# Why does the expectancy of a discrete random variable depend only on its distribution and not the r.v. directly?

In some lecture notes that I have on discrete probability, after defining expectancy, it says "the expectancy doesn't depend on the random variable directly; it depends only on its distribution", where with "distribution" the function $$W:X(\Omega) \rightarrow \mathbb{R},\ W(x)=P(X=x),$$ $X:\Omega \rightarrow \mathbb{R}$ being our random variable.

As an explanation for the above, the following line is given: $$\mathbb{E}X=\sum_{x\in X(\Omega) } x \cdot W(x)=\sum_{\omega \in \Omega} X(\omega) P(\omega).$$

Now I understand why the above holds, but I don't understand why this line entitles one to say that expectancy depends only on the distribution of a random variable. If I would change my random variable $X$ to $X'$, so that $X(\omega')\neq X'(\omega')$, for some $\omega'\in \Omega$, than by the above line, of course $\mathbb{E}X \neq \mathbb{E}X'$ .

For a better understand: Could someone provide me with an example of two different r.v.'s having the same distribution ?

-
Why don't you expand what r.d. stands for ? I, for one, have no idea what it is. – Sasha Apr 18 '12 at 18:37
(In my answer I assume that "r.d." is a typo for "r.v." and stands for "random variable"). – Henning Makholm Apr 18 '12 at 18:38
Yes, it was a typo, sorry – MyCatsHat Apr 18 '12 at 19:06

For example, let the experiment be to roll two fair 6-sided dice of different colors, and the $X$ be the random variable that gives the result of the red die and $Y$ be the random variable that gives the result of the green die.
Then $X$ and $Y$ are different random variables because they map $\Omega$ to $\{1,2,3,4,5,6\}$ in two different ways. However their distribution is the same, because for every $x\in \mathbb R$ it holds that $P(X=x)=P(Y=x)$, and therefore $\mathbb EX = \mathbb EY$.
More variable with the same distribution, but different from each other as well as from $X$ and $Y$, are $7-X$ and $((X+Y)\bmod 6) + 1$.
@nubis: But if the random variables $X$ and $Y$ were the same, then the event $X=Y$ would be the same as $X=X$, which occurs with probability $1$. And it clearly isn't. – Henning Makholm Apr 18 '12 at 18:57
I disagree. The random variable $X$ is not the same as the value obtained for a specif instance. Thus, $P(X=a,Y=a)=P(X=a,X=a)$ – nbubis Apr 18 '12 at 23:30