Would the ratio of Autocorrelation value and autocorrelation of noise for different data types be different? in Matlab
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 !!!