inequality exponential series with adjusted denominator The following problem might have something to do with the coefficients of the moment generating function. But I do not see how to prove it, nor do I have a counterexample.
Given that the positive random variable $X$ is first order stochastically dominated by $Y$ $(0\preceq X\preceq Y)$, we know
$$
E[\exp(-X)]\geq E[\exp(-Y)]
$$
In other words,
$$
\sum_{i=0}^\infty \frac{E[(-X)^i]}{i!} \geq \sum_{i=0}^\infty \frac{E[(-Y)^i]}{i!}
$$
Is the following true for any $k$ and $0\preceq X\preceq Y$
$$\sum_{i=0}^\infty \frac{E[(-X)^i]}{(k+i)!} \geq \sum_{i=0}^\infty \frac{E[(-Y)^i]}{(k+i)!}$$
One may assume that $k$ is even if needed.


Yes, for every $k\geqslant0$.

Consider $u_k(x)=(k-1)!\cdot\displaystyle\sum\limits_{i\geqslant0}\frac{(-x)^i}{(k+i)!}$, then it suffices to prove that the function $x\mapsto u_k(x)$ is nonincreasing.
To do so, note that $\displaystyle\frac{(k-1)!}{(k+i)!}=\int_0^1(1-t)^{k-1}\frac{t^i}{i!}\,\mathrm dt$ for every $i\geqslant0$ hence 
$$
u_k(x)=\sum\limits_{i\geqslant0}(-x)^i\int_0^1(1-t)^{k-1}\frac{t^i}{i!}\,\mathrm dt=\int_0^1(1-t)^{k-1}e^{-tx}dt.
$$
Since $x\mapsto e^{-tx}$ is decreasing for every $t$ in $(0,1)$, this proves the claim.
