How likely is it that the absolute value of a random variable will be greater than some specific value, say, ? 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 is a random variable with mean , and is a positive number, then

**Proof**

Let be an indicator variable such that

If we take samples from , namely , some of them will be and the rest will be . Let be the value of for the event . Therefore,

and the expected value of $I$ will be

.

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

- If , then and thus
- If , then and thus

Therefore, we have

Taking the expectation at both sides, we have

Proved

**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 3 goals today is **at most**

.

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

Comments

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 in addition to its mean.

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September 28, 2012 at 10:23 pm

Chebyshev’s Inequality « Saad Quader[…] Lvovich Chebyshev found an improvement over the upper bound suggested by Markov’s inequality. His inequality gives a tighter bound on the value of the random variable . While Markov’s […]