# 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.

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{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}

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.