How to show that an m.d.s is not independent? I have to prove that this Martingale Difference: $x_t = u_t u_{t-1}$ where $u_t \sim^{iid} (0, \sigma^2)$ is not serially independent, but am failing to do such thing.
I also have to prove that it's conditionally homoscedastic.
What I have tried so far is to show that $E[x_t^2|x_{t-1}^2] \neq E[x_t^2]$, and therefore it would not be serially independent because of the second moment, but in my view that would be proving that the process is conditionally heteroscedastic, instead of homoscedastic.
What am I doing wrong? How would you proceed?
Thanks
 A: Although comments to the question indicate the original problem assumes the $u_t$ have Normal distributions, it is of interest to determine the extent to which this distributional assumption is needed to prove the $x_t$ are not serially dependent.
Let's begin, then, with a clear statement of the problem.  It focuses on three variables $(u_{t-1}, u_t, u_{t+1}).$  Suppose they are identically and independently distributed (iid); and that this distribution has zero mean and a finite variance of $\sigma^2.$  (We drop any Normality assumption.)  Out of these variables we construct two others, $x_{t+1}=u_{t+1}u_t$ and $x_t = u_tu_{t-1},$ which are products of successive $u$'s.  Since $u_t$ is involved in the construction of both of the $x$'s, intuitively those $x$'s should not be independent.  But how to prove this?
This is trickier than it might appear, because in simulations with Normally distributed $u$'s, the scatterplot of $(x_t, x_{t+1})$ shows a complete lack of correlation:

Indeed, if we try to demonstrate lack of independence by showing there is nonzero covariance, we will fail (because the $u$'s are independent and have zero mean):
$$\begin{aligned}
\operatorname{Cov}(x_t, x_{t+1}) &= E[x_tx_{t+1}]-E[x_t]E[x_{t+1}]\\
&= E[u_{t-1}u_t\, u_tu_{t+1}] - E[u_{t-1}u_t]E[u_tu_{t+1}]\\
&=E[u_{t-1}]E[u_t^2]E[u_{t+1}] - E[u_{t-1}]E[u_t]\,E[u_t]E[u_{t+1}]\\
&= 0-0 = 0.
\end{aligned}$$
The simplest way I have found to demonstrate lack of independence is to consider the magnitudes of the $x$'s.  The logic will be this: if we can show that some function $f$ (in this case, the absolute value) when applied to both $x$'s, produces variables with nonzero covariance, then the $x$'s could not originally have been independent.  (A simple proof is at https://stats.stackexchange.com/questions/94872.)
So, let's carry out the preceding calculations using the absolute values.  We are permitted to do this because the finite variance of the $u$'s implies the expectations of their absolute values are finite, too (see Prove that $E(X^n)^{1/n}$ is non-decreasing for non-negative random variables).  Let this common expectation be $E[|u_t|]=\tau.$  Then
$$\begin{aligned}
\operatorname{Cov}(|x_t|, |x_{t+1}|) &= E[|x_t||x_{t+1}|]-E[|x_t|]E[|x_{t+1}|]\\
&=E\left[|u_{t-1}|\right]E\left[|u_t|^2\right]E\left[|u_{t+1}|\right] - E\left[|u_{t-1}|\right]E\left[|u_t|\right]\,E\left[|u_t|\right]E\left[|u_{t+1}|\right]\\
&= \tau \sigma^2 \tau - \tau^4\\
&= \tau^2(\sigma^2 - \tau^2).
\end{aligned}$$
Unless $\sigma^2=\tau^2$ or $\tau=0,$ this is nonzero and we are done.
Because squaring is a convex function, Jensen's Inequality implies $\tau^2\le\sigma^2$ with equality if and only if $u_t$ is almost surely constant.
The case $\tau=0$ occurs when $|u_t|$ is almost surely constant, whence $u_t$ itself is either $0$ or takes on the values $\pm\sigma$ with equal probability (a scaled Rademacher variable).  Both situations can be characterized as scaled Rademacher variables (with scaling factor $0$ in the first case).
That covers all the possibilities.  The result is,

When  $(u_{t-1}, u_t, u_{t+1})$ are iid with zero mean, $x_t = u_{t-1}u_t$ and $x_{t+1} = u_t u_{t+1}$ are independent if and only if the $u$'s each have a scaled Rademacher distribution.

(The "if" part can be shown with a direct calculation: tabulate the eight possible values of $(u_{t-1}, u_t, u_{t+1}),$ compute the corresponding values of $(x_t, x_{t+1}),$ and verify independence.)
To help the intuition, here is a scatterplot of the absolute values in the preceding figure (shown on fourth-root scales to remove the skewness from the marginal distributions; if you like, the function involved is $f(x) =\sqrt[4]{|x|}$).  The red line is the least-squares fit to these fourth roots.  The correlation is obviously nonzero.

