0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $\sigma(g(Y))\subseteq\sigma(Y)$.

$\endgroup$
  • $\begingroup$ 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! $\endgroup$ – AnonymousIGuess Nov 17 '17 at 11:53
  • $\begingroup$ 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 $Z_{2}$? $\endgroup$ – Ivan Feb 28 '18 at 4:36
  • $\begingroup$ @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. $\endgroup$ – Jack M Mar 2 '18 at 15:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.