I want to prove the following property :
Let $(X_n)$ and $(Y_n)$ be a martingale with respect to the filtration $(F_n)$.
Let's also assume that $E(X_n^2),E(Y_n^2)<+\infty$
Prove that $$E(X_{n-1}Y_{n-1})=\frac{E(X_nY_{n-1})+E(X_{n-1}Y_n)}{2}$$
Above fact is obvious when $(X_n)$ and $(Y_n)$ are independent due to fact that $X_n$ and $Y_n$ are marginals with respect to the filtration $(F_n)$
(Then $E(X_n)=E(X_{n-1}$),$E(Y_n)=E(Y_{n-1}),E(X_nY_n)=E(X_n)E(Y_n),E(X_{n-1}Y_{n-1})=E(X_{n-1})E(Y_{n-1}))$.
But I don't have idea in a general case when $X_n$ and $Y_n$ are not independent. I tried to use some martingales properties but i didnt't get anything crucial.
Note that $$ X_{n-1} Y_{n-1} = (X_{n-1} - X_n) Y_{n-1} + X_n Y_{n-1} $$ and $$X_{n-1} Y_{n-1} = X_{n-1} (Y_{n-1} - Y_n) + X_{n-1} Y_n. $$ For the right-hand side of the first equation we can use \begin{align*} E[(X_{n-1} - X_n) Y_{n-1}] &= E[E[(X_{n-1} - X_n) Y_{n-1}| F_{n-1}]] \\ &= E[Y_{n-1}\ E[(X_{n-1} - X_n) | F_{n-1}]] \\ &= 0, \end{align*} where the last equality follows by the martingale property for $\{X_n\}$. We can do the same for the second equation using the martingale property of $\{Y_n\}$. Adding the equations together, taking expectations, and dividing by 2 yields the desired result.