Let $X_1, ..., X_n$ be independent random variables and let $f:\mathbb{R}^k\rightarrow \mathbb{R}$ be some function. Denote $Y=f(X_1, ..., X_k)$. Show that $Y, X_{k+1}, ... , X_n$ are independent.

I believe that the definition that should be used is that $X_1, ..., X_n$ are independent if and only if for every sets $A_1, ..., A_n$, it holds that $P(X_1\in A_1, ..., X_n\in A_n)=P(X_1\in A_1)...P(X_n\in A_n) $.

It's probably easy, but other than the case $k=1$ I can't figure it out. Any ideas?


Let $A,\ A_{k+1},\ldots,A_{n}$ be Borel subsets of $\mathbb{R}$.

Write \begin{align} \mathbb{P}[f(X_1,\ldots,X_k)\in A, X_{k+1} \in A_{k+1}, \ldots, X_n \in A_{n}] = \mathbb{P}[(X_1,\ldots,X_k)\in f^{-1}(A), X_{k+1},\ldots,X_n) \end{align} Decompose $f^{-1}(A)\in \mathbb{R^k}$ to the cartesian product of $A_1,\ldots,A_k \in \mathbb{R}$, that is $f^{-1}(A) = A_1 \times \ldots \times A_k$.

Clearly \begin{equation} (X_1,\ldots X_k) \in f^{-1}(A) \Leftrightarrow X_1\in A_1, \ldots, X_k \in A_k \qquad (1) \end{equation}

Use (1) and the independence of $X_i,\ i = 1,\ldots n$ to deduce

\begin{align} \mathbb{P}[X_1 \in A_1, \ldots X_k \in A_k, X_{k+1}\in A_{k+1}, \ldots X_n \in A_n] &= \prod^k_{i=1} \mathbb{P}[X_{i} \in A_i] \cdot \prod^n_{i=k+1} \mathbb{P}[X_i \in A_i] \\ \end{align} Finally using again (1) write \begin{equation} \prod^k_{i=1} \mathbb{P}[X_{i} \in A_i] = \mathbb{P}[f(X_1,\ldots,X_k) \in A] \end{equation} proving that $f(X_1,\ldots X_k), X_{k+1}\ldots X_n$ are independent.

  • 1
    $\begingroup$ That was exactly my problem: I believe you can't just write the inverse image as a cartesian product. $f$ could be any function; what if it sends only vectors with all-entries-equal to $A$, for example? How would you decompose $f^{-1}(A)$ now? $\endgroup$ – 35T41 Dec 9 '16 at 6:03
  • $\begingroup$ @35T41 You're right. I cannot see how this proof could cover such cases. $\endgroup$ – vkonton Dec 9 '16 at 18:05

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