I want to solve this difficult problem, so I need your help,
I want to prove that if $X$ and $Y$ are independent real variables such that $X+Y$ and $X-Y$ are independent, then $X^2$ and $Y^2$ are integrable and $X-E(X)$ and $Y-E(Y)$ are identically distributed .
I don't know exactly how to begin. Thanks for your answers.
Assume that $X$ and $Y$ have zero mean. Let us consider the characteristic functions $f$ of $X$, $g$ of $Y$, $h$ of $X+Y$, $j$ of $X-Y$.
Now, $h(s)j(t)$ is the average of $e^{i(s+t)X+i(s-t)Y}$, thus $h(s)j(t)=f(s+t)g(s-t)$. In particular, $h(s)=f(s)g(s)$, $j(t)=f(t)g(-t)$.
Thus $f(s)f(t)g(s)g(-t)=f(s+t)g(s-t)$. (E)
Now write $s=u+v$, $t=u-v$, then $f(u+v)f(u+(-v))g(u-(-v))g(v-u)=f(2u)g(2v)$.
Applying (E) to the left-hand side yields: $f(2u)g(2v)=f(u)f(v)g(v)g(-u)f(u)f(-v)g(u)g(v)=f(u)^2g(v)^2|f(v)|^2|g(u)|^2$.
Thus $f(2u)=c_1f(u)^2|g(u)|^2$, $g(2v)=c_2|f(v)|^2g(v)^2$, for some constants $c_1,c_2$. Hence $c_1=c_2=1$ (with $u=v=0$).
In particular, $|f|=|g|$, and the set of zeroes of $fg$ is stable under division by $2$ and closed. Since it does not contain $0$, $|f|$ does not vanish.
In particular, there exists a unique differentiable function $\theta$ such that $f=ge^{i\theta}$, and $\theta(0)=0$.
The equations and uniqueness yield that $\theta(2t)=2\theta(t)$, which entails $\theta(t)=\theta’(0)t$. Now since $X$ and $Y$ have zero mean, $\theta’(0)=0$ hence $f=g$.
Edit: I realize I did not address the last part of the question. I am adding it now.
$X+Y$ and $X-Y$ are independent and integrable thus so is their product $X^2-Y^2$, thus, for some large $R>0$, so is $|X^2-Y^2|1(|Y| \leq R) \geq X^21(|Y| \leq R) -R^2$. Therefore, $X^21(|Y| \leq R)$ is $L^1$, hence, if $|Y| \leq R$ has positive probability, $X^2$ is integrable.