[Math] Inverse Markov’s Inequality

probability

Let Z be a random variable that takes values in [0,1]. Assume that $\mathbb{E}[Z]=\mu$. Then, for any $a\in(0,1)$, $\mathbb{P}[Z>1-a]\geq\frac{\mu-(1-a)}{a}$

I tried:

$\mathbb{P}[Z>1-a]=1-\mathbb{P}[Z\leq1-a]=1-\mathbb{P}[\mu-(1-a)\leq\mu-Z]$

I thought that maybe Markov's inequality could help me here, but obviously the problem now is that i don't know whether $\mu -(1-a)\\$ or $\mu-Z$ are positive.

Best Answer

There is a simple way. The usual trick for this type of question is to use indicator function.

Given the assumptions, We claim that the following inequality is true $$ aI(Z>1-a)\geq Z-(1-a).$$

Then we discuss in case that (i) $Z>1-a$ and (ii) $Z\leq 1-a$.

For (i) since $Z>1-a$, we have $a>Z-1+a$ which is true;

For (ii), the conclusion is obvious.

Then you take expectation for the inequality, you have $$ aP(Z>1-a)\geq \mu-(1-a).$$ Note that expectation of indicator is just probability. Hope this help :)

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