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.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.