To prove $\operatorname{Cov}(X,Y)=0$ with $X \sim {U}[-1,1]$ and $S$ is sampled from $\{{-1,1}\}$ with $p=0.5$ and $Y=XS$

Question

To prove: $\operatorname{Cov}(X,Y)=0$ with $X \sim {U}[-1,1]$ and $S$ is sampled from $\{-1,1\}$ with $p=0.5$ with $Y = XS$. There is no statement about the independence of $X$ and $S$.

Its not stated but if I assume $X$ and $S$ are independent.

In which case it is simple.

$E[X] = 0 $ since $X \sim U[-1,1]$, also

$E[XS]$ = $E[X].E[S]$ = 0 and similarily $E[XY] = E[X^2].E[S]$ = 0

$ \implies Cov(XY) = E[XY] - E[X]E[Y] = 0$

Else, If I do not assume since it is not mentioned,

$E[Y] = E[XS] = \Sigma_{s \in \{-1,1\}} \int xs P(X=x,S=s)dx$ (I have doubt on this step)

$ = \int x P(X=x,S=1)dx + \int (-x) P(X=x,S=-1)dx$,

$ = \int x P(X=x | S=1) p(S=1)dx + \int (-x) P(X=x | S=-1) p(s=-1)dx$,

Now what? we do not have joint distribution of $X$ and $S$...

Answer 1

$X$ and $S$ are independent, so $P(X=x|S=1)=P(X=x)$. But I think you meant to use the probability density function of $X$ there instead of $P()$. We do know the joint distribution because the two variables are independent.