TITLE: Chebyshev's inequality approximation for one sided case
QUESTION [0 upvotes]: Let's suppose I have a total of $1000$ measures of some qunatity X, with $\mathbb{E}X = 100$ and $VarX = 4^2$. I'd really like to use Chebyshev's inequality to answer two questions

a maximum number of measures with value higher than $112$
a maximum number of measures with value lower than $80$

Although Chebyshev's inequality states that
$$
\mathbb{P}(|X-\mathbb{E}X| \geq a) \leq \frac{VarX}{a^2}
$$
it is clear that
$$
\mathbb{P}(X-\mathbb{E}X \geq a) + \mathbb{P}(X-\mathbb{E}X \leq -a) \leq \frac{VarX}{a^2} 
$$
So, in the first case
$$
\mathbb{P}(X \geq 112) =  \mathbb{P}(X - 100 \geq 12) \leq \frac{4^2}{12^2} \approx 11\%
$$
Thus there are at most $112$ measures with value greater than $112$
But in the second case
$$
\mathbb{P}(X \leq 80) =  \mathbb{P}(X - 100 \leq 80 - 100) = \mathbb{P}(X - 100 \leq -20)
$$
I get $a < 0$. What am I doing wrong? I really don't want to use other approaches; simply get the result from this:
$$
\mathbb{P}(X-\mathbb{E}X \geq a) + \mathbb{P}(X-\mathbb{E}X \leq -a) \leq \frac{VarX}{a^2} 
$$

REPLY [0 votes]: One recommendation is that you be more precise with your statements, as sloppy language can lead to sloppy conclusions. It's possible that all 1000 measurements have values greater than 112 (but incredibly unlikely). But just a tangential suggestion not directly related to your question.
Chebyshev's inequality is derived as a special case of Markov's inequality which states
$$P(X>t)\le \frac{E[X]}{t}$$
for any non-negative random variable. As $Y=|X-E[X]|$ is non-negative it certainly fits the criterion. So when you take advantage of the fact that $X$ has bounded variance, you obtain Chebyshev's inequality (see https://www.math.uci.edu/~rvershyn/papers/HDP-book/HDP-book.html)
As Chebyshev's inequality is not derived as a union bound (adding the two probabilities of deviating from the mean), it is not going to decompose how you want it to based on your description. Certainly you could answer the top as $P(X\ge 112)$ and $p(X\le 80)$ using a Chebyshev bound with $a=12$. However, getting sharper bounds on the probability will require using other approaches beyond Chebyshev.
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