Exercise 4.1 - Uncorrelated does not imply independent

The mean for $Y$ is:

Calculate the covariance of $X$ and $Y$:

whose value is zero since we are intergrating an odd function in range $[-1,1]$, hence:

Independence is a much stronger condition than uncorrelation. The former exerts constraints on the $\sigma$-algebra that random variables generate while the latter only regulates the value of the expectation of a new random variable. Decomposition $p(X,Y)=p(X)\cdot p(Y)$ is sufficient for reducing the covariance to zero, but not necessary.