I am studying for an exam on stochastic order. I am struggling with a question on functional invariance of exponential order ($\leq_{\mathrm{e}}$), where for r.v.s $X$ and $Y$, $$X \leq_{\mathrm{e}} Y \stackrel{\mathrm{def}}{\iff}m_X(t) \leq m_Y(t), \qquad t>0,$$ where $m_X(t)$ is the moment generating function of $X$.
I need to prove that for any nondecreasing function $f(\cdot)$, $$ X \leq_{\mathrm{e}} Y \implies f(X) \leq_{\mathrm{e}} f(Y),$$ or give a counterexample.
My attempt was this:
Let $f:\mathbb{R} \to \mathbb{R}$ be a nondecreasing function, i.e. $$ \forall\, x,y \in \mathbb{R},\; x \leq y \implies f(x) \leq f(y). $$ Let $X$ and $Y$ be random variables such that $X \leq_{\mathrm{e}} Y$.
\begin{align} \operatorname{\mathbb{E}}[e^{tX}] \leq \operatorname{\mathbb{E}}[e^{tY}] & \implies \operatorname{\mathbb{E}}[e^{tf(X)}] \leq \operatorname{\mathbb{E}}[e^{tf(Y)}] \tag{$\ast$}\\ &\implies m_{f(X)}(t) \leq m_{f(Y)}(t)\\ &\implies f(X) \leq_{\mathrm{e}}f(Y). \end{align}
However I am not convinced that the implication in $(\ast)$ is actually true. The reasoning looks too ‘simple’. I am not even confident the proof holds at all. Can you help me on this?