$Z$ is a random variable that follows a standard normal distribution. Is $X = Z^2$ independent from $Y = Z^3$? I think the title is pretty clear about the problem. Should I try to find the joint probability of $X$ and $Y$ and decide if $\, f_{X,Y}(x,y) = f_X(x) \cdot f_Y(y)$? If so, how am I going to find a joint distribution like this? Or is there some shortcut to prove it?
 A: $EXY^{2} \neq EXEY^{2}$ so $X$ and $Y^{2}$ are not independent. This implies that $X$ and $Y$ are not independent. [Use a table of moments of $Z$ to see that $EXY^{2} \neq EXEY^{2}$. You can also argue analytically using that fact that 6-th moment is strictly smaller than the 8th moment].  Alternative proof: if $X$ and $Y$ are independent so are $X^{3}$ and $Y^{2}$. This makes $Z^{6}$ independent of itself which implies $Z^{6}$ is a constant random variable. This is obviously false. 
A: Because both $X,Y$ depend on a single random variable $Z$, intuitively $X,Y$ are not independent. To prove this, sketch the parametric curve $(t^2,t^3)$ over all real $t$:

(This is the semicubical parabola $y^2=x^3$.) The joint pdf is non-zero at some $(x,y)$ iff that point lies on this curve. If $X,Y$ were independent, any vertical line (pdf of $Y$ conditioned on constant $X$) would be a multiple of any other line, but taking the vertical lines $X=1$ and $X=2$ clearly shows that they are not multiples of each other, so $X,Y$ are dependent.
