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 $f (x_1, x_2)$ ?

migrated from mathoverflow.netJan 28 '16 at 19:54

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• $X^2$ is not normally distributed – William Stagner Jan 28 '16 at 19:55
• Besides the main question, you have to be very careful about the statement "It is well known that for two normal random variables, zero covariance implies independence.". See a counter example from en.wikipedia.org/wiki/… A more formal statement should be stating that "for two random variables that jointly follows a bivariate normal distribution, zero covariance implies independence." – BGM Jan 29 '16 at 3:08

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.