This is a problem refer to my answer to the following question, moments of random Fibonacci
What can be said about the equations that follows. Let $X_1,X_1$ be defined on $-1,1$ with symmetric distribution. The question is what can be said about the following difference equation. $$ f(s,t) = \text{E}(f(aX_1s+t,bX_2 s) = \frac{1}{4}(f(t+a,bs) + f(t-a,bs) + f(t+a,-bs) + f(t-a,bs)) $$ Is there specific $a,b$ for which this equation is solvable? To check at least the viability of this lets expand $f$ as $$ f(s,t) = f + f_s s + f_t t + f_{ss} s^2/2 + f_{tt} t^2/2 + f_{st}2st $$ The constant term is ok. The linear terms: $$ sf_s+tf_t = f_s t $$ which leads to $f_s=f_t = 0$, for the second order we get $$ f_{ss} s^2/2 + f_{tt} t^2/2 + f_{st}2st = f_{ss} E(t+X_1a)^2/2 + f_{tt} t^2/2E(bX_2)^2+ f_{st}2(E(bX_2(aX_1+t)) $$ which is the same as $$ f_{ss} s^2/2 + f_{tt} t^2/2 + f_{st}2st = f_{ss} (t^2+s^2a^2)/2 + f_{tt} (sb)^2/2 $$ we conclude that we must have $f_{st} = 0$ and $$ f_{ss} = b^2f_{tt} + a^2f_{ss} $$ and $$ f_{tt} = f_{ss} $$ which is solvable if $a^2 + b^2 = 1$. This means that there exists a nonzero solution. But can we produce a solution that is a characteristic function of a probability distribution function on $R^2$.
Take $\xi_1,\xi_2$ as the $X_1,X_2$ in the question. Here is a try. Let's go back to the refeered question and the random fibonacci. Then what we are after is the following recurrence which the steady state we are after that can be constructed from the original fibonacci by scaling and a limit distribution is saught. So the recurrence is $$ X_n = a\xi_{i,1} X_{n-1} + b\xi_{i,2} X_{n-1} $$
If we hypothesize that the marginal distribution is normal we see that we can build such a distribution out of normal bivariate distribution like what follows. Assume $X_u\in N(0,\Sigma_u)$ with $$ \Sigma_u=\begin{pmatrix} 1 & u \\ u & 1 \end{pmatrix}. $$ Note that given $u$, The transform is $(a\xi_1X_u + b \xi_2Y_u,X_u)$ and we see that the new sigma, due to $a^2+b^2=1$ is
$$ \Sigma_u'=\begin{pmatrix} 1 & a\xi_1 + b \xi u \\ a\xi_1 + b \xi u & 1 \end{pmatrix}. $$ The distribution of U must now be such that $U,V$ is equal in distribution where $$ V = a\xi_1 + b \xi_2U, $$ limited to the restriction $U\in [-1,1]$ or $$ \frac{|a|}{1-|b|} \leq 1 $$ using $a^2+b^2=1$ this means that $\frac{1}{\sqrt{2}}\leq |a| \leq 1$. Then the solution is $$ U=a\sum_{i=0} \xi_{i,1}b^i\prod_{j=1}^{i-1}\xi_{j,2} $$ because $\pm \xi_{i,1}$ is in distribution the same as $\xi_{i,1}$ we can simplify the expression to $$ U=a\sum_{i=0} \xi_{i,1}b^i $$