Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
3  
Imagine $X$ that is bimodal: it is about $0$ or $1$ with probability $1/2$ each (you can smear that to get continuous density). Imagine $Y=\pm 0.1$ with probability $1/2$ for each sign (again, smear a bit) and $Z$ smeared over a huge interval uniformly. Then the condition $B=b$ tells next to nothing about $X$ but when $A=0.1$, you know that $Y$ is $0.1$ (otherwise there is no chance to get there) and when $A=0.9$, you are pretty certain that $Y=-0.1$. Bad, isn't it? As to simple conditions, what terms are allowed? –  fedja Dec 13 '11 at 2:55
    
Thanks, fedja. Bimodality seems like a reasonably robust way to construct counterexamples. Perhaps a natural assumption would be to assume that all the density functions are log-concave (i.e., $x \mapsto -\log f(x)$ is a convex function). This would include Gaussian density functions (though not power-law). –  Tom LaGatta Dec 13 '11 at 17:35
add comment

1 Answer

up vote 1 down vote accepted

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.