TITLE: Mean independent
QUESTION [1 upvotes]: 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.

REPLY [0 votes]: 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 $Y$.
{"set_name": "stack_exchange", "score": 1, "question_id": 4165745}