Let $G$ be an abelian locally (separable?) compact group with Haar measure $\mu$. Inspired by the interesting proof of [A sum of two binomial random variables](https://mathoverflow.net/questions/357655/a-sum-of-two-binomial-random-variables) :
Let $X$ and $Y$ be $G$-valued independent random variables with $\mu$-density $f\_X$ resp. $f\_Y$. Let $f\_{X \pm Y}$ be the $\mu$-density of $X \pm Y$. Is it always true that
$$\int f\_{X-Y}^2 d\mu = \int f\_{X+Y}^2 d\mu \ ?$$
This holds true for $G = \mathbb{Z}$ (here both integrals are finite)
.



Comparing $X+Y$ and $X-Y$ for independent random variables with values in an abelian locally compact group
The answer is yes. Indeed, let $dx:=\mu(dx)$ for brevity. We have 
\begin{align}
I\_{X,Y}&:=\iiint f\_X(x'+y'-y)f\_Y(y)f\_Y(y')f\_X(x')dx'\,dy\,dy' \\ 
&=\iint f\_{X+Y}(x'+y')f\_Y(y')f\_X(x')dx'\,dy' \\ 
&=\iint f\_{X+Y}(t)f\_Y(t-x')f\_X(x')dt\,dx' \\ 
&=\int f\_{X+Y}(t)^2dt. 
\end{align}
Also,
\begin{align}
I\_{X,Y}&=\iiint f\_X(x'-u'+u)f\_Y(-u)f\_Y(-u')f\_X(x')dx'\,du\,du' \\ 
&=\iiint f\_X(x'+u-u')f\_{-Y}(u)f\_{-Y}(u')f\_X(x')dx'\,du\,du' \\ 
&=\iiint f\_X(x'+u'-u)f\_{-Y}(u)f\_{-Y}(u')f\_X(x')dx'\,du\,du' \\ 
&=I\_{X,-Y} =\int f\_{X-Y}(t)^2dt, 
\end{align}
by what was shown in the previous display. 


Thus, 
$$\int f\_{X+Y}(t)^2dt=\int f\_{X-Y}(t)^2dt,$$
as desired. 


