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})$?

Question

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...

Answer 1

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.