zero covariance but not independent - normally distributed random variable $X$ and $X^2$ This is one of my homework question, which the answer sheet has already been given out. However, I still don't understand it.
Exercise 1.1. It is well known that for two normal random variables, zero
covariance implies independence. Why does this not apply to the following
situation: 
$X ∼ N(0, 1), Cov(X, X2) = E[X^3] − E[X]$ x $E[X^2] = 0 − 0 = 0 $ 
but obviously $X^2$ is totally dependent on X?
So, I assume that I have to show :
$f (x_1, x_2) = f_{X_1} (x_1) * f_{X_2} (x_2) $
for the random variables $X$ and $X^2$ to be not independent. The problem is that I do not know how to compute the joint density function $f (x_1, x_2) $ for the two random variables $X$ and $X^2$.
First of all, is this correct?
$f_{X_1} (x) =$ $1\over{\sqrt{2\pi\sigma^2}}$ $e^{- x^2 \over 2\sigma^2}$
$f_{X_2} (x) =$ [$1\over{\sqrt{2\pi\sigma^2}}$ $e^{- x^2 \over 2\sigma^2}]^2$. 
Then, how do I compute the joint density function $f (x_1, x_2) $ ?
I am so confused. Please help!!!!!
 A: The direct intuition (as signified by the adverb "obviously")
is that we cannot know either $X$ or $X^2$ a priori,
but once we find out the value of $X$, we know the exact value of $X^2$.
This is something that never happens with two non-trivial
independent random variables $X$ and $Y$.
(The only way that we could know the value of the independent variable $Y$ immediately after knowing $X$ is if we already knew the value of $Y$,
for example if $Y$ is constant.)
If it does not quack, it is not a duck.

A more formal argument is as follows.
If $X$ and $Y$ are independent, then for any sets $A$, $B$ in the
respective ranges of $X$ and $Y$,
$$ P((X \in A) \text{ and } (Y \in B)) = P(X \in A)\, P(Y \in B). $$
This is a consequence of the joint density function you tried to
compute, but you do not necessarily need to know everything about
the joint density function in order to evaluate this statement.
In particular, let $X \sim N(0, 1)$, $Y = X^2$, $A = [-1, 1]$, and $B = [0, 1]$.
Then $X \in [-1, 1]$ if and only if $Y = X^2 \in [0, 1]$,
that is, $X \in A$ occurs if and only if $Y \in B$ occurs,
so $Y \in B$ occurs if and only if both $X \in A$ and $Y \in B$ occur.
Therefore
$$P(X \in A) = P(Y \in B) = P((X \in A) \text{ and } (Y \in B)).$$
But $P(X\in A) = \int_{-1}^1 f_X(x) dx \approx 0.683 < 1$.
Therefore 
$$ P(X \in A)\, P(Y \in B) < P(Y \in B).$$
But $P(Y \in B) = P((X \in A) \text{ and } (Y \in B))$, so
$$ P(X \in A)\, P(Y \in B) < P((X \in A) \text{ and } (Y \in B)).$$
Therefore $X$ and $Y = X^2$ are not independent.
