Show independence of $(aX,bX^2)$, $\,X\sim N(0,\sigma^2)$? How can we proof that $aX,bX^2$ are independent iff $b\cdot a=0$, when $\,X\sim N(0,\sigma^2)$?
I found that $X^2$ is Chi-Square distributed, and the correlation is:
$$\rho(bX,aX^2)=ba\rho(X,X^2)=ba\dfrac{\mathrm{Cov}(X,X^2)}{ \sigma_X \sigma_{X^2}} =ba\dfrac{E[(X-\mu_X)(X^2-\mu_{X^2})]}{ \sigma_X\sigma_{X^2}}\stackrel{!}{=}0$$
So we have $\rho=0$ when $ba=0$, but as we know, $\rho=0$ does not directly imply independence in general. We may need to argue why it does here, or use some other dependence measure to show independence.
 A: Let $U$ be any random variable, and let $V$ be equal to $k$ with probability $1$. Then $U$ and $V$ are independent. 
There are two cases. (i) Suppose that $k\not\in B$. Then $\Pr((U\in A) \cap (V\in B))=0$. This is equal to $\Pr(U\in A)\Pr(V\in B)$, since $\Pr(V\in B)=0$.  (ii) Suppose that $k\in B$. Then  $\Pr((U\in A) \cap (V\in B))=\Pr(U\in A)$. This is $\Pr(U\in A)\Pr(V\in B)$, since $\Pr(V\in B)=1$.
A: Sufficient condition is trival, since any constant is indepedent of any random variable. For the necessary condition, we can do something as following. 
Suppose that $ab \neq 0$. Without loss of generality, we can only consider the cases that $a, b$ are both positive. (For other cases, the proofs are similar.) We plan to show that $aX$ and $bX^2$ are not independent. 
Suppose they are independent. For any two (Borel) sets $A$ and $B$, we have $\mathrm{Pr}(aX\in A, bX^2 \in B) = \mathrm{Pr}(aX \in A)\mathrm{Pr}(bX^2 \in B)$. Now consider following two special cases for $A$ and $B$:
$$
A := \{x | -\delta_1 < x < \delta_1 \} \\
B := \{x | 0 \leq x < \delta_2 \}
$$
where $\delta_1, \delta_2 > 0$ and $\frac{\delta_1^2}{a^2} = \frac{\delta_2}{b}$. It's easy to check that $aX\in A$ if and only if $bX^2\in B$. Note that $0<\mathrm{Pr}(aX\in A) = \mathrm{Pr}(bX^2 \in B) < 1$. Therefore, we have
$$
\mathrm{Pr}(aX\in A, bX^2 \in B) = \mathrm{Pr}(aX\in A) > \mathrm{Pr}(aX \in A)\mathrm{Pr}(bX^2 \in B)
$$
The necessarity is followed by contradiction. 
