Is $E[Z E[Z^2\mid Y] ]$ positive or negative? Let $Y=X+Z$ where $X$ and $Z$ are independent, zero mean, finite variance r.v. Moreover, $Z$ is Gaussian. Is there are way to say wether 
\begin{align*}
E[Z \ E[Z^2 \mid Y] ]
\end{align*}
is positive or negative?
I was able to show that
\begin{align*}
E[Z \ E[Z^2 \mid Y] ]=E[Z^2 \ E[Z \mid Y] ]
\end{align*}
using orthogonality principle, however, that doesn't help.
Thanks for any help.
 A: Suppose $X$ and $Z$ are independent, and they both have symmetric densities, so for all $t \in \mathbb{R}$: 
$$ f_X(t) = f_X(-t) \: \: , \: \: f_Z(t)=f_Z(-t) $$
Suppose $Y=X+Z$. 
Claim 0: The vector $(X,Z)$ has the same distribution as the vector $(-X,-Z)$. 
Claim 1:  The vector $(Z,Y)$ has the same distribution as the vector $(-Z,-Y)$. 
Claim 2: $f_Y(y)=f_Y(-y)$ for all $y \in \mathbb{R}$. 
Claim 3: $E[Z^2|Y=y]=E[Z^2|Y=-y]$ for all $y \in \mathbb{R}$.
Claim 4: $E[Z|Y=y] =-E[Z|Y=-y]$ for all $y \in \mathbb{R}$. 
Claim 5: $E[ZE[Z^2|Y]]=0$. 

Proof of Claim 0: Since $f_X(x)$ is symmetric we know $X$ and $-X$ have the same distribution.  Likewise, $Z$ and $-Z$ have the same distribution.  Thus: $$Pr[X \leq x, Z\leq z] = Pr[X\leq x]Pr[Z\leq z] = Pr[-X\leq x]Pr[-Z\leq z] \: \: \Box$$
Proof of Claim 1:  Since $(X,Z)$ has the same distribution as $(-X,-Z)$, we know $(Z,Y)=(Z,X+Z)$ has the same distribution as $(-Z,-X+(-Z))=(-Z, -Y)$. $\Box$
Proof of Claim 2: This follows because $Y$ and $-Y$ have the same distribution. $\Box$
Proof of Claim 3: Since $(Z,Y)$ and $(-Z,-Y)$ have the same distribution: 
$$ E[Z^2|Y=y] = E[(-Z)^2|(-Y)=y] = E[Z^2|Y=-y] \: \: \Box$$
Proof of Claim 4: Since $(Z,Y)$ and $(-Z,-Y)$ have the same distribution: 
$$ E[Z|Y=y] = E[(-Z)|(-Y)=y] = -E[Z|Y=-y] \: \: \Box $$ 
Proof of Claim 5: We have: 
\begin{align} 
E[ZE[Z^2|Y]] &= \int_{-\infty}^0 E[ZE[Z^2|Y]|Y=y]f_Y(y)dy + \int_{0}^{\infty}E[ZE[Z^2|Y]|Y=y]f_Y(y)dy\\
&=\int_{-\infty}^0 E[Z^2|Y=y]E[Z|Y=y]f_Y(y)dy + \int_{0}^{\infty}E[Z^2|Y=y]E[Z|Y=y]f_Y(y)dy\\
&=\underbrace{\int_{0}^{\infty} E[Z^2|Y=-u]E[Z|Y=-u]f_Y(-u)du}_{\mbox{change of variables}} + \int_{0}^{\infty}E[Z^2|Y=y]E[Z|Y=y]f_Y(y)dy\\
&= \underbrace{-\int_0^{\infty}E[Z^2|Y=u]E[Z|Y=u]f_Y(u)du}_{\mbox{by claims 2-4}}+ \int_{0}^{\infty}E[Z^2|Y=y]E[Z|Y=y]f_Y(y)dy\\
&=0
\end{align} 

A simpler proof of Claim 5 uses the fact that $(Z,Y)$ and $(-Z,-Y)$ have the same distribution: Define $m = E[ZE[Z^2|Y]]$. Note that since $Y$ determines $-Y$,  for any random variable $W$ we have $E[W|Y]=E[W|-Y]$.  Thus: 
$$m=E[ZE[Z^2|Y]] = E[(-Z)E[(-Z)^2|(-Y)]] = -E[ZE[Z^2|-Y]] = -E[ZE[Z^2|Y]]=-m $$
It follows that $m = 0$. $\Box$
