3
$\begingroup$

Let $\big\{X_1, X_2, ..., X_n \big\}$ be $n$ jointly exchangeable Bernoulli random variables, i.e., exchanging the order of these random variables does not change the joint distribution. If we know that $$\sum_{i=1}^{n} X_i \leq m < n$$ holds for sure, does this imply that any $X_i$ and $X_j$ are negatively correlated? This is quite intuitive to me because these random variables are symmetric and their sum is bounded from above...

$\endgroup$
  • 3
    $\begingroup$ Consider the case when with probability $1/2$ random $m$ of them are $1$ and with probability $1/2$ random $n-m$ of them are $1$ where $m=n-1$, say. Then it looks to me like the correlations are rather positive for large $n$. $\endgroup$ – fedja Nov 18 '18 at 21:42
  • 1
    $\begingroup$ @IlyaBogdanov On the contrary. If $X_1=1$, it makes it almost certain that we are in case $1$, so the probability that $X_2=1$ gets close to $1$. More precisely, $P(X_1=1,X_2=1)=P(X_1=0,X_2=0)=\frac 12\frac{(n-1)(n-2)}{n^2}$, which is almost full correlation. $\endgroup$ – fedja Nov 19 '18 at 1:07
  • $\begingroup$ @fedja : Very nice example. In fact, the random variables are even more correlated: $P(X_1=X_2=1)=\frac12\,\frac{n-2}n$ for $n\ge2$, I think. $\endgroup$ – Iosif Pinelis Nov 19 '18 at 2:29
  • $\begingroup$ @fedja: Ah yes, you are right! $\endgroup$ – Ilya Bogdanov Nov 19 '18 at 5:05
  • $\begingroup$ @fedja Nice example! Thanks so much! $\endgroup$ – Stupid_Guy Nov 19 '18 at 5:34

Your Answer

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

Browse other questions tagged or ask your own question.