# Independence of $X+Y$ and $X-Y$

In a roll of die, if $$X$$ is the number on the first die and $$Y$$ is the number on second die, then determine whether the random variable $$X+Y$$ and $$X-Y$$ are independent.

The covariance between the two turned out to be $$\mathrm{Var}(X) + \mathrm{Var} (Y)$$. So zero covariance would mean that $$\mathrm{Var}(X)$$ and $$\mathrm{Var}(Y)$$ is zero, that is, no spread. But we also know that zero covariance does not imply independence. I really cannot think of a way to prove independence between the two.

• The contrivance is Var(X) - Var(Y). Also, do the dies have the same number of sides? – t.f Oct 21 '18 at 7:47
• Independence is P(XY)=P(X)P(Y). You can calculate all probabilities for 36 outcomes and show that the equation holds. – keiv.fly Oct 21 '18 at 7:59
• @keiv.fly Do we calculate X-Y and X+Y probabilities for different cases and then use P((X-Y)(X+Y))=P(X-Y)P(X+Y)? Isn't there a shorter, more formal method to do the same? – Shinjini Rana Oct 21 '18 at 8:12
• @t.f I'm not aware of the term 'contrivance' yet. Yes, they have same no of sides. – Shinjini Rana Oct 21 '18 at 8:13
• If your dice are known, e.g. with standard numbering from 1 to n, then (X+Y) and (X-Y) are not independent. A simple way of thinking about disproving independence is that you only need to show that there exists at least one outcome of X+Y such that X-Y is known with absolute certainty. if X+Y = 2, then (X-Y) is known and it has to be 0. – NofP Oct 21 '18 at 11:07

They're not: If $$X+Y=12$$ then both rolls were sixes, so $$X-Y=0$$. So you have:
$$1 = \mathbb{P}(X-Y =0|X+Y=12) \neq \mathbb{P}(X-Y =0) = \frac{1}{6}.$$