probabilty, random variable independent

Question

Let $X$ and $Y$ be independent Poisson random random variables with ($\lambda=1$). Are $X-Y$ and $X+Y$ independent? Justify

My attempt:

$X-Y$ => random variable is $0$.

$X+Y$=> Poisson of ($\lambda=2$).

yes? they are independent?

Answer 1

The unconditional probability that $X-Y=0$ is not $1$. The conditional probability that $X-Y=0$, given that $X+Y=0$, is obviously $1$. So our random variables are not independent.

Alternately, and almost equivalently, we have $\Pr((X+Y=0)\land (X-Y=0))=\Pr(X=0)\Pr(Y=0)=e^{-2}$.

But $\Pr(X+Y=0)=e^{-2}$ and $\Pr(X-Y=0)\lt 1$, and therefore $$\Pr((X+Y=0)\land (X-Y=0))\ne \Pr(X+Y=0)\Pr(X-Y=0).$$ Hence $X+Y$ and $X-Y$ are not independent.