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Suppose we have a time series $X_t$ s.t. $X_t \sim^{iid} (0,1)$.

How do you prove that if $ X_t \sim^{iid} (0,1) $, then $ E(X_t^{2}X_{t-j}^{2}) = E(X_t^{2})E(X_{t-j}^{2})$?

Or, I guess, if $X,Y\sim^{iid} (0,1)$ (which implies $E(XY)=E(X)E(Y)$), why then is it that $E(X^2Y^2)=E(X^2)E(Y^2)$?

This spins-off from another question wherein apparently "If the squares were dependent, there's a form of dependence among the unsquared values."

This makes sense, but how does one prove this exactly? My attempt:

Instead of dependence => dependence (which I think would involve probability distributions), I try to prove uncorrelatedness => uncorrelatedness as follows:

$E(X^2Y^2) \neq E(X^2)E(Y^2)$

$\implies E(X^2Y^2) \neq (Var(X)+E(X)^2)(Var(Y)+E(X)^2)$

$\implies Var(XY)+E(XY)^2 \neq (Var(X)+E(X)^2)(Var(Y)+E(Y)^2)$

$\implies Var(XY)+(E(X)E(Y))^2 \neq (Var(X)+E(X)^2)(Var(Y)+E(Y)^2)$

$\implies Var(XY)+(E(X)E(Y))^2 \neq Var(X)Var(Y)+Var(X)E(Y)^2+Var(Y)E(X)^2+(E(X)E(Y))^2$

$\implies Var(XY) \neq Var(X)Var(Y)+Var(X)E(Y)^2+Var(Y)E(X)^2$

$\implies ...$

$\implies E(XY) \neq E(X)E(Y) \ QED$

Ugh...

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Note that $X$ and $Y$ being independent is not equivalent to ${\rm E}[XY]={\rm E}[X]{\rm E}[Y]$ (the last is the definition of being uncorrelated) although independence implies uncorrelatedness. Instead $X$ and $Y$ are independent if $$ P(X\in A,Y\in B)=P(X\in A)P(Y\in B) $$ for all (Borel) sets $A,B\subseteq\mathbb{R}$, or equivalently, that $\sigma(X)$ and $\sigma(Y)$ should be independent under $P$.

If $X$ and $Y$ is independent, then any $f(X)$ and $g(Y)$ is also independent for any pair of (measurable) functions $f$ and $g$. This can be seen, for instance, by noting that

$$ \begin{align*} P(f(X)\in A,g(Y)\in B)&=P(X\in f^{-1}(A),Y\in g^{-1}(B))\\ &=P(X\in f^{-1}(A))P(Y\in g^{-1}(B))\\ &=P(f(X)\in A)P(g(Y)\in B) \end{align*} $$ for all (Borel) sets $A,B\subseteq \mathbb{R}$. If you are familiar with the definition of independency in terms of sigma-algebras, then this is an easy consequence of the fact that $\sigma(h(X))\subseteq \sigma(X)$ for all (Borel) functions $h$.

As an immediate consequence we have that if $f(X)$ and $g(Y)$ are not independent for some $f$ and $g$, then $X$ and $Y$ can not be independent either.

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  • $\begingroup$ thanks. Edited "that is" to "which implies". I am not yet familiar with independence(/y?) in terms of sigma-algebras, but I do understand the proof in the middle there. Additional question: The suggestion in the other question was to make use of the fact that dependence in the squares implies dependence in the originals and then deduce that independence in the originals implies independence in the squares. How does one prove that dependence in the squares implies dependence in the originals? $\endgroup$ – BCLC Jun 20 '14 at 12:00
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    $\begingroup$ I think that comment was meant as an intuitive explanation. I would definitely show that claim by showing the contrapositive (i.e. showing the independence implication). $\endgroup$ – Stefan Hansen Jun 20 '14 at 12:05
  • $\begingroup$ Okay just recently I learned independence in terms of sigma-algebras in Advanced Probability. The problem is that this came after our Advanced Statistics class where the problem originated (the problem luckily did not appear in our Adv Stat exam). So, is there anyway to show $X^2$ and $Y^2$ are independent or at least uncorrelated just because X and Y are independent w/o the above line of reasoning? I can't imagine anything else because these are precisely the definitions of independence. Even w/o sigma-algebras, this seems like the way I would go about doing this. $\endgroup$ – BCLC Aug 2 '14 at 20:24
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    $\begingroup$ Nah, this is the way to do it. $\endgroup$ – Stefan Hansen Aug 4 '14 at 5:28
  • $\begingroup$ Stefan, actually, I think there is a way to do this in basic probability language? Just call A and B intervals in $\mathbb{R}$? Of course, it's not right from an advanced viewpoint but some elem prob textbooks seem to use such language (i.e. "intervals" instead of "Borel sets") $\endgroup$ – BCLC Sep 19 '14 at 23:17

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