Impact of RV Independence of E[X^2Y] I just took an exam where one of our true false questions asked if the following statement is true when considering two independent RVs with mean 0:
$$E[(X + Y)^3] = E[X^3] + E[Y^3]$$
By expansion of the product, I arrived at the following:
$$E[(X + Y)^3] = E[X^3 + Y^3 + 3X^2Y + 3XY^2]$$
$$E[(X + Y)^3] = E[X^3] + E[Y^3] + E[3X^2Y] + E[3XY^2]$$
My question is whether we can gain any inside into the relationship of $$E[X^2Y]$$ from independence.  I'm aware that independence by definition states that $$E[XY] = E[X]E[Y]$$
If someone could provide some insight into this problem by proof or counterexample, that would go a long way in furthering my understanding of the problem.
 A: All you need is to notice that if $X$ and $Y$ are independent, so are $Z=X^2$ and $Y$.

Two random variable $X$ and $Y$ are independent, if the $\sigma$-field generated by $X$ and the $\sigma$-field generated by $Y$ are independent.  That is $\forall A \in \sigma(X), \, \forall B \in \sigma(Y)$, we have $\mathbb P(A \cap B)= \mathbb P(A)\cdot \mathbb P(B)$

Now $X$ and $Y$ are independent, thus $\sigma(X)$ and $\sigma(Y)$ are independent.
Then $Z=f \circ X$, where $f=x^2$ for $x \in \mathbb R$
$$\sigma(X)=\{X^{-1}(A)|A \in \mathcal B(\mathbb R)\}$$
$$\sigma(Z)=\{(f\circ X)^{-1}(A)|A \in \mathcal B(\mathbb R)\}=\{X^{-1} \circ f^{-1}(A)|A \in \mathcal B(\mathbb R)\}$$
Now since $f(x) = x^2$ is continuous, it is Borel measurable, and thus $\sigma(Z) \subset \sigma(X)$. So $X^2$ and $Y$ are also independent.
Then you could have $\mathbb E[X^2Y]=\mathbb E[X^2]\cdot \mathbb E[Y]=0$
A: If $X$ and $Y$ are independent, then so are $g(X)$ and $h(Y)$, where $g$ and $h$ are any measurable functions.
So, if $X$ and $Y$ are independent, then so are $X^2$ and $Y$.
This can be shown in several ways. I will show it via the fact that we can still factor the CDF of $g(X)$ and $h(Y)$ if $X$ and $Y$ are themselves independent.
$$     \mathbb{P}(X \leq x , Y \leq y ) = \mathbb{P} (X \leq x ) P (Y \leq y ) $$
is assumed.
From this it follows that for any Borel sets $B_1$ and $B_2$
$$     \mathbb{P}(X \in B_1 , Y \in B_2 ) = \mathbb{P} (X \in B_1 ) P (Y \in B_2 ) $$
holds. 
Now let $g,h$ be (Borel) measurable functions. We can factor the corresponding joint CDF into a function of $x$ and a function of $y$.
\begin{align*}
     F_{g(X), h(Y)} (x,y) =& \mathbb{P} ( g(X) \leq x, g(Y) \le y ) = \mathbb{P}( X \in g^{-1} (-\infty,x], Y \in h^{-1}(-\infty, y] ) \\\\
    =& \mathbb{P} (X \in g^{-1} (-\infty,x] ) \mathbb{P} ( \in h^{-1}(-\infty, y])\\\\ =& \mathbb{P} (g(X) \leq x ) \mathbb{P} (h(Y) \leq y ) \\\\
    =& F_{g(X)}(x) F_{h(Y)}(y)
\end{align*}
To tie it back into the actual question from your exam,
since $\mathbb{E}(X^2Y) = \mathbb{E}(X^2)\mathbb{E}(Y)$ (and similarly for $\mathbb{E}(XY^2)$), then we have 
$$\mathbb{E}([X + Y]^3) = \mathbb{E}(X^3) + \mathbb{E}(Y^3) + \mathbb{E}(X^2)\mathbb{E}(Y) + \mathbb{E}(X)\mathbb{E}(Y^2).$$ But since $\mathbb{E}(X) = \mathbb{E}(Y) = 0$, this just becomes $\mathbb{E}([X + Y]^3) = \mathbb{E}(X^3) + \mathbb{E}(Y^3)$.
