The extension of Chebyshev inequality 2

Theorem

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

Let $f(X)$ be a probability distribution function and $f(x)\leq f(\mu)$ hold for all $|x-\mu|\geq k\sigma$ .

Let $f(x)x\rightarrow 0$ as $|x|\rightarrow \infty$ .

Let $m(K)$ be $m(k)=\max(f(\mu+k\sigma), f(\mu-k\sigma) )$ .

Then, for any $r\geq \frac{1}{2}$ ,

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

For $0\leq r\leq \frac{1}{2}$ , if k satisfies the condition $\log(\frac{f(\mu)}{m(k)}) + 2r-1\geq 0$ , the same inequality holds.

The simple examples are $r=0$ , $r=\frac{1}{2}$ , $r=1$ .

We have

$Pr(|x-\mu|\geq k\sigma)\leq \frac{k\sigma \{f(\mu+k\sigma) + f(\mu-k\sigma)\}}{\log(\frac{f(\mu)}{em(k)})}$ .

$Pr(|x-\mu|\geq k\sigma)\leq [{\frac{\sigma \{f(\mu+k\sigma) + f(\mu-k\sigma)\}}{\log(\frac{f(\mu)}{m(k)})}]}^{\frac{2}{3}}$ .

$Pr(|x-\mu|\geq k\sigma)\leq [{\frac{\sigma \{f(\mu+k\sigma) + f(\mu-k\sigma)\}}{k\log(\frac{ef(\mu)}{m(k)})}]}^{\frac{1}{2}}$ .

Proof of Theorem

By assumption of log-concavity, the inequality $f(x+\lambda(y-x) )\geq f(x){(\frac{f(y)}{f(x)})}^{\lambda}$ follows on $0\leq \lambda\leq 1$ .

We subtract $f(x)$ , devide by $\lambda$ for this inequality and the limit $\lambda\rightarrow 0$ , we have

$(y-x)\frac{df(x)}{dx}\geq f(x)(\log(f(y) )-\log(f(x) )$ .

We put $y=\mu$ , then we have

$(\mu-x)\frac{df(x)}{dx}\geq f(x)(\log(f(\mu) )-\log(f(x) )$ eq(1)

By this inequality, $\frac{df(x)}{dx}\leq 0$ holds for x which satisfies $f(x)\leq f(\mu)$ and $x-\mu\geq 0$ . So $f(x)$ monotonically decreases on $x-\mu\geq 0$ .

In the same way, we find $f(x)$ monotonically increases on $x-\mu\leq 0$ .

Then, we have $\sup_{|x-\mu|\geq k\sigma} f(x)=\max(f(\mu+k\sigma), f(\mu-k\sigma) )$ .

We multiply $|x-\mu|^{-2r}$ and integrate eq(1) on $|x-\mu|\geq k\sigma$ , and apply integration by parts for LHS,

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

For RHS, we have

$\int_{|x-\mu|\geq k\sigma}dx |x-\mu|^{-2r}f(x)(\log(f(\mu) )-\log(f(x) )\geq (\log(f(\mu) )-\log(m(k) )\int_{x-\mu\geq k\sigma}dx (f(x) + f(-x) ){(x-\mu)}^{-2r}$

From these results, we have

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

Here, we put $Pr_k=Pr(|x-\mu|\geq k\sigma)$ .

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 the RHS term of eq(2).

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

We define as $s_k=\int_{(x-\mu)\geq k\sigma}dx \frac{f(x)+f(-x)}{Pr_k}(x-\mu)^2 dx$ .

$s_k$ satisfies

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

Then we have

$(k\sigma)^{1-2r}\{f(\mu+k\sigma) + f(\mu-k\sigma)\} \geq (\log(f(\mu) )-\log(m(k) +2r-1) \frac{Pr_k^{r+1}}{\sigma^{2r}}$

If the condition $\log(\frac{f(\mu)}{m(k)}) + 2r-1\geq 0$ holds, deforming this inequality, we have

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

趣味の研究

趣味の数学を公開しています。初めての方はaboutをご覧ください。

The extension of Chebyshev inequality 2