Suppose $X$ is some random variable. This question may be vague , but is it always possible to find a large deviation bound of the following form that would work for ALL $s\geq 0$.
i.e $P(X \geq \mathbb{E}[X] + s) \leq \left( \right)$
More importantly, the bound $\left( \right)$ must be something that decays exponentially (either sub-exponential or sub-Gaussian).
I have seen in literature that for certain distributions, existence of such a bound is possible only for $s \geq \mathbb{E}[X]$ (especially Chernoff bounds) so I'm not sure why those bounds can't be extended for all $s\geq 0$.
Any comments on this will be appreciated!
No, it is not always possible to get such a bound. Example: $X$ is a random variable with density
$$ f(x) = \frac{2}{x^3}1(x>1). $$
You can see for any $t >1$ $$ P(X\geq t) = \frac{1}{t^2}. $$ You can choose $t=s+\mathbb{E}(X)$ for any $s\geq 0$.