :

$$p(x) = \frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)$$

with $\mathbb{E}[x] = \mu = 0 \textrm{ and } \mathrm{var}(x) = \sigma^2 = \mathbb{E}[x^2] = 1$.

Thus the probabilites $\mathrm{Pr}(x &gt;= 0) = \mathrm{Pr}(x &lt; 0) = 0.5$.

The processed data $u = 1 \textrm{ if } x \ge 0 \textrm{ and } u=-1 \textrm{ otherwise}$. Thus $\mathbb{E}[u] = (1) \times \mathrm{Pr}(x &gt;= 0) + (-1) \times \mathrm{Pr}(x &lt; 0) = 0.$

Then $\mathrm{var}(u) = \mathbb{E}[(u - \mathbb{E}[u])^2] = \mathbb{E}[u^2] = (1)^2 \times \mathrm{Pr}(x &gt;= 0) + (-1)^2 \times \mathrm{Pr}(x &lt; 0) = 1$.

The means and variances are just coincidentally identical.

MATLAB/Octave code

octave:6&gt; N = 65536;
octave:7&gt; x=randn(1,N);
octave:8&gt; u=2*(randn(1,N)&gt;0) - 1;
octave:9&gt; var(x)
ans =  1.0020
octave:10&gt; var(u)
ans =  1.0000


The case of OP question $x$ is (0,1)-uniform (by rand()) and $u$ is binary:

$\mathrm{Pr}(x &gt;= 0.5) = \mathrm{Pr}(x &lt; 0.5) = 0.5$.

$\mathbb{E}[x] = 0.5$

$\mathrm{var}(x) = \int_0^1 (x-0.5)^2 \mathrm{d}x = \frac{2}{3}\times 0.5^3 \approx 0.8$

The processed data $u = 1 \textrm{ if } x \ge 0.5 \textrm{ and } u=0 \textrm{ otherwise}$. Thus $\mathbb{E}[u] = (1) \times \mathrm{Pr}(x &gt;= 0.5) + 0 \times \mathrm{Pr}(x &lt; 0.5) = 0.5$.

$\mathrm{var}(u) = \mathbb{E}[(u - 0.5)^2] = (1 - 0.5)^2 \times \mathrm{Pr}(x &gt;= 0.5) + (0 - 0.5)^2 \times \mathrm{Pr}(x &lt; 0.5) = 0.25$

MATLAB/Octave code

octave:13&gt; N = 65536;
octave:14&gt; x=rand(1,N);
octave:15&gt; u=rand(1,N)&gt;0.5;
octave:16&gt; var(x)
ans =  0.083424
octave:17&gt; var(u)
ans =  0.25000


The variances are different !!!