# Are four random variables independent if pairs and sums are?

Consider four random variables $$X_1, X_2, X_3, X_4$$. Let $$X_1$$ and $$X_2$$ be independent as well as $$X_3$$ and $$X_4$$. We also have that $$X_1+X_2$$ and $$X_3 + X_4$$ are independent. Does this imply that all four random variables are independent? If not, is there a simple counter-example?

I suspect that they are not independent but I have failed to come up with a counter-example.

If we only have that $$X_1+X_2$$ and $$X_3 + X_4$$ are independent then $$X_1 = -X_2 = X_3$$ is an example where $$X_1$$ and $$X_3$$ are not independent but $$X_ 1+X_2$$ and $$X_3 + X_4$$ are independent.

• @LeeDavidChungLin I completely agree but really my problem is failing to come up with a counter-example. – Arthur Apr 12 at 17:05

Here's what I imagine is the minimal counterexample: set $$X_4 = 0$$ to be a constant and forget about it.
Write $$X \sim B$$ if $$X$$ is either $$0$$ or $$1$$ with probability $$1/2$$.
Now choose $$X_1,X_2 \sim B$$ independently. Then choose $$X_3 \qquad \begin{cases} = X_1 & \text{if } X_1 \neq X_2 \\ \sim B \text{ (independently)} & \text{if } X_1 = X_2 \end{cases}$$
Now the pairs are independent by construction. Now consider $$X_1 + X_2$$. If this sum is anything even then necessarily $$X_1 = X_2$$, so $$X_3 \sim B$$. Otherwise, the sum is odd, so $$X_1 \neq X_2$$ and $$X_3 = X_1 \sim B$$ still. Hence $$X_1 + X_2$$ is indeed independent of $$X_3$$.
However, clearly $$X_3$$ is not independent from $$(X_1,X_2)$$. In fact, $$X_3$$ is not independent from $$X_1$$ alone: $$3/4$$ of the time they must agree.