How to calculate $\operatorname{Cov}(X + 3, XZ)$?

Question

How do I calculate $\operatorname{Cov}(X + 3, X\cdot Z)$ if I know that

$$ f_X(x) = \begin{cases} 4x^3 &, \text{if} \ 0 < x < 1 \\ 0 &, \text{otherwise} \end{cases}$$ where $Z \sim N(1,2)$ and $X$ and $Z$ are independent?

I started with $\operatorname{Cov}(X+3,X\cdot Z) = \operatorname{Cov}(X,X \cdot Z)$ and thought maybe I should use that $\operatorname{Cov}(X,Y) = E[XY] - E[X]E[Y]$ where I would set $Y = X \cdot Z$ but that would end up giving $0$ and I know the answer is $2/75$.

What property can I use?

Thanks.

Answer 1

You are almost where. $$Cov(X,X Z)=E[X^2Z]-E[X]E[XZ]=\left(E[X^2]-(E[X])^2\right)E[Z]$$

Answer 2

$$\text{Cov}(X+3,XZ)=\text{Cov}(X,XZ)=EX^2Z-EX\cdot EXZ=EX^2 \cdot EZ-(EX)^2\cdot EZ$$

Answer 3

Note that

$$ \begin{align*} \text{Cov}(X, XZ) &=E(X\cdot XZ)-(EX)(EXZ) \tag{0} \\ &=E(X^2)EZ-(EX)^2EZ\tag{1} \\ &=EX^2-(EX)^2\tag{2} \\ &=\text{Var}(X)\tag{3} \end{align*} $$ where in (0) we used the definition of covariance, in (1) we used the fact that $X$ and $Z$ are independent and hence so are $X^2$ and $Z$. In (2) we use the fact that $EZ=1$. You should be able to compute (3).