The title isn't quite representative of the whole problem but it is where I found myself stuck. For simplicity, I will post the entire problem and then explain where I ended up at.
Let $ \{Z_t\} $ be independent random variables with mean $0$ and variance $σ2$. Let $\{Y_t\}$ be a stationary sequence with a covariance function $γ_Y (h)$. Assume also that the sequences $ \{Z_t\} $ and $\{Y_t\}$ are independent from each other. Define $X_t = Y_tZ_t.$ Verify that for $h ≥ 1$ we have $Cov(X_t, X_{t+h}) = 0 $ and $Cov(X_t^2 , X_{t+h}^2) \neq 0 $ (at least not for all t and h).
The first result was quite easy to find just by writing down the definition of covariance. However, for the second result I get that $Cov(X_t^2 , X_{t+h}^2) = σ^2σ^2Cov(Y_t^2 , Y_{t+h}^2)$ by once again just writing out the definition. But I'm stuck at trying to show $Cov(Y_t^2 , Y_{t+h}^2) \neq 0 $ for all $h$. I think I can assume the cov function $γ_Y (h) \neq 0$ for atleast one $h$ as it is a function of $h$, but I'm not sure how to use this to prove the final result.
Any help would be greatly apreciated. If you need any additional info please ask.
In the above conditions, the statement is false. Let us consider the stationary process $Y_t=Y_0$ for all $t$, where $$ \mathbb P(Y_0=1)=\mathbb P(Y_0=-1)=\frac12. $$ Then $\gamma_Y(h)=\text{Var}(Y_0)=1\neq 0$ and $$ \text{Cov}(Y_t^2 , Y_{t+h}^2) = \text{Cov}(1 , 1) = 0. $$