Suppose $x \in \mathbb{R}^+$ is the variable and $Y$ is an independent arbitrary positive real RV. It is easy to verify that $$ e^{-y^2 + xy} \leq e^{x^2/4} $$ so that function $f(x)$ defined as $$ f(x) = E_{Y}\{ e^{-Y^2 + xY} \} \leq e^{x^2/4} $$ for any possible distribution of $Y$. I was wondering with $x \rightarrow \infty$, is there a tighter upper bound of it? I think there might be something like $e^{ax}$. Can anyone give a hint on any reference about the problem?
A slightly better solution: given an $x >> 0$, function $$e^{-y^2 + xy}$$ is convex in $(0, \frac{x - \sqrt{2}}{2})$ and $(\frac{x + \sqrt{2}}{2}, \infty)$ and concave in $(\frac{x - \sqrt{2}}{2}, \frac{x + \sqrt{2}}{2})$, so $f(x)$ (x >> 0) can be bounded as $$ f(x) = \int_0^{\infty} e^{-y^2 + xy} \text{d}F(y) = (\int_0^{\frac{x - \sqrt{2}}{2}} + \int_{\frac{x - \sqrt{2}}{2}}^{\frac{x + \sqrt{2}}{2}} + \int_{\frac{x + \sqrt{2}}{2}}^{\infty}) e^{-y^2 + xy} \text{d}F(y) $$ $$ \leq e^{x^2/4 - 1/2} + e^{-E^2Y + x EY} $$