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...
5

4$\begingroup$ Consider the case when with probability $1/2$ random $m$ of them are $1$ and with probability $1/2$ random $nm$ of them are $1$ where $m=n1$, say. Then it looks to me like the correlations are rather positive for large $n$. $\endgroup$– fedjaNov 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, $P(X_1=1,X_2=1)=P(X_1=0,X_2=0)=\frac 12\frac{(n1)(n2)}{n^2}$, which is almost full correlation. $\endgroup$– fedjaNov 19 '18 at 1:07

$\begingroup$ @fedja : Very nice example. In fact, the random variables are even more correlated: $P(X_1=X_2=1)=\frac12\,\frac{n2}n$ for $n\ge2$, I think. $\endgroup$– Iosif PinelisNov 19 '18 at 2:29

$\begingroup$ @fedja: Ah yes, you are right! $\endgroup$– Ilya BogdanovNov 19 '18 at 5:05

$\begingroup$ @fedja Nice example! Thanks so much! $\endgroup$– zxzx179Nov 19 '18 at 5:34
Add a comment
