If $X$ and $Y$ are i.i.d., and $N=X+Y$, is $\mathbb{P}(X, Y, X+Y) = \mathbb{P}(X, Y)$? I have a homework problem that asks:

Let $X$ and $Y$ be i.i.d. $\mathrm{Geom}(p)$, and $N = X+Y$. Find the joint PMF of $X,Y$, and $N$.

Based on advice from my professor, it seems like ultimately the probability of X, Y, and X+Y is the same as the probability of $X$ and $Y$ (which makes sense intuitively, because if you know $X$ and $Y$ that would determine their sum). But I don't know how to prove that and I feel like I might be missing something.
If $\mathbb{P}(X, Y, X+Y) = \mathbb{P}(X, Y)$, then: 
$\mathbb{P}(X=x, Y=y, N=x+y) = \mathbb{P}(X=x)\mathbb{P}(Y=y) = pq^x \cdot pq^y = p^2q^{x+y} = p^2q^n,$
where $q=1-p$. I do think that's the right final answer but I'm not sure about my logic.
 A: A few quibbles with the language and notation you're using:

ultimately the probability of X, Y, and X+Y is the same as the probability of X and Y

This doesn't quite make sense; we talk about the probabilities of events. You could discuss, for instance, "the probability that $X$ is $12$," but it doesn't make sense to just talk about "the probability that $X$."
Second: be careful with your notation in expressions like $\mathbb P(X, Y, X+Y)$; here, you've made the notation version of the same error as above. What (I think) you're referring to is the joint probability function $p$ for all three variables: that is,
$$p(x, y, n) = \mathbb P(X = x, Y = y, N = n).$$
Notice the distinction: capital letters represent the random variables (things that can change), and lower-case letters represent fixed values (things that don't change). On the other hand, another expression you wrote:

$P(X=x,Y=y,N=x+y)$

is perfectly good and sensible. This discusses a probability that three random variables will do something all at the same time.
Quibbles aside, the crux of your argument is good. The equation $\mathbb P(X=x,Y=y,N=x+y)=\mathbb P(X=x) \mathbb P(Y=y)$ holds because the event $\{X = x, Y = y\}$ implies (i.e. is a subset of) the event $\{N = x + y\}$, and anytime you are considering events $A$ and $B$ with $A \subset B$, it follows that $\mathbb P(A \cap B) = \mathbb P(A)$. Since $\mathbb P(X = x, Y = y, N = x+y) = \mathbb P(X = x, Y = y)$, and since $X, Y$ are independent, it does indeed follow that this is equal to $\mathbb P(X = x) \mathbb P(Y = y)$. The rest of your work follows from things you know about geometric random variables.
One more thing you should be sure of, though: remember that a joint pmf is a function of three variables. That means you need to consider cases like, for instance, $p(3, 6, 8)$ -- that is, $\mathbb P(X = 3, Y = 6, N = 8)$. (Consider the trouble with such an expression.) This needs to be encoded into your pmf formula in some specific way, and this is why it's not quite right to say that the joint pmf of all three is just equal to the joint pmf of $X$ and $Y$; remember, that joint pmf is just a function of two input variables instead of three.
