Let $\{X_{t}\}$ be a sequence of mean zero independent and identically distributed (iid) random variables. Now define the sequence $\{Y_{t}\}$ as $Y_{t} = X_{t}X_{t-1}$. Show that $Y_{t}$ is not necessarily an iid sequence.
Further Context:
My professor has suggested that this sequence is a white noise process, but is not always an iid process. I have been trying to prove that $Y_{t}$ and $Y_{t+1}$ are not independent but am having difficulty doing this. In particular, I do not know how to show that $\{Y_{t}\}$ is not a sequence of independent variables. I have not been able to find anything online or in my textbooks that deals with this particular problem.
It suffices to find only one example such that $Y_{t}$ is not an iid sequence.
Suppose $X_{t}\sim Bernoulli(p)$ where $P(X_{t}=1)=P(X_{t}=2)=\frac{1}{2}$. To show $Y_{t}$ is not an iid sequence it suffices to show that $P(Y_{t+1} = a | Y_{t} = b)\neq P(Y_{t+1}=a)$.
Indeed, $a$ and $b$ can take on values $4$, $2$ and $1$ where $Y_{t}$ takes on these values with probability $p^{2}$, $2p^{2}$ and $p^{2}$ respectively. It is easy to see that knowledge of the value of $Y_{t}=b$ will affect the probability of observing $Y_{t+1}=a$.
For example, if $Y_{t} =1$ then $P(Y_{t+1} = 4 | Y_{t} = 1)= 0 \neq P(Y_{t+1}=4)= \left(\frac{1}{2}\right)^{2}$.
Therefore $P(Y_{t+1} = a | Y_{t} = b)\neq P(Y_{t+1}=a)$ and thus $Y_{t}$ is not an iid sequence.