How to prove Markov's inequality $P_X(X\geq t) \leq \frac{\mathbb{E}[X]}{t}$?

Question

As the subject states, how can Markov's inequality $P_X(X\geq t) \leq \frac{\mathbb{E}[X]}{t}$ be proven?

Is the proof distribution-dependent or there is a general way to prove it?

Answer 1

Note you need to assume the random variable $X$ is never negative, otherwise the inequality fails. Perhaps the best way to see the proof is intuitively: When you take expectation, you will be adding up contributions of the form $xP(X=x)$ over all $x$. If $x \geq t$, then these contributions are greater than or equal to $t P(X = x)$. So one contribution to the expectation of $X$ is at least as great as $t P(X \geq t)$. (and there's no negative values to cancel out any of this positive contribution).