Help in example for Covariance zero doesn't always imply independence

Question

I have the following example for which I need help.

Suppose we first sample a real number x from a uniform distribution over the interval [−1, 1]. We next sample a random variable s. With probability 1/2, we choose the value of s to be 1. Otherwise, we choose the value of s to be −1. We can then generate a random variable y by assigning y = sx. Clearly, x and y are not independent, because x completely determines the magnitude of y. However, Cov(x, y) = 0.

Can someone show me full derivation of the covariance? Thanks in advance.

Answer 1

\begin{align} Cov(X,Y) = Cov(X,SX) &= \mathbb{E} \left[X SX\right] - \mathbb{E}[X] \mathbb{E} [SX] \\ & = \mathbb{E}[S] \mathbb{E} \left[X^2\right] - \mathbb{E} \left[X\right]\mathbb{E}[S]\mathbb{E}[X] \quad \text{(independence)} \\ & = 0 \times \mathbb{E} \left[X^2\right] - 0 \times 0 \times 0 \\ &= 0\end{align}

so you only need that for (measurable) functions of independent random variables $X$ and $Y$, say $g$ and $f$:

$$\mathbb{E}\left[g(X) f(Y)\right] = \mathbb{E} \left[g(X) \right] \mathbb{E} \left[ f(Y) \right] $$