Derivatives of conditional expectations Let $X$, $Y$ and $Z$ be independent, real-valued random variables, probably with continuous density functions.  Define $A = X + Y$ and $B = X + Z$.  Consider the regular conditional expectation $E_Y(a,b) = \mathbb E[Y|A=a, B=b]$.  
Is it the case that $$\frac{\partial}{\partial a} E_Y(a,b) > 0 \quad \mathrm{and} \quad \frac{\partial}{\partial b} E_Y(a,b) \le 0?$$ If there is a counterexample, then are there simple conditions which guarantee this statement to be true?
Edit:  fedja provides a counterexample in the case of bimodality.  Assume that all the density functions in question are unimodal, or perhaps even log-concave.  Under these assumptions, is it the case that the above partial differential inequalities hold?
 A: The wanted inequalities seem not to be possible outside of very 
'pathological' situations.
First, a general remark. If $U$ and $V$ are two r.vs, the condition that
$\mathbb{E}(U|V=v)$ is non-decreasing in $v$ is a kind of ``positive association'' 
between $U$ and $V$. This condition implies $\mathrm{Cov}(U, V) \ge 0$ 
(under suitable conditions of existence). The proof of this is quite simple.
Now, your hypothesis imply that $\mathbb{E}(Y|A=a)$ is non-decreasing 
in $a$. Indeed, $\mathbb{E}(Y|A=a)$ writes as an integral in $b$ involving
$E_Y(a, b)$. This integral can be derivated w.r.t. $a$ under the $\int$ sign, leading 
to a non-negative derivative $\partial_a \mathbb{E}(Y|A=a) \ge 0$. Then
$$
  \mathbb{E}(B|A=a) = \mathbb{E}(X+Z|A=a) 
  =  \mathbb{E}(A-Y+Z|A=a)= a - \mathbb{E}(Y|A=a) + \mathbb{E}(Z|A=a)
$$
But $\mathbb{E}(Z|A=a)=E(Z)$ since $Z$ and $A$ are independent. It thus appears 
that $\mathbb{E}(B|A=a)$
is non-increasing in $a$ since $\mathbb{E}(Y|A=a)$ is non-decreasing. So we find
a negative association between $A$ and $B$ (in the former sense),
 implying $\mathrm{Cov}(A, B) \le 0$ as claimed. But this is not possible 
since $\mathrm{Cov}(A, B) = \mathrm{Var}(X) \geq 0$.
