Suppose we have two finite dimensional random variables $(X_1,\ldots ,X_n),(Y_1,\ldots,Y_n)$. How do we show formally that if $$(X_1,X_2-X_1,\ldots ,X_n-X_{n-1})\stackrel{d}{=} (Y_1,Y_2-Y_1,\ldots ,Y_n-Y_{n-1})$$ then $$(X_1,\ldots ,X_n)\stackrel{d}{=}(Y_1,\ldots,Y_n)$$

Is this even true?


1 Answer 1


I will leave the explicit solution to you, after giving you this well known fact: Let $X,Y$ be random variables taking values in $\mathbb{R}^n$ such that $X\stackrel{D}{=} Y$. For any measurable function $f:\mathbb{R}^n\rightarrow \mathbb{R}^m$ it holds that $f(X)\stackrel{D}{=}f(Y).$

To prove this result we notice that for any $A\in \mathbb{B}^m$ we have $$P(f(X)\in A)=P(X\in f^{-1}(A))=P(Y\in f^{-1}(A))=P(f(Y)\in A)$$ Can you prove the result from here?

  • $\begingroup$ Right that makes sense. How is this fact proven? $\endgroup$
    – John
    Apr 2, 2019 at 20:13
  • $\begingroup$ See edit of answer $\endgroup$
    – Conformal
    Apr 2, 2019 at 20:24

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