# Consequences of identical distribution of random variables

Consider two real-valued random variables $Y_1: \Omega \rightarrow \mathbb{R}$ and $Y_2:\Omega \rightarrow \mathbb{R}$. Even if they have the same domain and codomain, $Y_1$ and $Y_2$ generate different $\sigma$-algebras, $\sigma(Y_1)$ and $\sigma(Y_2)$, and are defined in different probability spaces, $(\Omega, \mathcal{F}_{Y_1}, \mathbb{P}_{Y_1})$ and $(\Omega, \mathcal{F}_{Y_2}, \mathbb{P}_{Y_2})$. Suppose that $Y_1$ and $Y_2$ are identically distributed: $\mathbb{P}_{Y_1}(Y_1 \in E)=\mathbb{P}_{Y_2}(Y_2 \in E)$ $\forall E \in \mathcal{B}(\mathbb{R})$, i.e. $\mathbb{P}_{Y_1}(\omega \in \Omega | Y_1(\omega) \in E)=\mathbb{P}_{Y_2}(\omega \in \Omega | Y_2(\omega) \in E)$ $\forall E \in \mathcal{B}(\mathbb{R})$.

Could you help me in saying whether the following statements are true?

(1) Identical distribution does not imply that $\mathbb{P}_{Y_1}$ and $\mathbb{P}_{Y_2}$ are the same functions because the sets $\{\omega \in \Omega | Y_1(\omega) \in E \}$ and $\{\omega \in \Omega | Y_2(\omega) \in E \}$ could be different.

(2) Identical distribution does not imply that $Y_1$ and $Y_2$ are the same function, i.e. $Y_1(\omega)=Y_2(\omega)$ $\forall \omega \in \Omega$. Therefore, even if $Y_1$ and $Y_2$ are identically distributed, $\sigma(Y_1)$ and $\sigma(Y_2)$ are different.

(3) Identical distribution implies that the image of $Y_1$ and $Y_2$ is the same.

• This seems like routine bookwork, perhaps assigned work for some subject, or perhaps just for your own personal study. Either way, could you please add the self-study tag, and read the guidelines at the link I gave above, then edit your question to accord with those guidelines. – Glen_b Mar 20 '14 at 9:17