Given X, Y dependent, bounded random variables: $-1\leq X\leq 1$ and $0\leq Y\leq 1$. Via Cauchy Schwarz $E( (X - E X) (Y- E Y)^2 )\leq \sqrt{ {\rm Var}(X) E(Y-E Y)^4}$ and with $E(Y-E Y)^4 = {\rm Var}(Y-EY)^2 + {\rm Var}^2(Y) \leq {\rm Var}(Y) + {\rm Var}^2(Y) $ I can bound everything in terms of second moments: $E( (X - E X) (Y- E Y)^2 )\leq \sqrt{ {\rm Var}(X)\left({\rm Var}(Y) + {\rm Var}^2(Y) \right) }$ but the bounding ${\rm Var}(Y-EY)^2 \leq {\rm Var}(Y) $ is to loose, respectively on $E(Y-E Y)^4$, and I cannot compute directly higher moments. Is there another bounding of $E( (X - E X) (Y- E Y)^2 )$?
Edit:
E.g. can one show somthing like $E( (X - E X) (Y- E Y)^2 ) \leq \sqrt{{\rm Var}(X) {\rm Var^2}(Y) }$
I don't think you can do better than with $\sqrt{Var[X]Var[(Y-E(Y))^2]}$ on the RHS. Consider $1-2p$ probability to have $(-1,1/2)$, $p$ probability to have $(1,0)$, $p$ probability to have $(1,1)$. Then $E((X-E(X))(Y-E(Y))^2)=p(1-2p)$. On the other hand $Var(X)=8p(1-2p)$ and $Var[(Y-E(Y))^2]=p(1-2p)/8$ This achieves the inequality. While $Var^2(Y)=p^2/4$, this does not suffice to prove when $p$ is close to zero.
On the other hand, since $X-E(X)$ has mean zero, we can write $E((X-E(X))((Y-E(Y))^2))=E[(X-E[X])((Y-E[Y])^2-E[(Y-E[Y])^2])]$, which can have this $\sqrt{Var[X]Var[(Y-E(Y))^2]}$ upper bound.