Covariance of uniform distribution and it's square

Question

I have $X$ ~ $U(-1,1)$ and $Y = X^2$ random variables, I need to calculate their covariance. My calculations are: $$ Cov(X,Y) = Cov(X,X^2) = E((X-E(X))(X^2-E(X^2))) = E(X X^2) = E(X^3) = 0 $$ because $$ E(X) = E(X^2) = 0 $$ I'm not sure about the $X^3$ part, are my calculations correct?

Answer 1

$\begin{align} \mathsf {Cov}(X,Y) & = \mathsf{Cov}(X,X^2) \\[1ex] & = \mathsf E((X-\mathsf E(X))\;(X^2-\mathsf E(X^2))) \\[1ex] & = \mathsf E(X^3-X^2\mathsf E(X)-X\mathsf E(X^2)+\mathsf E(X)\mathsf E(X^2)) & \text{ by expansion} \\[1ex] & = \mathsf E(X^3)-\mathsf E(X^2)\mathsf E(X)-\mathsf E(X)\mathsf E(X^2)+\mathsf E(X)\mathsf E(X^2) & \text{ by linearity of expectation} \\[1ex] & = \int_{-1}^1 \tfrac 1 2 x^3\operatorname d x -\int_{-1}^1 \tfrac 1 2 x^2\operatorname d x\cdot\int_{-1}^1 \tfrac 1 2 x\operatorname d x & \text{ by definition of expectation} \\[1ex] & = 0 \end{align}$

Reason The integrals of the odd functions are both zero over that domain. $\;\mathsf E(X^3)=\mathsf E(X) = 0$.

Note that $\;\mathsf E(X^2) = \int_{-1}^1 \tfrac 12 x^2 \operatorname d x = \tfrac 1 3$

Answer 2

We know it is $E(X^3)$ so:

$$E(X^3)=\int_{-1}^{1}x^3f(x)=0$$

So it is correct