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Does Chernoff's bound work with negative values? In other words, it is defined for a SYMMETRIC random variable $Y$ with an MGF of $M(t)$ that exists and $a>0$, $t>0$ such that:

$$ P(Y\geq a) \leq e^{-at}M(t) $$

But suppose $a<0$, would it work as well? If not, is there an alterate version to this? Thanks!

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    If your variable is symmetric, then $M(t)\geq 1$ for all $t>0$, so Chernoff's bound becomes useless for $a<0$. – 2017-02-10

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Yes, but you can't get rid of the condition $t>0$. Since exponential function is positive and the sets $\{X\geq a\}$ and $\{e^{tX}\geq e^{ta}\}$ are equavalent for any $a$, you can apply Markov's inequality $\mathbb{P}(X \geq a) = \mathbb{P}(e^{t\cdot X} \geq e^{t\cdot a})\leq\mathbb{E}\left [e^{t\cdot X}\right]e^{-t\cdot a}.$