Does joint IID imply marginal IID? Suppose you have a 2 dimension random vector denoted $(X,Y)$ that is Independent and Identically Distributed (IID) for a sequence of draws $((X_1,Y_1),(X_2,Y_2),...,(X_n,Y_n))$. Does this imply that the marginal draws, $(X_1,X_2,...,X_n)$ and $(Y_1,Y_2,...,Y_N)$, are IID?
I think the answer is no. Here is a counterexample. Suppose $\epsilon$ is independent of $(X,Y)$ and assume $Y=X+\epsilon$. Then the distribution of $Y$ depends on $X$ and thus is Independent and Not Identically Distributed. 
 A: You are conflating a few different things here, which is leading you into some trouble.  The joint-IID condition implies that $X_1,\ldots,X_n$ is IID, and $Y_1,\ldots,Y_n$ is IID, but these two vectors are not necessarily independent of each other, since there is no specification of the relationship within the bivariate vector.

Theorem: If $(X_1,Y_1),\ldots,(X_n,Y_n)$ is IID then $X_1,\ldots,X_n$ is IID and $Y_1,\ldots,Y_n$ is IID.
Proof: From the joint-IID condition there exists a joint-distribution function $F_{X,Y}: \bar{\mathbb{R}} \rightarrow [0,1]$ such that:
$$\mathbb{P}(X_1 \leqslant x_1, Y_1 \leqslant y_1, \ldots, X_n \leqslant x_n, Y_n \leqslant y_n) = \prod_{i=1}^n F_{X,Y}(x_i,y_i),$$
for all argument values in the set of extended real numbers.  Now, if we define the marginal distribution $F_X: \bar{\mathbb{R}} \rightarrow [0,1]$ by $F_X(x) \equiv F_{X,Y}(x, \infty)$ we then have:
$$\begin{equation} \begin{aligned}
\mathbb{P}(X_1 \leqslant x_1, \ldots, X_n \leqslant x_n) 
&= \mathbb{P}(X_1 \leqslant x_1, Y_1 \leqslant \infty, \ldots, X_n \leqslant x_n, Y_n \leqslant \infty) \\[6pt]
&= \prod_{i=1}^n F_{X,Y}(x_i,\infty) \\[6pt]
&= \prod_{i=1}^n F_X(x_i), \\[6pt]
\end{aligned} \end{equation}$$
for all argument values in the extended real numbers.  This establishes that $X_1,\ldots,X_n$ is IID.  By analogous reasoning we can also show that $Y_1,\ldots,Y_n$ is IID.
