# How to prove the tightness of Markov's bound?

Show that Markov's inequality is as tight as it possible. Given a positive integer $k$, describe a random variable $X$ that assumes only non-negative values:

$$\Pr[X \geq k E[X] ] = 1/k.$$

Using Markov's bound, we can show at most $1/k$. But how to show equality?! My question to be exact what is the idea to prove the tightness of this bound!

Consider a random variable $X$ which takes the value $1$ with probability $1/k$ and $0$ with probability $1-1/k$.
• To @Davide : Maybe you meant "$X$ which takes the value $1$ with probability $1/k$"? – Arundo Donax Sep 28 '17 at 17:00