# If X ≤ Y, then E(X) ≤ E(Y)

1. Jul 8, 2009

### kingwinner

There is a theorem that says:
"Let X and Y be random variables. If X ≤ Y, then E(X) ≤ E(Y)."

But I don't really understand the meaning of "X ≤ Y". What does it mean?
For example, if X takes on the values 0,1,2,3, and Y takes on the values -1,2,5. Is X ≤ Y??

Any help is appreciated!

2. Jul 9, 2009

### Pere Callahan

The theorem assumes that X and Y are defined on the same probability space $\Omega$. $X\leq Y$ means $X(\omega)\leq Y(\omega),\quad \forall\omega\in\Omega$. Actually, it would be enough to have $X(\omega)\leq Y(\omega)$ for P-almost all $\omega\in\Omega$, where P is the probability measure on [itex\Omega][/itex].

3. Jul 9, 2009

### kingwinner

Thanks! Now I understand what X ≤ Y means in the theorem.

Consider a separate problem. How about X ≤ Y in the context of finding P(X ≤ Y)? In this case, do X any Y have to be defined as random variables with X($\omega$) < Y($\omega$) for ALL $\omega\in\Omega$?

Last edited: Jul 9, 2009
4. Jul 9, 2009

### Pere Callahan

No. In order to compute P(X ≤ Y), you have to take the probability of all omega such that X(omega) ≤ Y(omega). There might be other omega which do not satisfy this inequality but then they don't contribute to P(X ≤ Y).

$$P(X\leq Y)=P\left(\omega\in\Omega: X(\omega)\leq Y(\omega)\right)$$

You may have noticed that the probability measure P has strictly speaking two different meanings here. On the right hand side it is a function which takes as argument a subset of $\Omega$. While on the left hand side...well...it is only a shorthand for the right side

5. Jul 10, 2009

### kingwinner

6. Jul 11, 2009

### kingwinner

Two follow-up questions:

1) For P(X ≤ Y), do X and Y have to be defined on the SAME sample space $\Omega$?

2) In order statistics, when they say X(1)≤X(2)≤...≤X(n), they actually mean X(1)(ω)≤X(2)(ω)≤...≤X(n)(ω) for each and for all ω E $\Omega$ (or almost all), right??

7. Aug 8, 2009

### Focus

They need to be on the same probability space. Having the same sample space is not enough.
If they don't say a.e. or a.s. you can assume they mean for all $\omega \in \Omega$.