Chebychev and/or Markov

Let's call $A = \{ x \in \text{ sample space ;} \space X(x) \ge a\}$ , $a > 0$ and if $B$ is a set, let's define $I_{B} (x) = \begin{cases} 1 \space \text{ if x } \in B \\ 0 \space \text{ if x doesn't belong to } B \end {cases}$ , then $E(X) = \mu = E(X\cdot I_{A}) + E(X\cdot I_{A^{C}}) \ge E(X\cdot I_{A}) \ge E(aI_{A}) = aE(I_{A}) = aP(A) \Rightarrow \color{#D61F06}{\frac{E(X)}{a} \ge P( X \ge a)}.$ This is Markov's Inequality or general Chebychev inequality.

Hence, $P(| X - E(X) | \ge k) \leq \frac{E(|X - E(X) |^n)}{k^n}$ if $n \ge 1$ and $k > 0$ . Take $n = 2$ . The previous inequality remains $P(| X - E(X) | \ge k) \leq \frac{\text{Var(X)}}{k^2} \Rightarrow 1 - \frac{\text{Var(X)}}{k^2} \leq 1 - P(| X - E(X) | \ge k) = P(|X - E(X)| < k)$ Let's take now $k = 3\sigma$ , this implies that $1 - \frac{\text{Var(x)}}{9\sigma^2} = 1 - \frac{1}{9} = \frac{8}{9} \leq P(| X - E(X) | < 3\sigma) = P( \mu - 3\sigma < X < \mu + 3\sigma)$

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Note .- $\color{#69047E}{P( | X - \mu | \ge k \sigma) \leq \frac {1}{k^2} }$ is known as Chebichev's inequality . I still need to prove that there exists a random variable fulfilling $\frac{8}{9} = P(| X - E(X) | < 3\sigma) = P( \mu - 3\sigma < X < \mu + 3\sigma)$

Bonus.- Does there exist another similar inequality more accurate? or Can you give one example fulfilling the previous equality? or due to $Var(X)= \sigma^2 > 0$ and the proof for Markov's inequality, this inequality can not be improved?

I'm going to give almost one example.

Consider an unfair coin such that your chance to get face is 1/11, and the random variable $X$ having a Bernouilli distribution associated with this coin when it is launched once, then $\mu = E(X) = \frac{1}{11}$ and $\sigma^2 = Var(X) = \frac{10}{121} \Rightarrow \sigma = \frac{\sqrt{10}}{11}$ , then $P(\frac{1}{11} - \frac{3\sqrt{10}}{11} < X < \frac{1}{11} + \frac{3\sqrt{10}}{11} ) = P( X = 0) = \frac{10}{11} = 0.\hat{90} \approx 0.\hat{8} = \frac{8}{9}$