2
$\begingroup$

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 $\Omega^n$.

$\endgroup$
  • 1
    $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.