2
$\begingroup$

Let $\xi=(\xi_1,\ldots,\xi_n)$ be a sequence of independent random variables. Let us pick an index $\nu\in \{1,\ldots,n\}$, and replace the entry $\xi_\nu$ by a constant $c$. The rest of the $\xi_i$ remain unchanged.

Question: Is it true that the altered sequence remains independent, even if $\nu$ is random, possibly dependent on $\xi$?

Note: If $\nu$ is a fixed constant index, then the independence is clearly preserved. It is less clear, however, what happens if $\nu$ is random and dependent on $\xi$. For example, we may pick the first hit to some set, or apply any other data-dependent rule to pick the index $\nu$.

$\endgroup$
3
$\begingroup$

The answer is no. E.g., assume that $n=2$, $\xi_i=X_i$, and $X_1,X_2,\nu$ are any random variables (r.v's) each with values in the set $\{1,2\}$ and with $P(\nu=1)=p\in(0,1)$. With these conditions in place, the dependence between $X_1,X_2,\nu$ may be arbitrary. For instance, we may suppose that $\nu$ is independent of $X_1,X_2$, or we may suppose that $\nu=X_1$, or ... .

Take any $c\notin\{1,2\}$ and let $Y_1,Y_2$ be obtained from $X_1,X_2$ by replacing $X_\nu$ by $c$, so that
\begin{equation} (Y_1,Y_2)=\left\{ \begin{aligned} (c,X_2) &\text{ if }\nu=1,\\ (X_1,c) &\text{ if }\nu=2. \end{aligned} \right. \end{equation} Then $$P(Y_1=c,Y_2=c)=0\ne p(1-p)=P(\nu=1)P(\nu=2)=P(Y_1=c)P(Y_2=c). $$ So, $Y_1,Y_2$ are not independent.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.