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?



Derivatives of conditional expectations
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 $\mathrm{Cov}(A, B) = \mathrm{Var}(X) \geq 0$.


