Markov's Inequality

Definition

If $X$ is a non-negative random variable and $a > 0$, then the probability that $X$ is at least $a$ is at most the expectation of $X$ divided by $a.$

Mathematically

$$P(X\geq a) \leq \frac{\mathbb{E}[X]}{a}.$$

Proof

$$\begin{align*} \mathbb{E}[X]:&=\int_{-\infin}^{\infty}x f_X(x) dx\\ \end{align*}$$

since $X$ is non-negative

$$\begin{align*} \mathbb{E}[X]&=\int_{0}^{\infty}x f_X(x) dx\\ \end{align*}$$

for any $a>0$

$$\begin{align*} \mathbb{E}[X]&\geq \int_{a}^{\infty}x f_X(x) dx\\ \end{align*}$$

since $x>a$ in the integrated region

$$\begin{align*} \mathbb{E}[X]&\geq \int_{a}^{\infty}x f_X(x) dx\geq \int_{a}^{\infty}a f_X(x) dx\\ &\geq a\int_{a}^{\infty} f_X(x) dx\\ &\geq aP(X\geq a)\\ \implies P(X\geq a) &\leq \frac{\mathbb{E}[X]}{a}. \hspace{15px} \blacksquare \end{align*}$$

Note: We will use Markov's Inequality to prove Chebyshev's Inequality.

Example

Let $X$ is a random variable representing the income of an individual in the population. Since $X$ is income, it can not be negative.

Markov's inequality can be written as

$$\begin{align*} P(X\geq m \mathbb{E}[X]) &\leq \frac{\mathbb{E}[X]}{m \mathbb{E}[X]}\\ &\leq \frac{1}{m}. \end{align*}$$

Let $m=5$

$$\begin{align*} P(X\geq 5 \mathbb{E}[X]) &\leq 20\%\\ \end{align*}$$

This inequality states that the proportion of the population with an income more than $5$ times the average is at most $20\%$. In other words, no more than $20\%$ of the population can earn more than $5$ times the average income.