# Markov’s inequality and Chebyshev’s inequality

## Other

### Markov’s inequality and Chebyshev’s inequality

#### Lemma 1

$$E[I_{X\ge a}]=P(X\ge a)$$

Consider the indicator function.

$$I_{X\ge a}$$

This is equal to $$0$$ if $$X$$ is below $$a$$ and $$1$$ otherwise.

We can take expectations of this.

$$E[I_{X\ge a}]=P(X\ge a).1+P(X<a).0=P(X\ge a)$$

$$E[I_{X\ge a}]=P(X\ge a)$$

#### Lemma 2

$$aI_{X\ge a}\le X$$

While $$X$$ is below $$a$$ the left side is equal to $$0$$, which holds.

While $$X$$ is equal to $$a$$ the left side is equal to $$X$$, which holds.

While $$X$$ is above $$a$$ the left side is equal to $$a$$, which holds.

#### Markov’s inequality

$$P(X\ge a)\le \dfrac{\mu }{a}$$

From above:

$$aI_{X\ge a}\le X$$

We can take expectations of both sides:

$$E[aI_{X\ge a}]\le E[X]$$

$$aP(X\ge a)\le E[X]$$

$$P(X\ge a)\le \dfrac{\mu }{a}$$

#### Chebyshev’s inequality

We know from Markov’s inequality that:

$$P(X\ge a)\le \dfrac{\mu }{a}$$

Let’s take the variable $$X$$ to be $$(X-\mu )^2$$

$$P((X-\mu )^2\ge a)\le \dfrac{E[(X-\mu )^2]}{a}$$

$$P((X-\mu )^2\ge a)\le \dfrac{\sigma^2}{a}$$

$$P(|X-\mu | \ge \sqrt{a})\le \dfrac{\sigma^2}{a}$$

Take $$a$$ to be a multiple $$k^2$$ of the variance $$\sigma^2$$.

$$a=k^2\sigma^2$$

$$P(|X-\mu | \ge k\sigma )\le \dfrac{\sigma^2}{k^2\sigma^2}$$

$$P(|X-\mu | \ge k\sigma )\le \dfrac{1}{k^2}$$