Timeline for If all moment of X are greater than all moment of Y, can we said something about their probability?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Aug 9, 2022 at 22:53 | comment | added | Dieter Kadelka | 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 20:55 | comment | added | James Martin | 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 19:44 | comment | added | Christophe Leuridan | 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 19:37 | comment | added | En-Jui Kuo | oh, ok, thanks. let me think a little bit | |
| Aug 9, 2022 at 19:36 | comment | added | Christian Remling | 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 19:29 | history | asked | En-Jui Kuo | CC BY-SA 4.0 |