# If all moment of X are greater than all moment of Y, can we said something about their probability?

Consider a continuous probability distribution $$f$$ and two random variables $$X, Y$$ both are greater than equal to $$0$$ and they are not identical random variables. Let's say one can show that $$E[X]^k \geq E[Y]^k$$ for all $$k \geq 0$$ and suppose their moment generating function exists.

Can I conclude that $$X \geq Y$$ almost surely? This is my intuition

I can easily get $$Pr[X^k \geq Y^k] \neq 0$$ otherwise we get the contradiction. Or at least I want to find the lower bound of $$Pr[X^k \geq Y^k]$$.

Any comment is welcome and thanks for your help.

• Certainly not, because the moments only depend on the distribution while $X\ge Y$ is about pointwise values. For example $X(\omega_1)=1$, $X(\omega_2)=2$, with $P(\{\omega_j\})=1/2$, and then switch the two values to obtain $Y$. Aug 9, 2022 at 19:36
• oh, ok, thanks. let me think a little bit Aug 9, 2022 at 19:37
• The question would be less trivial if we ask whether $X \ge_{st} Y$ (stochastic order). But my intuition is that the answer will still be negative, even the distribution of $X$ and $Y$ are determined by their moments. Aug 9, 2022 at 19:44
• Let $M>1$. Suppose $Y$ takes value $1$ with probability $1$, and $X$ takes value $M$ with probability $1/M$ and value $0$ with probability $1-1/M$. Then $EY^k = 1 \leq M^{k-1} = EX^k$ for all $k\geq 1$. However, $P(X>Y)=1/M$, which you can make arbitrarily close to $0$ by taking $M$ large. This is a discrete example, but you can easily arrange a small perturbation to get an example where the random variables have continuous distributions, if that's what you need for some reason. Aug 9, 2022 at 20:55
• Assume that $X$ and $Y$ have the same distribution, that they are not concentrated on one point and that they are independent (for each distribution its always possible to construct such r.v.s). Then all moments are even equal but never $X \geq Y$ a.s. Aug 9, 2022 at 22:53