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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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  • $\begingroup$ Are the random variables mapping a (fair) coin toss (Head -> 0, Tails -> 1) and (Head -> 1, Tails -> 0) identically distributed? Are they the same functions? $\endgroup$
    – ziggystar
    Mar 19, 2014 at 10:59
  • $\begingroup$ Thank you! To be sure I have understood: let $X_1$ be the first r.v you defined and $X_2$ the second; I see that the two random variables are not the same function but they are identically distributed; in fact: we know that $P(\omega \in \Omega | X_1(\omega)=0)=P(head)$ and $P(\omega \in \Omega | X_2(\omega)=0)=P(tails)$; we can see that $P(head)=P(tails)$; similarly for $X_1(\omega)=X_2(\omega)=1$. Is that right? $\endgroup$
    – Star
    Mar 19, 2014 at 11:09
  • $\begingroup$ That's what I thought. :) But $X_1(\omega) \not = X_2(\omega)$. $\endgroup$
    – ziggystar
    Mar 19, 2014 at 11:23

1 Answer 1

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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 $X_2(\omega) = tails$.

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