How likely is it that the absolute value of a random variable $X$ will be greater than some specific value, say, $a$? The Russian mathematician Andrey Andreyevich Markov proved a simple, yet nice, result which enables us to answer the above question.

Markov’s Inequality states that if $X$ is a random variable with mean $\mu$, and $a>0$ is a positive number, then

$Pr[|X| \geq a] \leq \frac{|\mu|}{a}$

Proof

Let $I$ be an indicator variable such that

$I = \left\{\begin{array}{cc}1 & \text{if } |X| \geq a \\ 0 & \text{otherwise}\end{array}\right.$

If we take $n$ samples from $X$, namely $X_1, X_2, \cdots, X_n$, some of them will be $\geq a$ and the rest will be $< a$. Let $I_1, I_2,\cdots, I_n$ be the value of $I$ for the event $X=X_i, 1\leq i\leq n$. Therefore,

$\begin{array}{ccl}\sum_{i=1}^{n}{I_i}&=&0\times \text{Number of cases where }I_i=0\\&&+1\times \text{Number of cases where } I_i=1\\&=&\text{Number of cases where } I_i=1\end{array}$

and the expected value of $I$ will be

$\begin{array}{ccl} E[I] & = & \frac{\sum_{i=1}^n{I_i}}{n}\\ & = & \frac{\text{Number of cases where } I_i=1}{n}\\ & = & \Pr\left[|X| \geq a\right]\end{array}$.

Now, looking at $I$, we can make the following observations:

• If $|X| \geq a$, then $I = 1$ and thus $aI = a \leq |X|$
• If $|X| < a$, then $I = 0$ and thus $aI = 0 \leq |X|$

Therefore, we have

$aI \leq |X|$

Taking the expectation at both sides, we have

$\begin{array}{cl} & E[aI] \leq E[|X|]\\\implies & aE[I] \leq |\mu|\\\implies & a\Pr\left[|X| \geq a\right] \leq |\mu|\\\implies &\Pr\left[|X| \geq a\right] \leq \frac{|\mu|}{a}\end{array}$

Proved $\square$

What is the use?

We can use this inequality to derive an upper bound to a random variable.

For example, suppose that Lionel Messi scores 1 goal in every two matches he plays, on average. (Therefore, he scores 0.5 goals per match, on average). Today he is playing against your team, and you hope that he does not score many goals. You ask: what is the probability that he will not score a hat-trick today? Using Markov’s inequality, we can say, the probability that he scores $\geq$ 3 goals today is at most

$\frac{0.5}{3} = \frac{1}{6} \approx 0.16$.

Therefore, there is close to 84% chance that he will not score three goals today.

Markov’s inequality, while simple, gives rather loose bound. A much tighter bound can be found using Chebyshev’s inequality if we know the variance of $X$ in addition to its mean.