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# Chebychev’s Inequality

Let $$X$$ be a random variable and let $$g(x)>0$$. Then $$\forall r > 0$$,

$P(g(x) \geq r) \leq \frac{\mathbb{E}_x[g(x)]}{r}$
Proof \begin{align*} \mathbb{E}_x[g(x)] &= \int_x g(x) p(x) dx\\ &\geq \int_{x: g(x) \geq r} g(x) p(x) dx\\ &\geq r \int_{x: g(x) \geq r} p(x) dx\\ &= r P(g(x) \geq r) \end{align*}