Suppose I have two random variables defined on the probability space $(\Omega, \mathcal{F}, P)$: $X_1: \Omega \rightarrow R$ and $X_2: \Omega \rightarrow R$. Assume that $X_1$ and $X_2$ are identically distributed. Following the definition, this means that $P(X_1 \in E)=P(X_2 \in E)$ $\forall E \in B(R)$, i.e. $P(\omega \in \Omega | X_1(\omega) \in E)=P(\omega \in \Omega | X_2(\omega) \in E)$ $\forall E \in B(R)$. Does this means that for any $\omega$, $X_1(\omega)=X_2(\omega)$, i.e. $X_1$ and $X_2$ are the same function? Thanks!
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1 Answer
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You are correct that $X_1$ and $X_2$ are distributed the same if $P(X_1 \in E)=P(X_2 \in E) \forall E \in B(R)$. However, in general the inverse maps are not equal, $X_1^{-1}(E) \neq X_2^{-1}(E)$!
Say we have two fair coins, and our random variables $X_1$ and $X_2$ map a simultaneous coin flip to heads or tails. Since the coin is fair, $P(X_1 = heads) = P(X_2 = heads) = \frac{1}{2}$. However, for many $\omega \in \Omega$, $X_1(\omega) = heads$ and $X_2(\omega) = tails$.
(Head -> 0, Tails -> 1)and(Head -> 1, Tails -> 0)identically distributed? Are they the same functions? $\endgroup$