Chebyshev's Inequality

28 Feb 2018

Prologue To The Chebyshev's Inequality

The Chebyshev's inequality is the variation based on the Markov inequality, it uses the second moments, the variance instead of the mean of a random variable.

Chebyshev's Inequality

Let $Z$ be any random variable with $Var\lbrack Z\rbrack>\infty$, then
$\;\;\;\;P(Z\ge E\lbrack Z\rbrack+t\;or\;Z\le E\lbrack Z\rbrack-t)\le\frac {Var\lbrack Z\rbrack}{t^2}$
$\;\;\;\;\;\;$, for any $t\ge 0$

Proof:
➀$Z\ge E\lbrack Z\rbrack+t$ or $Z\le E\lbrack Z\rbrack-t$ is exactly the same as $\left|Z-E\lbrack Z\rbrack\right|\ge t$.
➁therefore, we have it that:
$P(Z\ge E\lbrack Z\rbrack+t\;or\;Z\le E\lbrack Z\rbrack-t)$
=$P(\left|Z-E\lbrack Z\rbrack\right|\ge t)$
=$P((Z-E\lbrack Z\rbrack)^2\ge t^2)$
➂by means of the Markov inequality, below inequality just holds.
$P((Z-E\lbrack Z\rbrack)^2\ge t^2)\le\frac {(Z-E\lbrack Z\rbrack)^2}{t^2}$
$\Rightarrow P(Z\ge E\lbrack Z\rbrack+t\;or\;Z\le E\lbrack Z\rbrack-t)\le\frac {Var\lbrack Z\rbrack}{t^2}$

Remember that we have a glance over the Chebyshev's inequality in Hoeffding Inequality v.s. Chebyshev’s Inequality, it has been proved by means of integral, now, distinct topic in the series of probability bounds would like to be chained together, said by using the Markov inequality in this proof.