Is $E[Z E[Z^2\mid Y] ]$ positive or negative?

Question

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.

Answer 1

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: $E[ZE[Z^2|Y]]=0$.

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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}

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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 $m = 0$. $\Box$