Mean independent

Question

A random variable $X$ is said to be mean independent of another random variable $Y$ if its conditional expectation given $Y$ is equal to its unconditional expectation, that is, $E[X \mid Y]=E(X)$.
My question is, for $\phi$ Borel function we can say that $E[\phi(X) \mid Y]=E[\phi(X)]$ ?
Thanks.

Answer 1

If $X\sim N(0,1)$ and $Y=X^{2}$ then $E(X|Y)=0$ (by symmetry of the distribution of $X$) but $E(X^{2}|Y)=X^{2} \neq EX^{2}$.

Answer 2

Let $R$ be the result of rolling a fair die.

Define the random variable $X$ by $$ X=(R\;\text{mod}\;3) \qquad\qquad\qquad\qquad\;\;\, $$ and define the random variable $Y$ by \begin{cases} Y=0&\text{if}\;R\in\{2,3,5,6\} \qquad\qquad\;\;\;\, \\[4pt] Y=1&\text{if}\;R\in\{1,4\}\\ \end{cases} Then we get $$ \left\lbrace \begin{align*} &E(X) = \frac{1}{3}{\,\cdot\,}0 + \frac{1}{3}{\,\cdot\,}1 + \frac{1}{3}{\,\cdot\,}2 = 1 \qquad \\[4pt] &E(X|Y=0) = \frac{1}{2}{\,\cdot\,}0 + \frac{1}{2}{\,\cdot\,}2 = 1 \\[4pt] &E(X|Y=1) = 1{\,\cdot\,}1 = 1 \\[4pt] \end{align*} \right. $$ so $X$ is mean independent of $Y$, but $$ \left\lbrace \begin{align*} &E(X^2) = \frac{1}{3}{\,\cdot\,}0^2 + \frac{1}{3}{\,\cdot\,}1^2 + \frac{1}{3}{\,\cdot\,}2^2 = \frac{5}{3} \\[4pt] &E(X^2|Y=0) = \frac{1}{2}{\,\cdot\,}0^2 + \frac{1}{2}{\,\cdot\,}2^2 = 2 \\[4pt] &E(X^2|Y=1) = 1{\,\cdot\,}1^2 = 1 \\[4pt] \end{align*} \right. $$ so $X^2$ is not mean independent of $Y$.