I am trying to find a tail bound for $X-Y$, where $X,Y$ are i.i.d. variables following Laplace distribution with parameter $(0,\lambda)$, i.e., the pdf is $P(X=x) = \frac{1}{2\lambda}e^{\frac{-|x|}{\lambda}}$.
Denote $Z=X-Y$, it is easy to calculate the pdf, $P(Z=z)=\frac{1}{4\lambda^2}(\lambda e^{\frac{-|z|}{\lambda}} + |z|e^{\frac{-|z|}{\lambda}})$.
Then I have such kind of tail probability and hope it be small, i.e, $P(Z \geq c \lambda) = \frac{1}{4} e^{-c}(2+c) \leq \delta$. When $c\geq 2$, we have $P(Z \geq c \lambda)\leq \frac{1}{2}c e^c \leq \delta$, where $0< \delta < 1$ is an arbitrary small constant.
A good kind of solution is that "as long as $c \geq f(\delta)$, then $ P(Z \geq c \lambda)\leq \delta$ always holds". Here $f(\delta)$ is a tail value, and we need to figure out what a particular $f(\delta)$ is.
Since directly inverse $\frac{1}{2}ce^{-c} = \delta$ is not easy, I turn to find an upper bound of $\frac{1}{2}ce^{-c}$, i.e., $\frac{1}{2}ce^{-c}\leq g(c)\leq \delta$, and $g(c)$ is easy to inverse. But I still have no idea about that.
Hope this is clear.