# Math Help - Chebyshev's inequality: "It is obvious ..."

1. ## Chebyshev's inequality: "It is obvious ..."

Having trouble internalizing the last sentence from my textbook's snippets:
For a binomially distributed random variable $X$, it can be shown, that the following holds:
$Var(X)=np(1-p)=npq$ (16)
[...]
Chebyshev: $t>0 \rightarrow P\big(E(X)-t < X < E(X)+t\big)\geq 1-\frac{D(X)}{t^2}$
[...]
From (16) it is obvious, that the probability of a complementary event, i.e. the random variable $X$ having value which differs from $E(X)$ for at least $t$, is at most $\frac{Var(X)}{t^2}$.
I would like to understand it visually (e.g.: probability distribution graph, $E(X)$ is at MAX., now look at the $E(X)\pm t$ points, ...).

BTW, I've also found a similar thread, but could not work out the proof for Theorem 4 from .pdf mr fantastic provided:
$Pr(|R-Ex(R)| \geq x) = Pr\big((R-Ex(R))^2 \geq x^2) \leq \frac{Ex\big((R-Ex(R))^2}{x^2} = \frac{Var[R]}{x^2}$. Why does the $\leq$ hold?

PS: Please, feel free to correct any grammatical mistake you find.

2. The inequality holds because you'rer splitting the integral/sum into two parts.
You throw one of those away and you obtain a lower bound on the other.
It's actually a very crude bound that pertains to all distributions hence it's not close in many situations.

3. ## Still can't internalize

I still don't intuitively know why
$Pr\Big(\text{that some random variable }X\text{ differs at least }t \text{ from }E(X)\Big) \leq \frac{Var(X)}{t^2}$.
I've put red color in original post.

It bothers me, because it says it is obvious from the fact that $Var(X)=npq$, so I should be able to intuit.
Wikipedia doesn't help the intuition (and I don't know anything about measure theory).