moments of random Fibonacci Consider random variables $X_{i,j}$ with $j \in \{1,2\}$, $i \in \{0,1,\ldots\}$ so that they are i.i.d with $EX_{i,j} = 0$ and $ \operatorname{Var}(X_{i,j}) = \sigma^2$ and assume all moments exists. Assume the random Fibonacci sequence $\{a_i\}_{i=0}^{\infty}$ so that $a_0 = X_{0,1}$, $a_1 = X_{1,1}$ and that else we have the recurrence
$$
 a_i = X_{i,1}a_{i-1} + X_{i,2}a_{i-2}
$$
The variance of $a_i$ follows
$$
E(a_i) = 0
$$ is trivial and
$$
 \operatorname{Var}(a_i) = \sigma^2( \operatorname{Var}(a_{i-1}) +  \operatorname{Var}(a_{i-2})),
$$ (which can be solved similarly like Fibonacci series).
which can be seen by
$$
\operatorname{Var}(a_i) = E(a_i^2) - (E(a_i))^2 = E(a_i^2)
$$
$$
E(a_i)^2 = E((X_{i,1}a_{i-1} + X_{i,2}a_{i-2})^2) = E(X_{i,1}a_{i-1})^2 + E(X_{i,2}a_{i-2})^2 +2E(X_{i,1}a_{i-1}X_{i,2}a_{i-2})
$$
and noting that $X_{i,*}$ is independent of $a_{i-1}a_{i-2}$. So
$$
E(X_{i,1}a_{i-1})^2 = E(X_{i,1})^2 E(a_{i-1})^2 = \sigma^2 E(a_{i-1})^2
$$ likewise for $E(X_{i,2}a_{i-2})^2$ and finally
$$
E(X_{i,1}a_{i-1}X_{i,2}a_{i-2}) = E(X_{i,1})E(X_{i,2})E(a_{i-1}a_{i-2}) = 0
$$
But what about the higher moments 3 and 4.
 A: As you note, the fourth moment has a mixed term:
$$E[a_i^4] = E[X_{i,1}^4] E[a_{i-1}^4] + 6E[X_{i,1}^2]E[X_{i,2}^2]E[a_{i-1}^2a_{i-2}^2]+ E[X_{i,2}^4]E[a_{i-2}^4]$$
So, set up a recursion for the mixed term as follows: multiply your initial recursion formula with $a_{i-1}$ to get
$$a_ia_{i-1} = X_{i,1}a_{i-1}^2 + X_{i,2}a_{i-1}a_{i-2}$$
Square this and take expectations
$$E[a_i^2a_{i-1}^2] = E[X_{i,1}^2]E[a_{i-1}^4] + E[X_{i,2}^2]E[a_{i-1}^2a_{i-2}^2]$$
This can be recast in a vector recursion for 
$$\left(\begin{array} \; E[a_i^2a_{i-1}^2] \\ E[a_i^4] \end{array}\right)$$
namely
$$\left(\begin{array} \; E[a_i^2a_{i-1}^2] \\ E[a_i^4] \end{array}\right) = \left(\begin{array} \; E[X_{i,2}^2] & E[X_{i,1}^2] \\ 6E[X_{i,1}^2]E[X_{i,2}^2] & E[X_{i,1}^4] \end{array}\right) \left(\begin{array} \; E[a_{i-1}^2a_{i-2}^2] \\ E[a_{i-1}^4] \end{array}\right) + \left(\begin{array} \; 0 & 0 \\ 0 & E[X_{i,2}^4] \end{array}\right)\left(\begin{array} \;  E[a_{i-2}^2a_{i-3}^2] \\ E[a_{i-2}^4] \end{array}\right)$$
A: Inspired by the answer I was wondering of we can say anything about the characteristic function
$$
\phi_n(s,t) = E\exp(isa_{n}+ita_{n})
$$ Use the recurrence and you will find
$$
\phi_n(s,t) = E\exp(is(X_{n,1}a_{n-1}+X_{n,2}a_{n-2})+ita_{n})=\exp(i(sX_{n,1}+t)a_{n-1}+isX_{n,2}a_{n-2})
$$ And because the indepens between $X_{n,*}$ and $a_{n-1},a_{n-2}$ we can write this as
$$
\phi_n(s,t) = \int K(s,t,s',t')\phi_{n-1}(s',t')ds'dt'
$$ So if one could find the eigenvalues of $K$ we should be able to effectively write calculate the characteristic function.
