Given a probability space $(\Omega,\Sigma,P)$. A random variable $X$ is a function mapping $\Omega \to \mathbf R$. On the other hand, we know that sum of two random variables is still a random variable.

Now, consider two random variables $X_1, X_2$ which are independent and identical. For a $\omega \in \Omega$, $(X_1+X_2)(\omega)=X_1(\omega)+X_2(\omega)$. Since $X_1$ and $X_2$ are identical, the equation would become $(X_1+X_2)(\omega)=2X_1(\omega)$. This result is obviously wrong! Can any one point out the mistake in the reasoning?

This problem appears when I am studying sum of i.i.d. random variables. For a sequence of i.i.d. random variables {$A_i$}. In the textbook, there is an event {$\omega\in\Omega\mid\sum_{i=0}^\infty A_i is finite.$}, but I think that it must be {$\omega\in\Omega^n\mid......$} or something like that. However, then there comes another problem why sum of random variables with domain $\Omega$ euals a random variable with domain $\Omega^n$.

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    $\begingroup$ By "identical," you probably mean identically distributed. Identically distributed does not mean that the random variables are equal. That is, $X_1 \overset{d}{=} X_2$ does NOT mean $X_1 = X_2$. $\endgroup$ – Clarinetist Mar 25 '16 at 13:22

You should be careful with the phrase "identical". i.i.d means "independent and identically distributed". So $X_1$ and $X_2$ are not the same functions, they share the same distribution function.


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