Sum of independent random variables is also independent Given that $X, Y$ and $Z$ are discrete independent random variables,
how can one show that $X+Y$ and $Z$ are independent as well?

So far, I tried using the definition of independent variables and simplifying (X+Y)'s probability function using discrete convolution. I'm not sure that's the best way though.
 A: Hint: The adding $X+Y$ is not necessarily important. It suffices to show $P((X,Y)|Z) = P(X,Y)$ and $P(Z|(X,Y)) = P(Z)$ (why?). So no need for convolutions. Note this original claim might not hold under just pairwise independence assumptions on $X,Y,Z$, but you said "independent" which is stronger and should suffice to show $(X,Y)$ is independent of $Z$.
A: $\begin{align} 
f_{X+Y,Z}(s,t) & = f_{X+Y}(s\mid t)f_Z(t)
\\ & = f_Z(t)\int_\Bbb R f_{X,Y\mid Z}(x,s-x\mid t)\operatorname d s 
\\ & \mathop{=}^{?} f_Z(t)\int_\Bbb R f_{X\mid Z}(x\mid t)f_{Y\mid Z}(s-x\mid t)\operatorname d s \tag{$\bigstar$}
\\ & = f_Z(t)\int_\Bbb R f_{X}(x)f_{Y}(s-x)\operatorname d s 
\\ & = f_Z(t) \int_\Bbb R f_{X,Y}(x,s-x)\operatorname d x
\\ & = f_Z(t)f_{X+Y}(s)
\end{align}$
$\star$ This is only true iff we have conditional independence ; $X\perp Y\mid Z$.  This is not necessarily the case. Independence does not guarantee that conditional independence will also hold.
A: 
Two discrete random variables $X,Y$ are independent iff for all $x\in X(\Omega)$ and $y\in Y(\Omega)$ we have $P(X=x, Y=y)=P(X=x)\cdot P(Y=y)$. (Here $\Omega$ denotes the discrete sample space)

We choose two arbitrary but fixed elements $u_0\in(X+Y)(\Omega)$ and $z_0\in Z(\omega)$. Then, we consider
\begin{align*}
&P((X+Y)=u_0,Z=z_0)\\
&=\sum\limits_{x\in X(\Omega), y\in Y(\Omega)\\\text{with }x+y=u_0}P(X=x, Y=y,Z=z_0)\\
&=\sum\limits_{x\in X(\Omega), y\in Y(\Omega)\\\text{with }x+y=u_0}P(X=x)\cdot P( Y=y)\cdot P(Z=z_0)=P(Z=z_0)\cdot \sum\limits_{x\in X(\Omega), y\in Y(\Omega)\\\text{with }x+y=u_0}P(X=x)\cdot P( Y=y)\\
&=P(Z=z_0)\cdot P((X+Y)=u_0).
\end{align*}
We have used the independence of $X,Y,Z$ at the second equal sign.
As $u_0$ and $z_0$ were chosen arbitrarily, $(X+Y)$ and $Z$ are independent.
