Markov Inequality

The probability of being bigger than the mean is very low. Not many things can be above the average, or else the average will be skewed.

Inequality

For any non-negative RV $X\ge 0$, right tail probability is bounded by mean $P(X\ge a)\le \frac{E(X)}{a}$

Proof

For RV $X>0$ $E(X)={\int }_{-\int f(x)dx}^{\infty }x\ast f_X(x)dx$ as $X>0$ $={\int }_0^{\infty }x\ast f_X(x)dx={\int }_0^{a}x\ast f_X(x)dx+{\int }_a^{\infty }x\ast f_X(x)dx$ $\ge 0+a\ast {\int }_a^{\infty }{f}_X(x)dx$