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How can i establish that for each $$p \in [0,1] $$ if $$X_{1},X_{2}...$$ are a coin runs to parameter p, with the propability P to cover up the confidence region $$R:=\left[\overline{X}_n-\frac{\sqrt{5}}{\sqrt{n}}, \overline{X}_n+\frac{\sqrt{5}}{\sqrt{n}}\right]$$ with Chebyshev inequality: $$P((\overline{X}_n-EX) \ge c\sigma) \le \frac{1}{c^2}$$ not less than .95!

I start with

$$=> $$ $$P((\overline{X}_n-EX) \le c\sigma) \ge 1 - \frac{1}{c^2} \ge 0.95 $$

$$P(\overline{X}_n-\frac{\sqrt{5}}{\sqrt{n}}\le p \le \overline{X}_n+\frac{\sqrt{5}}{\sqrt{n}}) = 0.95$$

$$1-\frac{1}{c^2} = 0.95$$ $$c=2\sqrt{5}$$

From here I am stuck...

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up vote 1 down vote accepted

Let $S = \frac{1}{n}\sum_{i=1}^n X_i$. Then, $E[S] = p$, $\text{var}(S) = \frac{p(1-p)}{n}$ and so by the Chebyshev inequality $$P\biggr\{p - c\sqrt{\text{var}(S)} < S < p + c\sqrt{\text{var}(S)}\biggr\} \geq 1 - \frac{1}{c^2}$$ where the right side has value $0.95$ when $c = \sqrt{20}$. But if $S$ does lie in the interval $$\left(p-\sqrt{20\cdot\text{var}(S)}, p+\sqrt{20\cdot\text{var}(S)}\right),$$ then $p$ must be in the interval $$\left(S - \sqrt{20\cdot\text{var}(S)}, S+\sqrt{20\cdot\text{var}(S)}\right),$$ that is, this is a $95\%$ confidence interval for the unknown parameter $p$. Since $\text{var}(S)$ has maximum value $\frac{1}{4n}$ when $p = \frac{1}{2}$, we have that $$\left(S - \sqrt{20\cdot\text{var}(S)}, S+\sqrt{20\cdot\text{var}(S)}\right) \subset \left(S - \sqrt{\frac{5}{n}}, S+\sqrt{\frac{5}{n}}\right),$$ and this superset is also a $95\%$ confidence interval for $p$.

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