Is the conditional expectation $E[ax^2-a\mid bx^2-b]=0$? where $x$ is a standard Gaussian variable.

Question

In my question, $a$ and $b$ are some constants. $E[]$ means the expectation.

My answer is that the conditional expectation $E[ax^2-a\mid bx^2-b]\not =0$.

Given the Gaussian random variable $X$, $E[ax^2-a]=a-a=0$.

For $E[ax^2-a\mid bx^2-b]$, $x$ is assumed to be fixed, thus $ax^2-a$ is not a random variable, hence $E[ax^2-a\mid bx^2-b]\not =0$.

Answer 1

Call $Y = bX^2 - b$. Notice that if $b \neq 0$ we have $$ aX^2 - a = b^{-1}abX^2 - b^{-1}ab = ab^{-1}(bX^2 - b) = ab^{-1}Y. $$ Thus, we have $$ E[aX^2 - a \mid Y] = E[ab^{-1}Y\mid Y] = ab^{-1}Y = aX^2 - a.$$ This should be already clear from the first line, since we proved $aX^2 - a$ is $Y$-measurable and for any measurable function $g: \mathbb{R} \to \mathbb{R}$ we have $E[g(Y) \mid Y] = g(Y)$.

Concluding, we have for $b \neq 0$ that $E[aX^2 - a\mid Y] = 0$ iff $a=0$. In the case $b = 0$, then $Y = 0$ is a constant and thus is independent of $X$. Thus, $$ E[aX^2 - a\mid Y] = E[aX^2 - a] = 0$$ for any value of $a$.