# If $(Y_i,X_i)$, $i=1,\ldots,n$, are iid, are $Y_i$, $i=1,\ldots,n$, iid?

If $(Y_i,X_i)$, $i=1,\ldots,n$, are iid, are $Y_i$, $i=1,\ldots,n$, iid? For example, if $n=2$ then the question is: if $Z_1=(Y_1,X_1)$ and $Z_2=(Y_2,X_2)$ are independent of each other with the same joint distribution, is it true that $Y_1$ and $Y_2$ are iid? I can find no formal counter-example and I cannot prove it either.

If $Z_1 = (X_1, Y_1)$ and $Z_2 = (X_2, Y_2)$ are equal in law as random vectors (their joint distributions are equal), then $X_1$ is equal in law to $X_2$ and $Y_1$ is equal in law to $Y_2$ (the converse is not true). Your claim follows from this.
Proof of my claim: say the random vectors $Z_i$ land in a space $\mathcal A\times \mathcal B$. Let $A$ be a measurable set of $\mathcal A$. We have to prove that $P(X_1\in A)=P(X_2\in A)$. But the set $\{X_i\in A\}$ is equal to the set $\{X_i\in A, Y_i\in\mathcal B\}$, ie to $Z_i^{-1}(A\times\mathcal B)$. The two probabilities are therefore equal by the assumption that $Z_1\sim Z_2$.
For independence, we have the more general fact that if $X$ and $Y$ are independent, then $f(X)$ and $g(Y)$ are independent for any functions $f$ and $g$. In your case we would have $f=g=\text{projection onto the first coordinate}$, so that $Y_i=f(Z_i)$.
How you would prove that might depend on precisely how you define independence of random variables, for me the definition is that for any $A\in\sigma(X)$, $B\in\sigma(Y)$, $A$ and $B$ are independent events. The claim then follows from $\sigma(f(X))\subseteq\sigma(X)$ and $\sigma(g(Y))\subseteq\sigma(Y)$.
• Okay, this sounds reasonable and is in line with first thought. My question was motivated by another question: why is a regression model of $Y_i$ on $X_i$ consistent with heteroskedastic errors? Some authors seems to suggest that, but that seems inconsistent with the assumption of $(Y_i,X_i),i=1,\ldots,n,$ being iid. But that is another question, thanks for your answer! – AnonymousIGuess Nov 17 '17 at 11:53
• Jack M, how can you prove that the converse is not true? I mean if $X_{1}$ is equal in law to $X_{2}$, and $Y_{1}$ is equal in law to $Y_{2}$, how can you prove that $Z_{1}$ is not equal in law to $Z_{2}$? – Ivan Feb 28 '18 at 4:36
• @Ivan That's not always the case. If $X_1\sim X_2$ and $Y_1\sim Y_2$, then $Z_1$ may either be or not be equal in law to $Z_1$, depending on the situation. – Jack M Mar 2 '18 at 15:58