# Effects of condensing a random variable to only 2 possible values

$X$ 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 $E[Y^4] \leq 2(E[Y^2])^2$?

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Can you provide some background for this question? –  Shai Covo Dec 21 '10 at 14:54

Here is a start. Let $f(x)$ be the probability density function of $X$. Let $p$ be the probability that $X<0, p=\int_{-\infty }^0f(x)dx$ $Y$ has only two values, $y_1=\frac{1}{p}\int_{-\infty }^0xf(x)dx$ with probability $p$ and $y_2=\frac{1}{1-p}\int_0^{\infty }xf(x)dx$ with probability $1-p$. The condition that $E[X]=0$ tells us that the integrals in $y_1$ and $y_2$ are the negative of each other, call them $-a$ and $a$. So $Y$ is $\frac{-a}{p}$ with probability $p$ and $\frac{a}{1-p}$ with probability $1-p$. $E[Y^2]=\frac{a^2}{p}+\frac{a^2}{1-p}=\frac{a^2}{p(1-p)}$. $E[Y^4]=\frac{a^4}{p^3}+\frac{a^4}{(1-p)^3}=\frac{a^4(p^3+(1-p)^3)}{p^3(1-p)^3}=\frac{a^4(1-3p+3p^2)}{p^3(1-p)^3}$
Wolfram Alpha says $E[Y^4]\leq 2(E[Y^2])^2$ if $\frac{5-\sqrt{5}}{10}\leq p \leq \frac{5+\sqrt{5}}{10}$ or about $0.276393 \leq p \leq 0.723607$ . Presumably the condition $E[X^4] \leq 2(E[X^2])^2$ can show us that, but I need to think more on it.
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 ${\rm E}(Y^4) > 2 [{\rm E}(Y^2)]^2$.