Is $X_t$ independent with $X_{t+k}^2$?

Question

Suppose {$X_t$} is a series of independent random variable, that is $X_1, X_2$, ... is independent. Further, suppose $X_t$ ~ $N(0,1)$.

In my book, it was stated that $Cov[X_t,X_{t+k}^2]=0$. Does this mean that $X_t$ is independent with $X_{t+k}^2$? Can anyone explain this?

Answer 1

If $X$ and $Y$ are independent so are $f(X)$ and $g(Y)$ for any measurable functions $f,g: \mathbb R \to \mathbb R$. This is easy to prove from definition of independence. Here you can take $f(x)=x$ and $g(x)=x^{2}$.

Of course,independence of $X_t$ and $X_{t+k}^{2}$ implies that the covariance is $0$.

Answer 2

The reasoning works the other way around. Since $X_t$ is independent of $X_{t+k}$, $X_t$ is also independent of $X_{t+k}^2$ hence $X_t$ and $X_{t+k}^2$ are uncorrelated.

In general, uncorrelatedness does not imply independence.