Calculating covariance, with multiplication by stochastic variable.

Question

As an exercise I'm supposed to calculate; $\text{cov}(X \cdot Y,X)$, where $X$ and $Y$ are independent discrete stochastic variables, with probability functions given by; $$ p\left(var\right) = \left\{ \begin{array}{ll} 0.1 & \text{ if } var = 0 \\ 0.4 & \text{ if } var = 1 \\ 0.5 & \text{ if } var = 2 \\ 0 & \text{ otherwise.} \end{array} \right. $$ However I'm unable to find a rule, which applies to this case. (That is covariance, with multiplication by stochastic varibles).

I've tried to substitute $X \cdot Y$, with a new stochastic variable $Z = X \cdot Y$, with $p_{Z}\left(z\right) = p_{X}\left(z\right) \cdot p_{Y}\left(y\right)$, however this does not seem to result in the correct answer.

So I'd love to get a pointer, on how to proceed.

The answer is supposed to be $0.616$, $EX = 1.4$, $Var\left(X\right) = 0.44$.

Answer 1

Note that:

1. $cov(X,Y)=E[XY] - E[X]E[Y]$
2. For independent variables $E[AB]=E[A]E[B]$
3. $var(X)=E[X^2]-E[X]^2$

So $cov(XY,X) =E[X^2Y]-E[XY]E[X]=(E[X^2]-E[X]^2)E[Y]=var(X)E[Y]=0.44*1.4=0.616$

Answer 2

In general, if $Y$ independent of $Z$ and $X$. We can write $\mathrm{Cov}(X,YZ)=E[Y] \;\mathrm{Cov}(X,Z)$. Here $Z=X$.