Chebyshev's Inequality and Binomial Distribution

Question

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?

Answer 1

$$P(X \geq n \alpha) = P(X - np \geq n \alpha - np) \leq P(|X - np| \geq n \alpha - np) \leq \frac{np(1-p)}{n^2(\alpha-p)^2} = \frac{p(1-p)}{n(\alpha-p)^2} \leq \frac{p(1-p)}{(\alpha-p)^2}$$

as $n \geq 1$

Answer 2

This is not exactly an answer to the question, but I feel tempted to apply Markov's inequality: if $X$ is a nonnegative random variable with finite expectation, then $$ P(X \geq a) \leq \frac{E(X)}{a} $$ for any positive number $a$. If $X$ is a binomial random variable with parameters $n$ and $p$, and $\alpha > 0$, we obtain $$ P(X \geq n\alpha) \leq \frac{np}{n\alpha} = \frac{p}{\alpha}. $$