TITLE: Effects of condensing a random variable to only 2 possible values
QUESTION [6 upvotes]: $X$ is a random variable, which is not constant. $E[X]=0$. $E[X^4] \leq 2(E[X^2])^2$.
Let $Y$ be given by: $P(Y=E[X|X \geq 0]) = P(X \geq 0)$ and $P(Y=E[X|X \lt 0]) = P(X \lt 0)$.
Do we necessarily have $E[Y^4] \leq 2(E[Y^2])^2$?

REPLY [3 votes]: No. Here is a counterexample. Define $X$ as follows. ${\rm P}(X=a) = p_1$, ${\rm P}(X=0) = p_2$, and ${\rm P}(X=\frac{{ - ap_1 }}{{1 - p_1  - p_2 }}) = 1-p_1-p_2$,
where $a$ is a positive constant. Then, ${\rm E}(X)=0$. Denote ${\rm E}(X^4)-2 [{\rm E}(X^2)]^2$ by $\xi$. Then,
$$
\xi = a^4 p_1  + \Big(\frac{{ap_1 }}{{1 - p_1  - p_2 }}\Big)^4 (1 - p_1  - p_2 ) - 2\Big[a^2 p_1  + \Big(\frac{{ap_1 }}{{1 - p_1  - p_2 }}\Big)^2 (1 - p_1  - p_2 )\Big]^2.
$$
To find ${\rm E}(Y^4)-2 [{\rm E}(Y^2)]^2$, which we denote by $\eta$, we first find 
$$
{\rm E}[X|X \ge 0] = a{\rm P}(X = a|X \ge 0) = a\frac{{{\rm P}(X = a)}}{{{\rm P}(X \ge 0)}} = \frac{{ap_1 }}{{p_1  + p_2 }}
$$
and
$$
{\rm E}[X|X < 0] = \frac{{ - ap_1 }}{{1 - p_1  - p_2 }}.
$$
Hence, by definition, ${\rm P}(Y = \frac{{ap_1 }}{{p_1  + p_2 }}) = p_1  + p_2 $ and ${\rm P}(Y = \frac{{ - ap_1 }}{{1 - p_1  - p_2 }}) = 1 - p{}_1 - p_2$. 
Thus, 
$$
{\rm E}(Y^i) = \Big(\frac{{ap_1 }}{{p_1  + p_2 }}\Big)^i (p_1  + p_2) + \Big(\frac{{ap_1 }}{{1 - p_1  - p_2 }}\Big)^i (1 - p_1  - p_2),
$$
from which we get an explicit expression for $\eta$. Now, to furnish a counterexample, it suffices to find a triple $(a,p_1,p_2)$ for which $\xi \leq 0$ and $\eta > 0$. This is most easily done using a computer. Here is a concrete example.
Letting $a=4$, $p_1=0.36$, and $p_2=0.4$, we have $-ap_1/(1-p_1-p_2) = -6$. Here,
$$
a^4 p_1  + \Big(\frac{{ap_1 }}{{1 - p_1  - p_2 }}\Big)^4 (1 - p_1  - p_2 ) = 403.2
$$
and
$$
2 \Big[a^2 p_1  + \Big(\frac{{ap_1 }}{{1 - p_1  - p_2 }}\Big)^2 (1 - p_1  - p_2 )\Big]^2 = 414.72,
$$
so ${\rm E}(X^4) \leq 2 [{\rm E}(X^2)]^2$; on the other hand,
$$
\Big(\frac{{ap_1 }}{{p_1  + p_2 }}\Big)^4 (p_1  + p_2) + \Big(\frac{{ap_1 }}{{1 - p_1  - p_2 }}\Big)^4 (1 - p_1  - p_2) \approx 320.835
$$
and
$$
2\Big[\Big(\frac{{ap_1 }}{{p_1  + p_2 }}\Big)^2 (p_1  + p_2) + \Big(\frac{{ap_1 }}{{1 - p_1  - p_2 }}\Big)^2 (1 - p_1  - p_2)\Big]^2 \approx 258.482,
$$
so ${\rm E}(Y^4) > 2 [{\rm E}(Y^2)]^2$.