This question seems so obvious that I wonder if I am just having a brain-melt moment... if so, apologies in advance! :(

The difference between pairwise independence and mutual independence is well known, and is often demonstrated with this classic set of $3$ events: "$1$st coin = H", "$2$nd coin = H", and "even number of Hs". (All coins are fair and independent. $0$ is even.)

This can be generalized to $K > 2$ coins, with the $(K+1)$th event still being "even number of Hs". In this set of $K+1$ events, any subset of $K$ events is mutually independent, but the full set is not. (If you wish, read more here.)

However, I have a hard time adding just one more event. Specifically, I seek: For some $K\ge 2$:

  • a set of $K+2$ events,

  • where any $K$-subset is mutually independent,

  • but any $(K+1)$-subset is not mutually independent.

I would prefer a small example using coins, but will accept any probability space. If this is impossible, a proof would also be great.

Attempts: If I add another independent coin, then some $(K+1)$-subset will be mutually independent. OTOH if I just add another Boolean formula (or $\mathbb{Z}_2$ formula) on the existing $K$ coins, then all the formulas that come to mind have some $K$-subset being mutually dependent.

I have also tried models where the actual coins are hidden (i.e. hidden variables), and the events (i.e. observables) are various combinations of the coins, but nothing I tried worked yet.

Further thoughts: In the classic example ($K\ge 2$ coins), the full $(K+1)$-sized set is not only dependent, but in fact conditioned on knowledge of any $K$-subset the remaining event is determined (not just dependent). This is not a requirement of my problem. However, if we add this as a requirement (making the problem harder but providing more structure that may inspire construction of an example), then this (vaguely) reminds me of various aspects of coding/decoding, which is why I am adding that tag.


I do not have a solution over $\mathbb Z_2$, but there is a solution with larger finite fields. More generally, given $n\ge k>0$, we can find $n$ random variables where any $k$ are independent, but any $k+1$ are dependent.

Let $P$ be a random polynomial of degree less than $k$ over the finite field of size $q$, where $q\ge n$. Specifically, $P$ is uniformly chosen among all $q^{k}$ such polynomials. Then writing the elements of this field as integers between $0$ and $q-1$, among the random variables $$ P(0),P(1),\dots, P(n-1), $$ any $k$ are independent, but any $k+1$ are dependent. This is the essence of the Reed-Solomon error correcting code.

To show $P(0),P(1),\dots,P(k-1)$ are independent, for example, you need to show that the probability that $$P(0)=y_0,P(1)=y_1,\dots,P(k-1)=y_{k-1}$$ is equal to $q^{-k}$, for any $y_0,\dots,y_{k-1}$ in the field. Since there are $q^k$ possible polynomials, this is equivalent to showing that there is a unique polynomial with degree less than $k$ which satisfies the above. This is indeed true, and can be found with Lagrange interpolation.

  • $\begingroup$ OK, I feel better now re: why I can't find a coin-based ($\mathbb{Z}_2$) solution. :) Anyway, it is obvious that any $k+1$ are dependent. But why is any $k$ independent? Sure a $(k-1)$-subset cannot determine the $k$th value, but can't it influence its distribution? Or is it true that, conditioned on the $(k-1)$-subset, the $k$th value is still uniform? If so, is there an obvious proof / argument why? $\endgroup$ – antkam Apr 10 at 21:43
  • $\begingroup$ @antkam Forgot that part, see edit. Still thinking about a $\mathbb Z_2$ solution to no avail... $\endgroup$ – Mike Earnest Apr 10 at 22:27
  • $\begingroup$ Thanks for the additions re: uniform distribution - so obvious in hindsight! Re: $\mathbb{Z}_2$, I tried using $q=2^m$, but couldnt get it to work. If I look at each $P(i)$ values as $m$ bits, for $nm$ bits total, then it is not obvious that any $km$ bits are independent. OTOH if I restrict each $P(i)$ to e.g. $1$ bit (e.g. the MSB in some encoding) then it is obvious that any $k$ bits are independent but it is not obvious that any $k+1$ bits are dependent. (We certainly cannot determine the $(k+1)$th bit from the other $k$. I cannot prove dependence either.) $\endgroup$ – antkam Apr 11 at 1:12

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.