@BenDerrett and Michael. Thanks for the solutions. This is precisely what I was looking for (sorry for the late reaction by me)... I'm torn to who to give the bounty. Is there a way to split the bounty?

Try keeping the question open for a bit. Ben's sufficient condition is not clear to me...I would like to know if the necessary and sufficient conditions can be matched. You can give the bounty to Ben eventually.

@BenDerrett are you sure $f(Z)=E[Y|Z]$ is suffficient? We need to prove $Cov(Y-E[Y|Z], X)=Cov(X,Y|Z)$, which reduces to proving $E[XY] - E[XE[Y|Z]] = E[XY|Z] - E[X|Z]E[Y|Z]$.

The most general sufficient condition I can currently muster is if $Cov(X,Y|Z)$ does not depend on $Z$, and if there is a function $g(Z)$ such that $Cov(X, g(Z))\neq 0$, in which case we define $U_1=ag(Z)$ for a suitable $a$.

I think what @BenDerrett is saying is: Assume that Cov$(X,Y|Z)=b$ i.e. doesn't depend on $Z$. Define $U_1=E[Y|Z]$ and $U_2=Y-E[Y|Z]$. Then $Y=U_1+U_2$ with Cov$(X,U_1|Z)=0$ and Cov$(X,U_2|Z)=$Cov$(X,U_2)$.

@user103828 Yes, but that is exactly what I cannot prove. It reduces to trying to prove $Cov(X, Y-E[Y|Z])=b$.

Is it not just $Cov(X,U_1|Z)=Cov(X,f(Z)|Z)=0$ and $Cov(X,U_2|Z)=Cov(X,Y-U_1|Z)=Cov(X,Y|Z)-Cov(X,U_1|Z)=Cov(X,Y|Z)$ which is constant by assumption... thanks, fixed the typo just in time. (In your notation $Cov(X,Y-E[Y|Z]|Z)=Cov(X,Y|Z)$ because $E[Y|Z]$ is a function in $Z$).

Actually I thought the second half of your answer was exactly the proof.

Everything you said is already in my answer. So How do you prove $Cov(X,U_2|Z)=Cov(X,U_2)$? In general, if $Cov(A,B|C)$ does not depend on $C$, that does not mean it is equal to $Cov(A,B)$.

Take $C$ as any random variable with nonzero variance and define $A=B=C$. Then $Cov(A,B|C)=0$ for all $C$, but $Cov(A,B)=Var(C)\neq 0$.

Yes, but for sufficiency aren't we assuming $Cov(X,Y|Z)=Cov(X,Y)$

No. We are assuming $Cov(X,Y|Z)$ does not depend on $Z$. Take my counterexample above to see these are different things.

oh... maybe I'm confused I thought saying $cov(X,Y|Z)$ does not depend on $Z$ is the same thing as saying $cov(X,Y|Z)=constant$

Yes, I thought so at first but it is kind of counter-intuitive.

oh i see.

I'll put that in a comment since it is not obvious and woudl be a common misconception.

Is it possible to have $Cov(X,Y|Z)=constant$ and the constant is not equal to zero?

The constant might have to be zero.

Okay obviously yes.

I'll think about this a little and put anything else I think about in a comment.

Hi there

Hi Ben, I'm (we're) a little confused by why $Cov(X,Y|Z)$ does not depend on $Z$ is also sufficient.

If $Cov(X,Y|Z)=Cov(X,Y)$ then it's straightforward (based on Michael's answer)

I'm about to post an edit

but $Cov(X,Y|Z)$ does not necessarily equal $Cov(X,Y)$... okay

See what you think...

it looks good.... just to confirm, in the "IF" part the line $E[Cov(X,Y|Z)]=Cov(X,Y|Z)$ is because we are assuming $Cov(X,Y|Z)$ is constant?

yes

thanks very much... there's something which is just lingering in my mind... i'll see if i can see what exactly it is.

Regardless, I'll mark your answer correct with the bounty tomorrow.

Great. I hope it helps you (and any future readers).