Let $X$ be a sub-gaussian random variable, and $K_i>0$ are parameters.
Its tail satisfy $P(|X|\geq t)\leq 2\text{exp}(-t^2/K_1^2)$ where $t\geq 0$.
I am confused on the case that $t=0$. If $t=0$, then the tail
$P(|X|\geq 0)\leq 2\text{exp}(0)=2$, which is not a meaningful bound. Am I making some obvious mistakes? or Is it possible to know the probability of a sub-gaussian $X$ is non-negative?
The bound $$P[|X|\geq t]\leq 2\exp(-t^2/K^2) \quad \forall t \geq 0$$ implies the tighter bound $$P[|X|\geq t]\leq \min[1, 2\exp(-t^2/K^2)] \quad \forall t \geq 0$$ However, the former bound does not have a min[,] and is often easier to work with.
We need the factor of 2 to enable random variables with bounded support to be subGaussian. Specifically if we use $X=c$ with prob 1, for some $c\neq 0$, then it can be shown this is subGaussian, but there is no $K>0$ for which $P[|X|\geq t] \leq \exp(-t^2/K^2)$ for all $t\geq 0$.
It can be shown that $X \sim Uniform[a,b]$ is subGaussian for any finite $a<b$, so we can have $P[X\geq t]$ be any value in $[0,1]$.
