1 What is the Markov Inequality?
The Markov inequality gives an upper bound on the probability that a non-negative random variable is greater than or equal to some positive constant. Markov’s inequality is tight in the sense that for each chosen positive constant, there exists a random variable such that the inequality is in fact an equality.
If \(X\) is a non-negative random variable, then for every \(\epsilon>0\) we have that \[ \mathbb{P}(X>\epsilon) \leq \frac{1}{\epsilon}\cdot \mathbb{E}X. \]
Proof: From the definition of expectation we have that \[ \begin{align} \mathbb{E}X & = \int _{-\infty}^\infty xf(x) \, dx \geq\int _{0}^\infty xf(x)\, dx =\int _{0}^\epsilon xf(x) \, dx +\int _{\epsilon}^\infty xf(x) \, dx \\ & \geq \int _{\epsilon}^\infty xf(x) \, dx \geq \int_{\epsilon}^\infty af(x) \, dx =\epsilon\mathbb{P}(X> \epsilon). \end{align} \] \(\square\)
More intuitively, we notice that \[ \mathbb{E}X = \mathbb{E}X \mathbb{1}_{X>\epsilon } + \mathbb{E}X \mathbb{1}_{X\leq \epsilon} \geq \mathbb{E}X \mathbb{1}_{X>\epsilon }\geq \epsilon \mathbb{E}\mathbb{1}_{X>\epsilon }=\epsilon \mathbb{P}(X>\epsilon ). \]
2 Markov Inequality for Non-Decreasing Functions
The Markov inequality for non-decreasing functions is stronger result that defines the bound by some nondecreasing non-negative function.
If \(\varphi\) is a nondecreasing non-negative function, \(X\) is a random variable (not necessarily nonnegative) and \(\varphi(a)>0\) then we have that \[ \mathbb{P}(X\geq a)\leq \frac{{\mathbb{E}(\varphi(X))}}{\varphi(a)}. \]
Corollary (Higher Moments) Immediately from above, using higher moments of \(X\) supported on values larger than \(0\) is \[ \mathbb{P}(|X|\geq a)\leq \frac{{\mathbb{E}|X|^n}}{a^n}. \]