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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 ?

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    Yes, it was a typo, sorry2012-04-18

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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 $((X+Y)\bmod 6) + 1$.

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    Throw a fair die $n$ times and let $X$ be the number of even numbers shown on top (so $1$ or $0$ even numbers if you throw it once). Toss a fair coin $n$ times and let $Y$ be the number of heads shown on top (so $1$ or $0$ heads if you toss it once). These are different random variables (they even have different units) but as counts they have the same distribution.2015-04-02