Wikipedia says:
$$P[X-\mu \geq a] \leq \frac{\sigma^2}{\sigma^2+a^2}$$ for $a > 0$
and
$$P[X-\mu\geq a] \geq 1- \frac{\sigma^2}{\sigma^2+a^2}$$ for $a <0$.
Is the second inequality (for $a<0$) correct?
If yes, do you know a proof? Or do you know a paper where it is referenced, as I don't trust this Wikipedia site (it's just a stub).
I'm not sure whether it is the same or whether there's a contradiction, but here is another paper where These inequalities are listed.
This paper says for $a>0$:
$$P[X \geq \mu + a] \leq \frac{\sigma^2}{\sigma^2+a^2}$$
$$P[X \leq \mu - a] \leq \frac{\sigma^2}{\sigma^2+a^2}$$
When I take the second formula of the paper and try to transform it into the Wikipedia equation, I get a Problem for discrete functions, as:
$$Pr[X-\mu \geq a]=Pr[X \geq \mu + a]=1-Pr[X \leq \mu + a -1] =1-Pr[X \leq \mu -(-a + 1)] \geq 1 - \frac{\sigma^2}{\sigma^2+(-a+1)^2}$$
Is my Derivation or one of These equations wrong?
Thank you for helping me!
The second inequality can be deduced from the first by replacing $X$ with $-X$.
Note that
$$P[X-\mu\geq a] \geq 1- \frac{\sigma^2}{\sigma^2+a^2}$$
for $a <0$ implies that
$$P[X\geq \mu + a] \geq 1- \frac{\sigma^2}{\sigma^2+a^2}$$
implies that (since the distribution is continuous, with no discrete part)
$$P[X\leq \mu + a] \leq \frac{\sigma^2}{\sigma^2+a^2}$$
implies that for $b = - a$ (and so now $ b > 0 $),
$$P[X\leq \mu - b] \leq \frac{\sigma^2}{\sigma^2+b^2}$$