Can the correlation of a random variable $X$ and $g(X)$ be $0$? Question: Can the correlation of a random variable $X$ and $g(X)$ be $0$?
My attempt:
I don't believe it can because they are dependent by definition therefore $Cov(X,g(X)) \ne 0$ which means the correlation cannot be $0$.
 A: It is true that independent variables have correlation zero, but variables that are not independent can also have correlation zero. For example let $X$ be uniform on the three numbers $-1, 0, 1$. Under the mapping $g(x):=x^2$, you can check that the variables $X$ and $Y:=g(X)$ have correlation zero. Indeed, the variables $(X,Y)$ are uniform on the three pairs $(-1,1)$, $(0,0)$, and $(1,1)$. Then $$E(XY)=\frac13[(-1)\cdot(1)+0\cdot0+1\cdot1]=0,$$ while $$E(X)=\frac13(-1+0+1)=0,$$ which is enough to conclude $\operatorname{cov}(X,Y):=E(XY)-E(X)E(Y)=0$.
A: Yes it can be. For an example:
Let $X$ be a random variable with $EX=0$  and $EX^3=0$, (for example a standard normal random variable).  Take $g(X)=X^2$. Clearly $X$ and $g(X)$ are related but:
$$cov(X,g(X))=E\left[Xg(X)\right]-E\left[X\right]\cdot E[g(X)]=EX^3=0.$$
A: Regardless of the distribution of X, let g(X) = c, where c is a constant (any constant will do). Then g(X) is independent of X, and hence g(X) is uncorrelated with X.  This is because a random variable which is constant with probability one is independent of every random variable, including itself.  Hence the OP's assertion that X and g(X) are dependent by definition is false (quite apart from the further erroneous claim, as refuted in other answers, that dependence implies non-zero covariance).
