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Markov's Inequality

If $x$ takes only Nonnegative values, then

P(x \geq a) \leq {\left\langle{x}\right\rangle{}\over a}.

To prove the theorem, write

\langle x\rangle = \int^\infty_0 xf(x)\,dx = \int^a_0 xf(x)\,dx+\int^\infty_a xf(x)\,dx.

Since $P(x)$ is a probability density, it must be $\geq 0$. We have stipulated that $x \geq 0$, so
$\displaystyle \left\langle{x}\right\rangle{}$ $\textstyle =$ $\displaystyle \int_0^a xf(x)\,dx+\int_a^\infty xf(x)\,dx$  
  $\textstyle \geq$ $\displaystyle \int^\infty_a xf(x)\,dx \geq \int^\infty_a af(x)\,dx$  
  $\textstyle =$ $\displaystyle a\int^\infty_a f(x)\,dx = aP(x \geq a),$  

Q. E. D.

© 1996-9 Eric W. Weisstein