Lets $X$ ~ $Binomial(n,p)$. We want to find an upper bound for $P(X \geq n\alpha)$ using Chebyshev's Inequality. note that $p<\alpha<1$. It should be independent from $n$. Find upper bound for $p = 0.5 , \alpha=\frac{3}{4}$.
I tried to use Chebyshev's one-way inequality:
$P(X \geq n \alpha) = P(X – np \geq n \alpha – np) \leq \frac{\sigma ^2}{\sigma ^2+ a^2}$
and I got:
$\frac{np(1-p)}{np(1-p) + (n \alpha – n p)^2}$
for upper bound.
The problem is that is not independent from $n$. So I can't find the answer for second part without having $n$.
Is there anything else I can do using Chebyshev's inequality?
Best Answer
$$P(X \geq n \alpha) = P(X - np \geq n \alpha - np) \leq P(|X - np| \geq n \alpha - np) \leq \frac{np(1-p)}{n^2(\alpha-p)^2} = \frac{p(1-p)}{n(\alpha-p)^2} \leq \frac{p(1-p)}{(\alpha-p)^2}$$
as $n \geq 1$