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Let $X$ be any random variable (maybe symmetric or asymmetric). Let $Z = (X - E(X))/Var(X)$. Is the following inequality always true ? $$ P(Z\le -a) \le 2P(Z\ge a), $$ for any $a>0$. Thank you for your answer.

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  • $\begingroup$ Suppose you have a distribution that satisfies this inequality. What happens when you reflect that distribution about zero? $\endgroup$ – probably_someone Apr 3 '18 at 22:26
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Suppose your inequality is true. Let $X$ be a random variable with $E(X)=0$. Let $X'$ be the random variable such that $P(X'=y)=P(X=-y)$ for all $y$. In other words, let $X'$ be the random variable drawn from the distribution of $X$ after reflecting it about zero.

Since $E(X)=0$, we also have that $Z=X/Var(X)$. Let $Z'=X'/Var(X')$ be the analogous quantity for $X'$. We immediately have the following four equalities (noting that $Var(X')=Var(X)$):

$$P(Z\leq -a)=P(X\leq -a\cdot Var(X))$$ $$P(Z\geq a)=P(X\geq a\cdot Var(X))$$ $$P(Z'\leq -a)=P(X'\leq -a\cdot Var(X'))=P(X'\leq -a\cdot Var(X))$$ $$P(Z'\geq a)=P(X'\geq a\cdot Var(X'))=P(X'\geq a\cdot Var(X))$$

But we also have that, because of our choice of $X'$,

$$P(X\geq a\cdot Var(X))=P(X'\leq -a\cdot Var(X))$$ $$P(X\leq -a\cdot Var(X))=P(X'\geq a\cdot Var(X))$$

which overall means that

$$P(Z'\leq -a)=P(Z\geq a)$$ $$P(Z'\geq a)=P(Z'\leq -a)$$

If the inequality is true for $X$, then

$$P(Z\leq -a)\leq 2P(Z\geq a)$$

but from the above equalities we must also have that

$$P(Z'\geq a)\leq 2P(Z'\leq -a)$$

which contradicts the original equality for $X'$,

$$P(Z'\leq -a)\leq2P(Z'\geq a)$$

so the statement cannot be true for arbitrary distributions.

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