1
$\begingroup$

Let $n$ be any positive integer larger than $1$. I want to construct random variables $X_1, X_2, ... X_n$such that any strict subset of $\{ X_1, X_2, ... X_n\}$ is independent but $X_1, X_2, ... X_n$ are not independent. Any hints or reference are appreciated.

$\endgroup$
1
  • $\begingroup$ The case $n=2$ is trivial. $\endgroup$
    – J.G.
    Commented Sep 16, 2018 at 8:28

1 Answer 1

6
$\begingroup$

Let $X_1$, ... , $X_{n-1}$ be i.i.d random variable taking values in $\lbrace -1, 1 \rbrace$ uniformly and let $$X_n = \prod_{i=1}^{n-1} X_i$$ Clearly they are not independent but if you take any subset of those of size at most $n-1$, they are independent:

Let $A\subsetneq \lbrace 1, \dots , n\rbrace$, if $n\notin A$ then independence is trivial so suppose $n\in A$. In that case there is $i\in[1:n-1]$ such that $i\notin A$ but the distribution of all the variables can be rewritten, by symmetry, as $\lbrace X_j \rbrace_{j\neq i}$ are i.i.d random variable uniformly distributed in $\lbrace -1, 1 \rbrace$ and $X_i=\prod\limits_{j\neq i} X_j$ and so the subset is independent.

$\endgroup$
7
  • $\begingroup$ Slight quibble. If the values are in $\{-1,1\}$ rather than $\{0,1\}$ they are not Bernoulli. If you wish, $X_j = 2 Y_j - 1$ where $Y_j$ is Bernoulli. $\endgroup$ Commented Sep 16, 2018 at 8:03
  • $\begingroup$ Thank you, I just edited. $\endgroup$
    – P. Quinton
    Commented Sep 16, 2018 at 8:15
  • $\begingroup$ You could use the term Rademacher. $\endgroup$
    – J.G.
    Commented Sep 16, 2018 at 8:28
  • $\begingroup$ I guess something is not clear here.. How $X_i$ takes values in $\{-1,1\}$ and later you say it takes values in $\{0,1\}$? $\endgroup$
    – Shashi
    Commented Feb 2, 2019 at 19:12
  • $\begingroup$ @Shashi Thank you, it should be fixed now $\endgroup$
    – P. Quinton
    Commented Feb 2, 2019 at 19:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .