The independent distributions $A_{1}...A_{n}$, and $B_{1}...B_{n}$ have expectations $0$ and variances $1$.
$C_{n} = A_{n} + \frac{-1}{12}A_{n-1}$
$D_{n} = B_{n} + \frac{3}{7}B_{n-1}$
What is the value of $cov(C_{n}, D_{n})$ ?
Any help would be appreciated. Here is my working so far:
$cov(C_{n}, D_{n}) = E(C_{n}D_{n}) - E(C_{n})E(D_{n})$
$ E(C_{n}) = E(D_{n}) = 0$
$ \therefore cov(C_{n}, D_{n}) = E(C_{n}D_{n})$
$ cov(C_{n}, D_{n}) = E((A_{n} + \frac{-1}{12}A_{n-1}) (B_{n} + \frac{3}{7}B_{n-1})) $
$ cov(C_{n}, D_{n}) = E(A_{n}B_{n} + \frac{-1}{12}A_{n-1}B_{n} + \frac{3}{7}A_{n}B_{n-1} + \frac{-1}{28}A_{n-1}B_{n-1}) $
$ cov(C_{n}, D_{n}) = $
$ E(A_{n}B_{n}) + \frac{-1}{12}E(A_{n-1}B_{n}) + \frac{3}{7}E(A_{n}B_{n-1}) + \frac{-1}{28}E(A_{n-1}B_{n-1}) $
But I am not sure how to calculate the rest.
All the four terms in $ E(A_{n}B_{n}) + \frac{-1}{12}E(A_{n-1}B_{n}) + \frac{3}{7}E(A_{n}B_{n-1}) + \frac{-1}{28}E(A_{n-1}B_{n-1}) $ are zero by independence. For example
$E(A_{n}B_{n})=(EA_n)(EB_n)=(0)(0)=0$.
In fact, $C_n$ and $D_n$ are independent so it is no surprise that the covariance is $0$.
[Assuming that the entire collection $\{A_1,A_2,..,A_n,B_1,B_2,...,B_n\}$ is independent the covariance is $0$. If you do not assume that $A_i$'s are independent of $B_j$'s then the answer is just what you have arrived at; you cannot go further than that].