$\hat A_n=Q(A_n)$,Are the processes $U_n$ and $\hat A_n$ the same? Assume $X_n$ is an iid Gaussian random process with zero mean and variance $\sigma^2$, and $U_n$ be an iid binary random process with $P_r\{ U_{n}=1\}=P_r\{U_n=-1\}=0.5$, and $\{U_n\}$ is independent of $\{X_n\}$, now let $A_n=X_n+U_n$
$\hat A_n=Q(A_n), \forall n$,where $Q(r)=\begin{cases}
+1,  r \ge 0 \\
-1, r \lt 0 
\end{cases}$
Are the processes $U_n$ and $\hat A_n$ the same? That is ,do they have the same process distributions? 
I think yes,but i don't know how to prove it
 A: Yes, they have the same process distributions: To prove this, you just need to show that the values in the process $\{ \hat{A}_n \}$ are independent with probabilities $\mathbb{P}(\hat{A}_n=1) = \mathbb{P}(\hat{A}_n=-1) = \tfrac{1}{2}.$  Actually, you don't even need $X_n$ to be normally distributed, since you get the same results with any continuous symmetric distribution.
Mutual independence of the $\hat{A}_n$ values follows directly from the independence of the underlying $X_n$ and $U_n$ processes.  The marginal probabilities can be found from the law-of-total-probability:
$$\begin{equation} \begin{aligned}
\mathbb{P}(\hat{A}_n=1) 
&= \mathbb{P}(X_n+U_n \geqslant 0) \\[6pt]
&= \mathbb{P}(X_n+1 \geqslant 0 | U_n=1) \mathbb{P}(U_n=1) + \mathbb{P}(X_n-1 \geqslant 0 | U_n=1) \mathbb{P}(U_n=-1) \\[6pt]
&= \mathbb{P}(X_n \geqslant -1 | U_n=1) \cdot \frac{1}{2} + \mathbb{P}(X_n \geqslant 1 | U_n=1) \cdot \frac{1}{2} \\[6pt]
&= \frac{1}{2} \Bigg[ \mathbb{P}(X_n \geqslant -1 | U_n=1) + \mathbb{P}(X_n \geqslant 1 | U_n=1) \Bigg] \\[6pt]
&= \frac{1}{2} \Bigg[ \mathbb{P}(X_n < 1 | U_n=1) + \mathbb{P}(X_n \geqslant 1 | U_n=1) \Bigg] = \frac{1}{2}, \\[10pt]
\mathbb{P}(\hat{A}_n=-1) 
&= \mathbb{P}(X_n+U_n < 0) \\[6pt]
&= \mathbb{P}(X_n+1 <0 | U_n=1) \mathbb{P}(U_n=1) + \mathbb{P}(X_n-1 <0 | U_n=1) \mathbb{P}(U_n=-1) \\[6pt]
&= \mathbb{P}(X_n < -1 | U_n=1) \cdot \frac{1}{2} + \mathbb{P}(X_n < 1 | U_n=1) \cdot \frac{1}{2} \\[6pt]
&= \frac{1}{2} \Bigg[ \mathbb{P}(X_n < -1 | U_n=1) + \mathbb{P}(X_n < 1 | U_n=1) \Bigg] \\[6pt]
&= \frac{1}{2} \Bigg[ \mathbb{P}(X_n \geqslant 1 | U_n=1) + \mathbb{P}(X_n < 1 | U_n=1) \Bigg] = \frac{1}{2}. \\[6pt]
\end{aligned} \end{equation}$$
(The fact that $X_n$ has a continuous symmetric distribution is used in the step to the last lines if each derivation.  Actually, you can make the requirements even weaker: all you need is continuity of the distribution at the points $x = \pm 1$.)
