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Let $X_1,X_2,\cdots,X_n$ be random variables. Is it true that the $\sigma$-algebra generated by all sets of the form $$\{X_1\leq x_1, X_2\leq x_2,\cdots, X_n\leq x_n\} $$ is the same as the $\sigma$-algebra generated by all sets of the form $$ \{X_1 - X_2\leq x_1^\prime, X_2 - X_3 \leq x_2^\prime, \cdots, X_{n-1} - X_n \leq x_{n-1}^\prime, X_n \leq x_n^\prime \}?$$ Thanks, Phanindra |
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The map $T\colon(t_1,\ldots,t_n)\mapsto (t_1-t_2,t_2-t_3,\ldots,t_{n-1}-t_n,t_n)$ is linear and bijective. The set $$\{X_1-X_2\leq x_1',X_2-X_3\leq x_2',\ldots,X_{n-1}-X_n\leq x'_{n-1},X_n\leq x_n'\}$$ is equal to $\{(X_1,\ldots,X_n), T(X_1,\ldots,X_n)\in\prod_{j=1}^n(-\infty,x'_j]\}$ hence to $T^{-1}\left(\prod_{j=1}^n(-\infty,x'_j]\right)$. Denote $\cal C_1$ the first class and $\cal C_2$ the second one. If $\cal B_j$ is the $\sigma$-algebra generated by $\cal C_j$, in fact we can show that $\mathcal{B}_1=\{\{(X_1,\ldots,X_n)\in B\}, B\in \cal B(\Bbb R^n)\}$ and $\mathcal{B}_2=\{\{T(X_1,\ldots,X_n)\in B\}, B\in \cal B(\Bbb R^n)\}$, but $B$ is a Borel set if and only if so is $T(B)$. |
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