I wonder if it is possible to estimate $\mathbb{P}(X<t)$ with the moment generating function?
This question popped up when I tried to proof this estimate $\mathbb{P}(T_a<t)\leq e^{-\frac{a^2}{2t}}$, since I know that $\mathbb{E}(e^{-uT_a})=e^{-a\sqrt{2u}}$, with $T_a$ being the hitting time of $a$ of a brownian motion.
For any $\lambda>0$ we have
$$\mathbb{P}(X<t) = \mathbb{P}(e^{\lambda \cdot X}<e^{\lambda \cdot t}) \leq \mathbb{E} \left( \frac{e^{\lambda \cdot t}}{e^{\lambda \cdot X}} \right) = e^{\lambda \cdot t} \cdot \mathbb{E}e^{-\lambda \cdot X}$$
Applying this to the hitting time $X=T_a$ yields
$$\mathbb{P}(T_a<t) \leq e^{\lambda \cdot t} \cdot e^{-a \cdot \sqrt{2\lambda}}$$
as $\mathbb{E}(e^{-\lambda T_a}) = e^{-a \sqrt{2\lambda}}$. Choose a suitable $\lambda$ to obtain the estimate
$$\mathbb{P}(T_a<t) \leq \exp \left( - \frac{a^2}{2t} \right)$$