Comparison of tail behaviour of two (bounded) random variables given their moments Given: two positive scalar (bounded) random variables $X$ and $Y$ with the following conditions to hold:
$$ E(X)=E(Y),\ E(X^k)\ge E(Y^k), \forall k>1$$
How to show (whether it is possible to show) that for cumulative distribution functions $F_X,F_Y$:
$$\exists a_0:F_X(a)\le F_Y(a), \forall a>a_0$$
If this statement is not correct, what additional assumptions are required.
Hypothetically this condition is equivalent to (assuming pdfs exist)
$$\int_0^{a}[f_Y(z)-f_X(z)]dz \ge 0,\forall a\ge a_0$$
or, using characteristic functions:
$$F_Y(a)-F_X(a)=$$ $$=\int_0^{a} \frac{1}{2\pi} \int_{R} e^{-itz}\sum_{k=0}^\infty \frac{(it)^k m_k^Y}{k!}dtdz-\int_0^{a} \frac{1}{2\pi} \int_{R} e^{-itz}\sum_{k=0}^\infty \frac{(it)^k m_k^X}{k!}dtdz$$
$$F_Y(a)-F_X(a)=\frac{1}{2\pi}\sum_{k=0}^\infty \frac{(m_k^Y-m_k^X)}{k!} \int_0^{a}  \int_{R} e^{-itz}(it)^kdtdz$$
where $m_k^Y-m_k^X\le 0, \forall k$. 
I am stuck here. 
Any help would be highly appreciated.
 A: First, here is a counterexample to a simpler statement with $a_0$ fixed at $0$: Let $Y$ be the constant random variable $2$. Let $X$ be $1$ with probability $1/2$, and $3$ with probability $1/2$. Then $F_Y(2) = 1 \gt F_X(2) = 1/2$ and $E[X]=E[Y]=2$, but for all $k \gt 1$, $E[X^k] \gt E[Y^k]$ by Jensen's inequality. We can use this create a counterexample for all $a_0$ simultaneously.
Let $Z$ be an unbounded random variable so that all moments exist and $Z$ is supported on powers of $4$, and let $Z$ be independent of $X$ and $Y$. Consider $X'=XZ$ and $Y'=YZ$. If $P(Z = 4^n) \gt 0$ then $P(Y' \le 2 \times 4^n) \gt P(X' \le 2 \times 4^n)$ because $P(Y' \le 2 \times 4^n) = P(Z \le 4^n)$ while $P(X' \le 2 \times 4^n) = P(Z \lt 4^n) + \frac{1}{2}P(Z=4^n)$. Further, $E[(XZ)^k] = E[X^k]E[Z^k] \gt E[Y^k]E[Z^k] = E[(YZ)^k]$. So, although all higher ($\gt 1$) moments of $X'$ are greater than the corresponding moments of $Y'$ and the expected values are the same, there is no $a_0$ so that for all $a\gt a_0, F_{X'}(a) \lt F_{Y'}(a)$. The distribution functions switch infinitely often so they are not asymptotically comparable. 
