If all moments of a non-negative random variable X are larger than those of Y, is P(X>x) larger than P(Y>x)? $X$ and $Y$ are two non-negative continuous random variables. The moments of $X$ are $\mu_i$ while that of $Y$ are $\nu_i$.
We know that $\mu_1=\nu_1$ and $\mu_i \ge \nu_i$ for $i=2,3,\ldots$
Can one then show that $P(X>x) \ge P( Y>x)$ for $x\ge \mu_1$?
We do know that
$$ \mu_i=i\int_{0}^{\infty} x^{i-1} P(X>x) dx \ge  i\int_{0}^{\infty} x^{i-1} P(Y>y) dx = \nu_i$$ for all $i$.
It holds for the special case where $\nu_i=0$ for $i=2,3,\ldots$

The higher moments say relatively little about the behaviour of the distribution at small values. The tails have a lot of influence on a distribution.
You can see this in particular with distributions that have infinite higher moments.
For example compare a Frechet distribution with location = 1, scale = 1 and shape = 1.5 (this has finite mean but infinite variance) to an exponential distribution with equal mean.

