I Need some help:
Let $Y=h(X)$ be a real square integrable function and X has a density function $f$. Show:
If $f$ and $h$ are even functions then $X$ and $Y$ are uncorrelated (but generally not Independent).
I know that two random variables are uncorrelated if only if $Corr(X,Y)=0$.
If the function $f$ is even then the function $x\mapsto xf(x)$ is odd hence its integral is zero, that is, $E(X)=0$. If the functions $f$ and $h$ are both even then the function $fh$ is even hence the function $x\mapsto xf(x)h(x)$ is odd hence its integral is zero, that is, $E(Xh(X))=0$. Thus, if the functions $f$ and $h$ are both even, then $\mathrm{Cov}(X,h(X))=E(Xh(X))-E(X)E(h(X))=0-0\cdot E(h(X))=0$.
To show that the random variables $X$ and $h(X)$ are not independent in general, assume that the random variable $h(X)$ is non degenerate and square integrable, then $E(h(X)^2)\ne E(h(X))^2$ since the difference between the LHS and the RHS is the variance of the random variable $h(X)$. Hence the random variables $h(X)$ and $h(X)$ are not independent, which implies that the random variables $X$ and $h(X)$ are not independent since, conversely, every random variables that are measurable functions of independent random variables are themselves independent.