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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.

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    $\begingroup$ The new version where you assume $X$ and $Y$ are bounded is trivial. $\endgroup$ Commented Dec 29, 2016 at 23:03
  • $\begingroup$ Thank you for your prompt replies. Analysing your counterexample, it seems like there are two sufficient conditions for the relationship to hold: 1. finite RVs; 2. RVs with a particular tail behaviour, say constant signs of the second derivative of pdf functions. However, I have difficulties to prove it. Would it be possible to point me in the right direction? Thanks! $\endgroup$
    – Puzzled
    Commented Dec 30, 2016 at 15:58

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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.

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  • $\begingroup$ OK, got it. $P(Y'\le 4^n)<P(X'\le 4^n)$, $P(Y'\le 2\times 4^n)>P(X'\le 2\times 4^n)$, $P(Y'\le 3\times 4^n) = P(X'\le 3\times 4^n)$. For this counterexample to hold we need to have unbounded random variable. I will add a word bounded in the original post. $\endgroup$
    – Puzzled
    Commented Dec 29, 2016 at 22:16

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