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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.

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

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