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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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    Why don't you expand what r.d. stands for ? I, for one, have no idea what it is.2012-04-18
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    (In my answer I assume that "r.d." is a typo for "r.v." and stands for "random variable").2012-04-18
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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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    I'd venture to say that if the distribution of to r.v.'s is same, they **are** the same.2012-04-18
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    @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.2012-04-18
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    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)$2012-04-18
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    @nbubis: You disagree that if two things "are the same", then one can be replaced by the other in an expression without changing the result?2012-04-19
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    @ Nope. I'm just trying to stress the difference between two random variables (which are the same if their distributions are) and two random events generated by the same random variables which are of course not the same.2012-04-19
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    @nbubis: What you're "just trying to stress" is _wrong_. A random variables _is not_ the same as its distribution, _because_ it is not valid to replace one random variable in an expression with a different random variable of the same distribution. Things that cannot be exchanged with each other are not the same! (Otherwise, you are using "the same" in a really, confusingly nonstandard way).2012-04-19
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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