Correlation of product of uncorrelated This seems like it should be pretty simple but I'm not seeing it. Suppose I have two r.v.s X and Y with means of zero and variances of one, such that cov(X,Y)=0. What is cov(X,XY)?
 A: This is a fun problem for testing the development code in the next version of mathStatica. 
Note that $Cov(X, XY) = \mu_{1,1}(X, XY)$ (i.e. the covariance operator is the {1,1} central moment), which is why I am requesting the {1,1} CentralMoment of {X, X Y} ... when the variables are {X,Y}:

(source: tri.org.au) 
If $X$ and $Y$ are independent (information not stated in the problem), then the answer simplifies further:

(source: tri.org.au) 
A: The covariance $Cov(X,XY) =0$. Here is the line of reasoning;
use the conditional covariance, or also called the law of total covariance,
$$Cov(X,Y) =  \mathbb{E}_Z(Cov(X,Y|Z)) + Cov_Z(\mathbb{E}(X|Z), \mathbb{E}(Y|Z)).$$
We can condition on $X$ and calculate: 
$$Cov(X,XY)= \mathbb{E}(Cov(X,YX|X)) + Cov(\mathbb{E}(X|X),\mathbb{E}(YX|X))$$
which would give us
$$\mathbb{E}(Cov(X,YX|X)) + Cov(\mathbb{E}(X|X),\mathbb{E}(YX|X)) = 0+ Cov(X, X\mathbb{E}(Y))$$ 
where the first term is $0$ because $
Cov(X,YX|X)$ is the covariance of a constant and a constant times the random variable $Y$. However, the covariance of a constant and a random variable is $0$. Now let's consider the second term, we are given that $\mathbb{E}(Y) =0$ so $X\times 0=0$ and we are again left with $Cov(X,0)=0$ because the covariance of a constant and a random variable is $0$. Here I used the assumption that $\mathbb{E}(Y|X)=0$ which may or may not be something the OP can assume. To the extent that $\mathbb{E}(Y|X)\neq 0$ and that $\mathbb{E}(Y|X)=f(X)$ for some function $f(\cdot)$ then you would have a non-zero covariance. Unless of course the function $f(\cdot)$ magically gave you a $0$ covariance. 
A: Since you know that $Cov(X,Y)=0$, $\mathbb{E}[X]=0$, and $\mathbb{E}[Y]=0$, you know that
$$\begin{align*}
\mathbb{E}[XY] &= \mathbb{E}\left[ (X-0)(Y-0) \right] \\
&= \mathbb{E}\left[ (X-\mathbb{E}[X])(Y-\mathbb{E}[Y]) \right] \\
&= Cov(X,Y) \\
&= 0
\end{align*}$$
Since this looks like a homework problem, I'll just leave it at that. Combining this with the definition of $Cov(X,XY)$ should be enough to get the answer.
