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So let $X,Y$ be real random variables on common probability space $(\Omega, \mathcal{F}, P)$, the measures on Borel $(\mathbb{R},\mathcal{B}_{\mathbb{R}})$ induced by $X$ and $Y$ are equal, that is for all $A \in \mathcal{B}_{\mathbb{R}}$.

$$ P_{X}(A) = P_{Y}(A)$$ where $$P_{X}(A) := P(X^{-1}(A))$$ and

$$P_{Y}(A) := P(Y^{-1}(A))$$

Do $X,Y$ generate the same $\sigma-$algebra? I feel that it might not be necessarily the case but was not able to construct a counter example.

Any help would be appreciated

them
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    Your feeling is right, try $\Omega=[0,1]^2$ with the Borel sigma-algebra and the Lebesgue measure and the random variables $X$ and $Y$ defined by $X(x,y)=x$ and $Y(x,y)=y$. – Did Apr 10 '16 at 16:57
  • @Did, Thanks, that answers my question. – them Apr 10 '16 at 17:02

1 Answers1

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Take a probability space $\left(\Omega,\mathcal F, \mathbb P\right)$ with two independent identically distributed random variables $X$ and $Y$ which are not almost surely constant. Then the law are equal but the the generated $\sigma$-algebras can not the same, since they would be independent and would contain an event whose probability is neither $0$ nor $1$.

I rewrite the example given by Did: let $\Omega=[0,1]^2$ with the Borel $\sigma$-algebra and the Lebesgue measure and the random variables $X$ and $Y$ defined by $X(x,y)=x$ and $Y(x,y)=y$.

Overall, we could not define an independent identically distributed sequence of non-constant random variable if having the same distribution would imply having the same generated $\sigma$-algebra.

Davide Giraudo
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