Distributions of two random variable remain the same after some operations I guess that the titles don't say much I'll be clearer here: 
Say we have two random variables with the same distributions and parameters, and we make the exact same math operation on both of them (multiplication, adding, subtracting etc..) I would've guessed that they both will have identical distributions afterwards, means the distribution itself can be changed but they both will have the same exact one.
Is there any proof for that assumption?
Thanks.
 A: Here is an unnecessarily intricate answer:

If a random variable $X : \Omega \to \Sigma$ is given on a probability space equipped with the probability measure $\mathbb{P}$, then its distribution is simply the pushforward measure $X_*\mathbb{P}$ defined by
$$ (X_* \mathbb{P})(A) = \mathbb{P}(X^{-1}(A)) = \mathbb{P}(\{\omega : X(\omega) \in A\}). $$
Now if $X$ and $Y$ have the same distribution, meaning that $X_*\mathbb{P} = Y_*\mathbb{P}$, and if $f$ is a transform, then we are interested in whether distributions of $f(X)$ and $f(Y)$ are the same, i.e. whether the equality $(f\circ X)_* \mathbb{P} = (f\circ Y)_* \mathbb{P}$ holds.
Now it is easy to show that pushforward is covariant, meaning that $(f\circ X)_* = f_* \circ X_*$. Thus
$$ (f \circ X)_*\mathbb{P} = f_*(X_* \mathbb{P}) = f_*(Y_* \mathbb{P}) = (f\circ Y)_* \mathbb{P}, $$
and so, $f(X)$ and $f(Y)$ have the same distribution.

In more friendly wording, we want to know whether the following two different paths commute:
$$
\require{AMScd}
\begin{CD}
X
@>\text{induce distribution}>>
X_*\mathbb{P}
\\
@V{\text{transform by } f}VV
@VV{\text{transform by } f_*}V
\\
f(X)
@>>\text{induce distribution}>
(f\circ X)_*\mathbb{P} \stackrel{?}{=} f_*(X_*\mathbb{P})
\end{CD}
$$
A: You have one distribution and one transformation of the variable. So there's only one transformed distribution. (Note that there is nothing "random" in the analytical process.)
A: Here is a proof.
Assume $X_1$ and $X_2$ are two real random variables with the same distribution.
For every (measurable) subset $A\subset\mathbb R$,  $P(X_1\in A) = P(X_2\in A$). 
Choose any (measurable) function $f:\mathbb{R}\rightarrow\mathbb{R}$.  Let $$Y_1=f(X_1)\quad\mathrm{ and }\quad Y_2=f(X_2).$$
Let $C_{Y1}$, and $C_{Y2}$ be the CDFs of  $Y_1,$ and $Y_2$ respectively.  
Fix any $y\in\mathbb R$.  
Let $A=\{x\in\mathbb R | f(x)<y\}.$
Now
$$
\begin{align}
C_{Y1}(y) &= P( Y_1<y) = P( f(X_1)<y) = P(X_1\in A) =  P(X_2\in A)\\
&=P( f(X_2)<y) = P( Y_2<y)= C_{Y2}(y).
\end{align}
$$ 
This proves that $Y_1$ and $Y_2$ have the same distribution.
