Extension of Chebyshev Inequality

Theorem1

Let $X \in \mathcal{R}$ be a random variable with expected value $\mu$ and variable $\sigma^2$ .

Let $f(X)$ be a probability distribution function.

Let $m(k)$ be $m(k)=\sup_{|x-\mu|\geq k\sigma} f(x)$ .

Then, for any r $r\in (\frac{1}{2},\infty)$ and k>0,

$Pr(|x-\mu|\geq k\sigma)\leq\frac{1}{k^2}{(\frac{2\sigma k^3}{2r-1}m(k) )}^{\frac{1}{r+1}}$

$r\rightarrow \infty$ , inequality in continuous case coincides with Chebyshev inequality.

Theorem 2

Let $X\in\mathcal{Z}$ be a random variable, $p(X)$ be a probability mass function.

Let m be $m=\sup_{\{\lfloor {x-\mu} \rfloor\geq k\sigma\}\cup\{\lceil {x-\mu} \rceil\leq -k\sigma\}} p(x)$ .

Let $\alpha=1+\frac{1}{12\sigma^2}$

Then, for any r $r\in (\frac{1}{2},\infty)$ and k>0,

$Pr(|x-\mu|\geq k\sigma)\leq\frac{\alpha}{k^2}{(\frac{2\sigma k^3}{(2r-1)\alpha}m(k) )}^{\frac{1}{r+1}}$

Proof of Theorem 1

Consider the function

$F(k)=\int_{|x-\mu|\geq k\sigma}dx \frac{f(x)}{Pr_k}\frac{1}{|x-\mu|^{2r}}$ .

Here,

$Pr_k=Pr(|x-\mu|\geq k\sigma)$

By definiton of m(k), $F(k)\leq 2\frac{m(k)}{Pr_k}\int_{x-\mu\geq k\sigma} dx(x-\mu)^{-2r}=\frac{2m(k)}{(2r-1)Pr_k {(k\sigma)}^{2r-1}}$ eq(1)

By deforming

$F(k)=\int_{|x-\mu|\geq k\sigma}dx\frac{f(x)}{Pr_k}\frac{1}{|x-\mu|^{2r}}=\int_{(x-\mu)\geq k\sigma}dx\frac{f(x)+f(-x)}{Pr_k}\frac{1}{{(x-\mu)}^{2r}}$ eq(2)

By definition of Pr, $\int_{(x-\mu)\geq k\sigma}\frac{f(x) + f(-x)}{Pr_k}=1$ , and

$\frac{1}{x^r}$ is convex function for $x\geq 0$ , and $\int_{(x-\mu) \geq k\sigma }dx (x-\mu)^2 \frac{f(x)+f(-x)}{Pr_k}\geq {(k\sigma)}^2$ .

We can apply Jesen's inequality for eq(2).

$\int_{(x-\mu)\geq k\sigma}dx\frac{f(x)+f(-x)}{Pr_k}\frac{1}{(x-\mu)^{2r}}\geq\frac{1}{s_k^r}$ .

Here,

$s_k=\frac{1}{Pr_k}\int_{(x-\mu)\geq k\sigma}dx (f(x) + f(-x) )(x-\mu)^2 dx=\frac{1}{Pr_k}\int_{|x-\mu|\geq k\sigma}dx f(x){(x-\mu)}^2 dx\leq\frac{\sigma^2}{Pr_k}$