Determine covariance of X+Y vs X-Y

Question

This challenge question was posed in class and I don't know the answer. Can anyone explain?

Answer 1

First note that X and Y are independent because they are the outcomes of two fair coins tossed independently. For each of them you can write: $$ \mathbb{E}[X] = \mathbb{E}[Y] = 3.5 $$ $$ \text{Var}(X) = \text{Var}(Y) = \frac{35}{12} $$ Then you can calculate Cov: $$ \text{Cov}(X + Y, X - Y) = \text{Cov}(X, X) + \text{Cov}(X, -Y) + \text{Cov}(Y, X) + \text{Cov}(Y, -Y) $$ $$ = \text{Cov}(X, X) - \text{Cov}(Y, Y) = \text{Var}(X) - \text{Var}(Y) = 0 $$ So $X + Y$ and $X - Y$ have zero covariance. But note that they are not independent. For example consider the following events. $$ X + Y = 12\quad,\quad X - Y = 1 $$ $$ \mathbb{P}[X - Y = 1] = \mathbb{P}[X = 2, Y = 1] + \mathbb{P}[X = 3, Y = 2] + \dots + \mathbb{P}[X = 6, Y = 5] > 0 $$ And Also: $$ \mathbb{P}[X - Y = 1 | X + Y = 12] = 0 $$ So: $$ \mathbb{P}[X - Y = 1 | X + Y = 12] \neq \mathbb{P}[X - Y = 1] $$ Which means that $X + Y$ and $X - Y$ are dependent. You can conclude that if two random variables are independent their covariance equals zero but the other way doesn't always hold.